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Projective

Geometry

Lecture Notes

W. D. GILLAM

Bogazici University 2014

Contents

Introduction iv

1 Transformations of the Plane 11.1 Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Orthogonal linear transformations . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Affine transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 Classification of isometries of the plane . . . . . . . . . . . . . . . . . . . . 16

1.6 Algebraic automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Parametric Plane Curves 20

2.1 Reparametrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Length and the arc length parametrization . . . . . . . . . . . . . . . . . . 22

2.3 Curvature and the Frenet formulas . . . . . . . . . . . . . . . . . . . . . . 23

2.4 Plane curves up to isometry . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Affine Algebraic Plane Curves 28

3.1 Polynomials and zero loci . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Smooth and singular points . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.3 Change of variables and affine equivalence . . . . . . . . . . . . . . . . . . 33

3.4 Five points determine a conic . . . . . . . . . . . . . . . . . . . . . . . . . 363.5 Plane conics up to affine equivalence . . . . . . . . . . . . . . . . . . . . . 38

3.6 Invariants of affine conics . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4 Projective Spaces 45

4.1 Motivation and construction . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Linear subspaces of projective space . . . . . . . . . . . . . . . . . . . . . 484.3 Real and complex projective spaces . . . . . . . . . . . . . . . . . . . . . . 48

4.4 Projective transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

ii

Contents iii

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 Projective Plane Curves 555.1 Homogeneous polynomials and zero loci . . . . . . . . . . . . . . . . . . . 565.2 Smooth and singular points . . . . . . . . . . . . . . . . . . . . . . . . . . 575.3 Linear change of variables and projective equivalence . . . . . . . . . . . . 585.4 Lines and linear functionals . . . . . . . . . . . . . . . . . . . . . . . . . . 595.5 Invariants of affine conics revisited . . . . . . . . . . . . . . . . . . . . . . 615.6 Classification of projective conics . . . . . . . . . . . . . . . . . . . . . . . 625.7 The cross ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675.8 Intersection multiplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.9 The theorems of Pascal, Pappus, and Desargues . . . . . . . . . . . . . . . 705.10 Cubic curves and the group law . . . . . . . . . . . . . . . . . . . . . . . . 725.11 Nodal and cuspital cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6 Appendix 826.1 Groups and group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . 826.2 Fields and division rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.3 The Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . . . . . 866.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Bibliography 88

Introduction

These notes are meant to serve as the text for an undergraduate course in elementaryprojective geometry. The current notes were written to accompany the course of thesame title given at Bogazici University in the fall of 2014. Some parts of these noteswere recycled from notes I wrote to accompany a course called Fundamental Problemsof Geometry, which I taught at Brown University in the spring of 2012. Both courseswere intended for junior and senior mathematics majors. The second incarnation of thiscourse was particularly aimed at students seeking a career in mathematics instruction.To this end, I made considerable effort to try to “survey” a variety of basic geometricmaterial, with a focus on the geometry of the plane.

In principle linear algebra is the only formal prerequisite, though there are manyplaces where I find it convenient to use the language of group theory, though no actualgroup theory is really used—it is enough to know the definition of a group, a subgroup,and perhaps a normal subgroup. Also, the idea of a group acting on a set arises (at leastimplicitly) at many points, so it might be helpful to have an idea what this means—it isa completely elementary concept. Both times I taught the course, I considered devotinga lecture to the basic notions of group theory and group actions (orbits and stabilizers,in particular), but in the end I just mentioned the necessary concepts as they arose, orrelegated them to the exercises—that seemed sufficient. At some point I may add anappendix to these notes covering this material.

Let me make some remarks about “level of generality” and the like, mostly intendedfor the instructor. The usual issues about fields are skirted in these notes in much thesame way they are skirted in a typical linear algebra class; that is, we pretty much thinkof a field as either R or C and we don’t much emphasize the difference even betweenthese two fields, except that we can’t get away with this quite as long as we can in linearalgebra because we will quickly consider non-linear equations like x2 + 1 = 0 which haveradically different behaviour over these two fields. For the sake of exposition, I make moststatements over R (instead of writing F or some such thing to denote a general field),but most of these are true over an arbitrary field, except where (I think) it should bereasonably clear from context that the particular field considered is of central importanceto the statement of the result. In fact, I stubbornly insisted on using R as the “base field”throughout most of the notes, even though most of the “algebro-geometric” content ofthe notes (of which there is very little, by the way) would probably be easier and more

iv

Introduction v

natural over C.Although projective geometry and, in particular, the projective plane RP2, are the

main subject matter of these notes, a large part of the text is actually devoted tovarious geometric considerations in the usual “affine” plane R2. Without some of this“background” material, much of the projective geometry would seem unmotivated. I alsowanted to emphasize the interplay of several different points of view on this subjectmatter coming from linear algebra, differential geometry, algebraic geometry, and classicalaxiomatic geometry.

Chapter 1 is devoted to defining and studying various transformation groups, such asthe group of invertible linear transformations, the group of isometries, and the group ofaffine transformations. The latter, in particular, is important because of its relationshipwith the group of projective transformations (the projective general linear group), whichwe discuss in the chapter on projective geometry. The material from this chapter is usedthroughout the rest of the text.

In the first incarnation of the course, I didn’t say anything about isometries, butthe group of isometries is so closely related to the group of affine transformations thatit seemed strange to discuss one but not the other. Having discussed isometries in theProjective Geometry course, I couldn’t resist fleshing out this material to show howthe group of isometries is used in a “real life” geometric study: To this end, I haveincluded Chapter 2 on parametric curves in the plane. I thought it might be useful forpedagogical reasons to have the students return to a study of parametric curves, familiarfrom calculus courses, equipped with an understanding of the meaning of isometry andthe group of isometries of the plane, as the idea of isometry is certainly lurking implicitlyin the treatments of parametric curves one sees in any elementary calculus text! Tomake sure that this chapter contains some new and interesting content, I have includeda proof of the (fairly simple) fact that two parametric curves in the plane are relatedby an orientation preserving isometry iff they have the same length and curvature. Mytreatment of this is taken from Do Carmo’s book Differential Geometry of Curves andSurfaces.

The final two chapters consist of some elementary algebraic geometry of affine andprojective plane curves. We introduce the general projective space RPn, but focus almostexclusively on RP2. We define the zero locus Z(f) of a polynomial f (or, rather, ahomogeneous polynomial in the projective setting), and what it means to be a singularpoints of Z(f). We explain what it means for polynomials to be “affine equivalent”and for homogeneous polynomials to be “projectively equivalent.” To make this moreconcrete, we mention the classification of degree two polynomials in two variabes up toaffine equivalence, though we only give a complete proof of the projective analog of thisclassification. At the very end we give a brief study of cubic curves in RP2, giving atleast the rough idea of the group law on the set of smooth points of an irreducible cubic.

Chapter 1

Transformations of the Plane

Let us agree that a transformation (of the plane) is a bijective function f : R2 → R2.Among all transformations we can single out various classes of transformations preservingvarious additional structures possessed by R2. In this chapter we will define and studythe following types of transformations:

1. homeomorphisms

2. invertible linear transformations

3. isometries

4. orthogonal linear transformations

5. affine transformations

6. algebraic automorphisms

One can consider these types of transformations more generally in the case of Rn, orover other fields. Instead of Rn, one could work with an abstract vector space, or witha vector space equipped with some extra structure. We will make some remarks aboutthese more general settings, but our main focus will be on the case R2, as this will bemost important in the later chapters. The set of transformations of each type forms agroup. In some cases, we will say something about the structure of this group.

Although detailed definitions will be given in the sections of this chapter, we will givea summary now for the convenience of the reader. A homeomorphism is a continuousbijection f : R2 → R2 with continuous inverse f−1 : R2 → R2. Homeomorphisms arethe most general type of transformation that will be of any use to us—every other typeof transformation we consider will be a homeomorphism. Homeomorphisms preserveproperties of subsets of R2 such as connectedness, being open, or being compact (closedand bounded).

An invertible linear transformation is a bijection f : R2 → R2 that “commuteswith scalar multiplication and vector addition.” The importance of invertible lineartransformations results mainly from their usage in linear algebra. An isometry of R2 is abijection f : R2 → R2 that preserves distance. Isometries can be useful when studying

1

2 Chapter 1 Transformations of the Plane

various geometric properties of subsets of R2 such as lengths, areas, angles, and so forth.One can define an orthogonal linear transformation to be a map f : R2 → R2 whichis both an isometry and a linear transformation. We will describe all of these. Anaffine transformation is a map f : R2 → R2 expressible as a composition of a lineartransformation and a translation. Affine transformations arise naturally when studyingmore “algebraic” subsets of R2 (zero loci of polynomials) and their properties. They arealso closely related to the projective transformations (§4.4) that we will study when weintroduce projective geometry in Chapter 4. An algebraic automorphism is a bijectionf : R2 → R2 such that the coordinates of both f and its inverse are given by polynomialfunctions of two variables. These are the most subtle kind of transformation that wewill consider—we won’t have too much to say about them, but it would be amiss not tomention these along with the others!

One reason to be interested in these sorts of transformations is the following: If one isinterested in calculating some quantity Q(A) associated to a subset A ⊆ R2 (or perhapsto several subsets A1, . . . , An ⊆ R2), it is often convenient to calculate the correspondingquantity Q(f(A)) for f(A) for some transformation f for which it is known a priorithat Q(A) = Q(f(A)). Of course this seems silly when expressed in such abstract terms,but imagine, for example, the problem of computing the circumference of a circle in R2,or, more generally, the length `(A) of some other curve A in R2. Because this probleminvolves doing significant computation (integration), it is usually helpful to move A tosome other position via an isometry f (for example: move the circle so it is centeredat the origin), which won’t effect the result of the computation (because one knows “apriori” that length is isometry invariant), but which will make the computation easier toactually carry out. Similar considerations are common in linear algebra, where variousoperations are naturally defined and studied in one basis; one then uses invertible lineartransformations to re-express things in the standard basis. We will make use of thevarious kinds of transformations for purposes such as this in the other chapters of thebook.

1.1 Linear transformations

Although I assume the student has some background in linear algebra, let me brieflyreview some of the more relevant material. Throughout, we work with a fixed field K,often called the base field. The reader unfamiliar with the general notion of a field shouldsimply focus on the case where K is the field Q of rational numbers, the field R of realnumbers, or the field C of complex numbers. Let

Kn := {(x1, . . . , xn) : x1, . . . , xn ∈ K}

be the set of ordered n-tuples of elements of K. We will use symbols like x, y, etc. todenote elements of Kn, writing x1, . . . , xn for the coordinates of x.

From the point of view of linear algebra, the set Kn is equipped with two importantstructures: (vector) addition and scalar multiplication. For x, y ∈ Kn, we let

x+ y := (x1 + y1, . . . , xn + yn)

1.1 Linear transformations 3

be the “vector sum” of x and y. Given λ ∈ K and x ∈ Kn, we set

λx := (λx1, . . . , λxn).

The vector λx is called a scalar multiple of x; the term rescaling of x is reserved for avector of the form λx with λ ∈ K∗ := K \ {0}. Vector addition and scalar multiplicationsatisfy various properties, which one is led to investigate and “axiomatize”: for example,scalar multiplication “distributes over vector addition.” After thinking about theseproperties, one is eventually led to define a vector space (or a vector space over K if K isnot clear from context) to be a set V equipped with two functions, written

V × V → V

(u, v) 7→ u+ v

R× V → V

(λ, v) 7→ λv,

and called, respectively, (vector) addition and scalar multiplication satisfying variousaxioms that we will not repeat here (see any linear algebra textbook).

It is customary in linear algebra to write the elements x of Kn as column vectors,but in most of the inline text of these notes I will usually write these as row vectorsx = (x1, . . . , xn) for aesthetic reasons. I could put some kind of “transpose” notation into be precise, but I think it will always be clear when I am using row vectors or columnvectors. I usually use u, v, etc. for elements of an abstract vector space, which I usuallydenote U , V , or W . In linear algebraic context, I try to use Greek letters like λ, µ, andso forth for elements of K, also called scalars.

Definition 1.1.1. If V and W are vector spaces, a linear transformation from V to Wis a function f : V →W satisfying f(λv) = λf(v) and f(u+ v) = f(u) + f(v) for everyλ ∈ K, u, v ∈ V . Informally, one says that f “respects addition and scalar multiplication.”Let HomK(V,W ) be the set of linear transformations f : V →W .

It is easy to check that a composition of linear transformations is a linear trans-formation, so that usual composition of functions defines a “composition of lineartransformations”

HomK(V,W )×HomK(U, V ) → HomK(U,W )

(f, g) 7→ fg

for vector spaces U, V,W . It is also easy to see that, if f : V → W is a bijective lineartransformation, then its inverse f−1 : W → V is also a (bijective) linear transformation.

Definition 1.1.2. An m× n matrix A = (Ai,j) consists of elements Ai,j ∈ K, called theentries of A, one for each

(i, j) ∈ {1, . . . ,m} × {1, . . . , n}.

We write Mat(m× n) for the set of m× n matrices; if there is any ambiguity about K,then we write Mat(m× n,K).


One thinks of the entries Ai,j of a matrix A as being arranged in a rectangular arraysuch that Ai,j is in row i and column j. Thus, for example, a 2× 3 matrix A is writtenin the form

A =

(A1,1 A1,2 A1,3

A2,1 A2,2 A2,3

).

It is common to drop the comma in the subscript, writing Aij instead of Ai,j , thoughthis is a bit sloppy since “ij” might be mistaken for the product of i and j. The entriesof a 2× 2 matrix are almost always called a, b, c, d rather than A1,1, A1,2, A2,1, and A2,2,so that a typical 2× 2 matrix is written

A =

(a b

c d

). (1.1)

The fundamental operation with matrices is the (matrix) product : Given an m× nmatrix A and an n× k matrix B, one defines an m× k “product” matrix AB by setting

(AB)i,j :=n∑p=1

Ai,pBp,j .

To each square matrix A ∈ Mat(n × n), one can associate a real number detA, calledthe determinant of A. If A is 1× 1, one not surprisingly has detA = A1,1 equal to theunique entry of A. If A is a 2× 2 matrix as in (1.1), then detA = ad− bc. In general, inan elementary linear algebra class, one defines detA by some kind of inductive procedure(“expansion by minors”), though there are more satisfying ways of doing this. In anycase, the most important property of the determinant is that

det(AB) = (detA)(detB)

for all n× n matrices A,B.

The “matrix product” is motivated by linear algebraic considerations as follows:

For i ∈ {1, . . . , n}, let ei ∈ Kn be the point with ith coordinate 1 and all othercoordinates zero. For example, e1 = (1, 0) and e2 = (0, 1) in K2. If f : Kn → Km isa linear transformation, we define an m × n matrix A = Af by letting the ith column(i = 1, . . . , n) of A be f(ei) ∈ Kn (viewing this as a column vector). For example, iff : R2 → R2 and f(e1) = (a, c), f(e2) = (b, d), then

Af =

(a b

c d

).

Given an m× n matrix A, we define a function fA : Kn → Km by letting fA(x) be thematrix product Ax defined above, viewing x as a column vector (n × 1 matrix). Thefollowing standard result of linear algebra says that “matrices are the same thing aslinear transformations and matrix multiplication is the same thing as composition:”

1.1 Linear transformations 5

Proposition 1.1.3. The maps f 7→ Af and A 7→ fA defined above are inverse bijectionsbetween HomK(Kn,Km) and Mat(m× n). Under these bijections, matrix multiplication

Mat(m× n)×Mat(n× k) → Mat(m× k)

corresponds to composition of linear transformations

HomK(Kn,Km)×HomK(Kk,Kn) → HomK(Kk,Kn).

That is, fAB = fAfB and Afg = AfAg.

Recall the following standard result of linear algebra, which, among other things,explains the importance of the determinant:

Proposition 1.1.4. For a linear transformation f : Kn → Kn, the following are equiva-lent:

1. f(e1), . . . , f(en) is a basis for Kn.

2. f is surjective.

3. f is injective.

4. f is bijective.

5. Ker f = {0}.

6. There is a linear transformation g : Kn → Kn such that gf(x) = fg(x) = x for allx ∈ Rn.

7. The determinant detAf of the matrix Af associated to f is non-zero.

Definition 1.1.5. A linear transformation f : Kn → Kn satisfying the equivalentconditions of the above proposition is called an invertible linear transformation.

It is clear that a composition of invertible linear transformations is an invertible lineartransformation. The set of invertible linear transformations f : Kn → Kn forms a groupunder composition, denoted GLn(K), and often called the general linear group. Blurringthe distinction between a linear transformation and the corresponding matrix, an elementof GL2(K) can be viewed as a 2× 2 matrix

A =

(a b

c d

)

where detA = ad− bc 6= 0.


1.2 Isometries

Recall that a metric d on a set X is a function

d : X ×X → R

satisfying the properties:

1. d(x, y) ≥ 0 for all x, y ∈ X with equality iff x = y.

2. d(x, y) = d(y, x) for all x, y ∈ X.

3. d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

A metric space (X, d) is a set X equipped with a metric d. We usually just write Xto denote a metric space, leaving d implicit; we often write dX for the metric on X ifthere is any chance of confusion (for example, if we are considering more than one metricspace).

For example, it is not hard to check that

d : Rn × Rn → Rn

d(x, y) :=

√√√√ n∑i=1

(xi − yi)2

is a metric on Rn, called the standard metric. By default, we regard Rn as a metricspace by using the standard metric unless some other metric is explicitly mentioned.In general, if d is a metric on a set X and A is a subset of X, then by restriction ddetermines a metric on A, also abusively denoted d. In particular, the standard metricon Rn determines a metric on any subset A ⊆ Rn; we always use this metric to view Aas a metric space, unless explicitly mentioned to the contrary.

One defines continuity for maps between metric spaces by using the familiar “epsilon-delta definition:”

Definition 1.2.1. A function f : X → Y between metric spaces X,Y is called continuousiff for every x ∈ X and every positive real number ε, there is a positive real numberδ = δ(x, ε) such that dY (f(x), f(x′)) < ε whenever x′ is a point of X with dX(x, x′) < δ.A homeomorphism f : X → Y is a continuous bijection whose inverse f−1 : Y → X isalso continuous.

It is easy to see that a composition of continuous maps between metric spaces iscontinuous, hence a composition of homeomorphisms is a homeomorphism. In particular,the set Homeo(Rn) of homeomorphisms f : Rn → Rn is a group under composition. Itis also not hard to check that any linear transformation f : Rn → Rn is continuous.It follows that the group GLn(R) of invertible linear transformations is a subgroup ofHomeo(Rn).

Definition 1.2.2. A function f : X → Y between metric spaces X and Y is calleddistance preserving iff dX(x, x′) = dY (f(x), f(x′)) for all x, x′ ∈ X. An isometry f : X →Y is a distance preserving bijection.

1.3 Orthogonal linear transformations 7

The following facts are readily established:

1. A composition of distance preserving functions is distance preserving.

2. A distance preserving function is continuous (one can take δ = ε) and one-to-one(injective).

3. The inverse of an isometry is an isometry.

4. Every isometry is a homeomorphism.

5. A composition of isometries is an isometry.

In particular, the set Isom(Rn) of isometries f : Rn → Rn is a subgroup of the groupHomeo(Rn).

The definitions above yield notions of continuity, isometry, etc. for functions betweensubsets of Rn. We will see later that any distance preserving function R2 → R2 isautomatically bijective (i.e. is an isometry). (This is true, more generally, for Rn.)Indeed, we will completely classify the isometries of R2, and this will fall out as aconsequence—the reader may wish to try to directly prove this assertion.

Example 1.2.3. (Translations) Let K be a field. Fix t ∈ Kn. Define a functionft : Kn → Kn by ft(x) := t + x. When K = R, the function ft : Rn → Rn preservesdistance because

d(ft(x), ft(y)) =

√√√√ n∑i=1

((ti + xi)− (ti + yi))2

=

√√√√ n∑i=1

(xi − yi)2

= d(x, y).

A function f : Kn → Kn equal to ft for some t ∈ Kn is called a translation—the t willbe unique since we can recover t from ft by the obvious formula t = ft(0). Notice thatthe composition fsft of two translations is the translation fs+t. The set of translationsKn → Kn is a group under composition of functions; this group is isomorphic to Kn

(under vector addition) via the maps t 7→ ft, f 7→ f(0). The inverse of the translation ftis the translation f−t. In particular, every translation ft : Rn → Rn is an isometry.

1.3 Orthogonal linear transformations

Given a real vector space V (vector space over the field of real numbers R), recall fromlinear algebra that an inner product (more precisely: positive definite inner product) onV is a function V × V → R, usually written (u, v) 7→ 〈u, v〉, satisfying the conditions

1. 〈u, v〉 = 〈v, u〉


2. 〈u, u〉 ≥ 0 with equality iff u = 0

3. 〈u+ u′, v〉 = 〈u, v〉+ 〈u′, v〉

4. 〈λu, v〉 = λ〈u, v〉

for all u, u′, v ∈ V , λ ∈ R.For example,

〈 , 〉 : Rn × Rn → R

〈x, y〉 :=n∑i=1

xiyi

defines an inner product on Rn, called the standard inner product.An inner product space is a real vector space V equipped with an inner product. An

inner product 〈 , 〉 on V gives rise to a metric d on V defined by

d(u, v) := 〈u− v, u− v〉.

Two vectors u, v in an inner product space V are called orthogonal iff 〈u, v〉 = 0. Moregenerally, one defines the angle θ ∈ [0, π] between two non-zero vectors u, v in V by theformula

cos θ =〈u, v〉|u||v|

(1.2)

where the symbol

|u| := d(u, 0)

=√〈u, u〉

denotes the magnitude of u. (One checks that the right hand side of (1.2) is in [−1, 1] sothat this makes sense.)

Unless mentioned to the contrary, we always regard Rn as an inner product space byequipping it with the standard inner product. The standard inner product on Rn givesrise to the standard metric on Rn. Let us check that the notion of the angle between twovectors defined by the standard inner product is the usual angle in Rn. Recall that theLaw Of Cosines says that in any triangle with sides of length a, b, c, we have

c2 = a2 + b2 − 2ab cosC,

where C is the angle opposite the side of length c. If we apply this to the “triangle”whose vertices are 0, and two non-zero vectors x, y ∈ Rn, letting c be the side connectingx to y, we find

〈u, v〉 = |u|2 + |v|2 − 2|u||v| cos θ.

If we expand out the inner product on the left, cancel the |u|2 + |v|2 appearing on bothsides and divide by −2, we find that the angle θ between u and v is the same as the


“abstractly defined” angle θ. In particular, x, y ∈ Rn are orthogonal in the usual sense iff〈x, y〉 = 0.

One says that a list of n vectors v1, . . . , vn in an n-dimensional inner product spaceV is an orthonormal basis for V iff 〈vi, vj〉 = δi,j is zero when i 6= j and one when i = jfor all i, j ∈ {1, . . . , n}. An orthonormal basis is, in particular, a basis: By linear algebra,it suffices to check that the vectors v1, . . . , vn are linearly independent. Indeed, if wehave

∑ni=1 λivi = 0, then taking 〈 , vj〉 and using properties of the inner product and

the orthogonality assumption on the vi, we see that λj = 0.

The standard inner product on Rn gives rise to a nice relationship between the linearalgebra of Rn and the isometries of Rn:

Proposition 1.3.1. For a linear transformation f : Rn → Rn, the following are equiva-lent:

1. 〈f(x), f(y)〉 = 〈x, y〉 for all x, y ∈ Rn.

2. The columns of the matrix A = Af associated to f form an orthonormal basis forRn.

3. f preserves distance

4. f is an isometry of Rn.

Proof. To see that (1) implies (2), just note that the ith column of Af is f(ei), so wehave to show that 〈f(ei), f(ej)〉 = δi,j . But by (1) we have

〈f(ei), f(ej)〉 = 〈ei, ej〉

and it is clear from the definition of the standard inner product on Rn that this is δi,j .

To see that (2) implies (1), write x =∑n

i=1 xiei, y =∑n

i=1 yiei for all real numbersx1, . . . , xn and y1, . . . , yn. Then f(x) =

∑ni=1 xif(ei) =

∑ni=1Ai, where Ai = f(ei) is the

ith column of A = Af . Using this, the analogous formula for f(y), the hypothesis on A,and properties of 〈 , 〉, we compute

〈f(x), f(y)〉 =

⟨n∑i=1

xiAi,n∑i=1

yiAi

⟩

=

n∑i,j=1

xiyj〈Ai, Aj〉

=n∑

i,j=1

xiyjδi,j

=n∑i=1

xiyj

= 〈x, y〉.


To see that (1) implies (3), we compute, for any x, y ∈ Rn that

d(f(x), f(y)) = 〈f(x)− f(y), f(x)− f(y)〉= 〈f(x− y), f(x− y)〉= 〈x− y, x− y〉= d(x, y)

using linearity of f for the second equality and (1) for the third equality.

To see that (3) implies (2), let Ai = f(ei) denote the ith column of A = Af . Firstnote that

〈Ai, Ai〉 = 〈Ai − 0, Ai − 0〉= d(Ai, 0)

= d(f(ei), f(0))

= d(ei, 0),

so we see that each column of A has magnitude one. We still have to show that any twodistinct columns Ai, Aj of A are orthogonal, so fix distinct i, j ∈ {1, . . . , n} and let θ bethe angle between Ai and Aj . Since f preserves distance, we have

d(Ai, Aj) = d(f(ei), f(ej))

= d(ei, ej)

=√

2.

Now the Law Of Cosines gives

2 = d(Ai, Aj)2

= |Ai|2 + |Aj |2 − 2|Ai||Aj | cos θ

= 2− 2 cos θ,

so we must have cos θ = 0, hence Ai and Aj are orthogonal, as desired.

To see that (3) implies (4), we just need to show that a linear transformation f : Rn →Rn which preserves distance is surjective. (Recall that every distance preserving map isinjective and that an isometry is a distance preserving bijection.) Since f is injective,we can just use the general linear algebra fact that an injective linear transformationbetween vector spaces of the same (finite) dimension is bijective. Alternatively, we coulduse the fact that (3) implies (2) to establish bijectivity of f .

Obviously (4) implies (3).

Definition 1.3.2. A linear transformation f : Rn → Rn satisfying the equivalent con-ditions of Proposition 1.3.1 is called an orthogonal linear transformation, or simply anorthogonal transformation. Similarly, a matrix A ∈ Mat(n×n) is called an orthogonal ma-trix iff the corresponding linear transformation fA is an orthogonal linear transformation(iff the columns of A form an orthonormal basis for Rn).


Since a composition of linear transformations is a linear transformation and a com-position of isometries is an isometry, we see that a composition of orthogonal lineartransformations is an orthogonal linear transformation. The group of orthogonal lineartransformations Rn → Rn is denoted On. The group On may be viewed as a subgroup ofboth GLn(R) and Isom(Rn). Indeed, the proposition above shows that

On = GLn(R) ∩ Isom(Rn),

viewing all of these groups as subgroups of Homeo(Rn).The orthogonal transformations of Rn can be explicitly described:

Proposition 1.3.3. Suppose f : R2 → R2 is an orthogonal transformation, with asso-ciated matrix A = Af . Then the determinant of A is ±1. If detA = 1, then there is aunique angle θ ∈ R/2πZ such that

A =

(cos θ − sin θ

sin θ cos θ

).

If detA = −1, then there is a unique angle θ ∈ R/2πZ such that

A =

(cos θ sin θ

sin θ − cos θ

).

Proof. If f is an orthogonal transformation, then f(e1) and f(e2) (the columns of A)have magnitude one, hence lie on the unit circle

S1 := {(x, y) ∈ R2 : x2 + y2 = 1},

so we can write f(e1) = (cos θ, sin θ) for a unique θ ∈ R/2πZ. Since f(e2) must also be apoint of the unit circle orthogonal to f(e1), the second column of A is either

(cos(θ + π/2), sin(θ + π/2)) = (− sin θ, cos θ)

or

(cos(θ − π/2), sin(θ − π/2)) = (sin θ,− cos θ).

Remark 1.3.4. Proposition 1.3.3 characterizing the orthogonal transformations of R2

in particular shows that each such orthogonal transformation has determinant ±1. It istrue more generally that any orthogonal transformation of Rn has determinant ±1. Tosee this, observe that for any square matrix A with real entries, we have

(AtrA)ij = 〈Ai, Aj〉,

where Atr denotes the transpose of A and Ai denotes the ith column of A. In particular,we see that A is an orthogonal matrix iff AtrA = Id. Since detA = detAtr by generaltheory of determinants, this shows that (detA)2 = det Id = 1.


Example 1.3.5. (Rotations) If f : R2 → R2 is an orthogonal transformation withdetAf = 1 (also called a special orthogonal transformation) and we write f as in the aboveproposition, then, geometrically, f is given by counter-clockwise rotation around theorigin through the angle θ. Indeed, one checks readily that d(0, f(x) = d(0, x) for x ∈ R2,and that if η is the counter-clockwise angle between the positive x-axis and x ∈ R2 \ {0},then the counter-clockwise angle between the positive x-axis and f(x) ∈ R2 \ {0} is η+ θ.The group SO2 of special orthogonal transformations is a normal subgroup of O2. Onechecks easily that the map taking θ ∈ R/2πZ to the first matrix A = Aθ in the aboveproposition (or rather, to the corresponding orthogonal transformation fA) defines anisomorphism of groups

(R/2πZ,+) ∼= SO2 .

One can also define an isometry f : R2 → R2 by rotating around an arbitrary point ofR2. The reader can check that such an isometry can be expressed as a composition oftranslations (Example 1.2.3) and a rotation around the origin.

Example 1.3.6. (Reflections) The matrix

A :=

(1 0

0 −1

)

is clearly an orthogonal matrix, so the associated linear transformation f = fA : R2 → R2

is an orthogonal linear transformation, hence, in particular, an isometry. Note thatf(x, y) = (x,−y), so that f can be described geometrically by reflecting over the x-axis.Since (

cos θ sin θ

sin θ − cos θ

)=

(cos θ − sin θ

sin θ cos θ

)(1 0

0 −1

),

the classification of orthogonal transformations of R2 in Proposition 1.3.3 shows thatevery such transformation is either a rotation around the origin, or the reflection acrossthe x-axis followed by such a rotation.

More generally, one can define an isometry rL : R2 → R2 by reflecting across anarbitrary line L ⊆ R2. The reader can check that such an isometry can be expressedas a composition of rotations, translations, and the reflection over the x-axis discussedpreviously. In particular, it is a standard linear algebra exercise to show that reflectionover a line passing through the origin is an orthogonal linear transformation of determinant−1.

1.4 Affine transformations

Let K be a field, Kn the set of ordered n-tuples of elements of K, as in §1.1.

Definition 1.4.1. An affine transformation is a function f : Kn → Kn expressible asthe composition of a linear transformation (Definition 1.1.1) followed by a translation(Example 1.2.3).

1.4 Affine transformations 13

The first thing to observe is that the expression of an affine transformation f = ftAas a linear transformation A followed by a translation ft is unique because we can recovert from f via

f(0) = ft(A(0)) = ft(0) = t

and then we can recover A from f and t by the obvious formula A = f−tf . In other words,if we fix A ∈ GLn(K) and t ∈ Kn, then we get an affine transformation [A, t] : Kn → Kn

by setting [A, t](x) := Ax + t. Every affine transformation is of the form [A, t] for aunique A ∈ GLn(K) and t ∈ Kn, so that the set of affine transformations is in bijectivecorrespondence with the set GLn(K)×Kn.

Affine transformations form a group under composition, denoted Aff(Kn). It isimportant to understand that, although Aff(Kn) is in bijective correspondence withGLn(K) × Kn via the bijection described above, this bijection is not an isomorphismof groups between Aff(Kn) and the product group GLn(K) × Kn. To see why this isthe case, let us calculate the composition of two affine transformations [A, t] and [A′, t′].Given x ∈ Kn, we compute

([A, t] ◦ [A′, t′])(x) = [A, t](A′x+ t′)

= A(A′x+ t′) + t

= (AA′)x+ (At′ + t),

so the composition of affine transformations is given by

[A, t][A′, t′] = [AA′, At′ + t]. (1.3)

It follows from (1.3) that the inverse of the affine transformation [A, t] is given by

[A, t]−1 = [A−1,−A−1t]. (1.4)

If we view Aff(Kn) as the set of pairs [A, t] ∈ GLn(K)×Kn with composition law (1.3),then the group of translations Kn is identified with the subgroup of Aff(Kn) consisting ofthe pairs [A, t] where A = Id, and the group GLn(K) of invertible linear transformationsis identified with the subgroup of Aff(Kn) consisting of the pairs [A, t] where t = 0.

Theorem 1.4.2. The group of translations (Kn,+) is a normal subgroup of Aff(Kn) andevery element of Aff(Kn) can be written uniquely as a composition of a translation andan invertible linear transformation. In other words, Aff(Kn) is a semi-direct product of(Kn,+) and GLn(K).

Proof. The only statement not proved in the discussion above is the normality of (Kn,+).We calculate the conjugate of a translation [Id, s] by an arbitrary affine transformation[A, t] by using the formulas (1.3) and (1.4):

[A, t][Id, s][A, t]−1 = [A, t][Id, s][A−1,−A−1t]= [A,As+ t][A−1,−A−1t]= [Id, A(−A−1t) +As+ t]

= [Id, As].

This is a translation, as desired.


At present, the consideration of the groups Aff(Kn) may seem somewhat unmotivated.In fact, these groups are perhaps the most important groups that arise in geometry—hereare some reasons why:

1. They are used to define and study (affine) linear changes of coordinates and(affine) linear changes of variables, hence they play a role in problems of classifyingpolynomials and (affine) plane curves.

2. The group Aff(Kn) is closely related to the projective general linear group, andhence to the geometry of the projective space KPn and the inclusion Kn ↪→ KPn,which we shall study later.

3. As we shall see in §1.5, the group of affine transformations Aff(R2) is closely relatedto the group of isometries of R2.

4. We will see in Theorem 1.4.5 that affine transformations arise naturally whenconsidering the geometry of lines in R2.

Proposition 1.4.3. Suppose P0, . . . , Pn ∈ Kn are (n+1) points of Kn in general position,meaning they are not all contained in a translate of some linear subspace V ⊆ Kn ofdimension < n. Then there is a unique affine transformation [A, t] ∈ Aff(Kn) such that[A, t](0) = P0 and [A, t](ei) = Pi for i = 1, . . . , n.

Remark 1.4.4. When n = 2, the condition in Proposition 1.4.3 that P,Q,R ∈ K2 be in“general position” just says that P,Q,R are not collinear (are not contained in a line).

Proof. Since the Pi are in general position, the vectors P1 − P0, . . . , Pn − P0 must belinearly independent, hence by linear algebra, they form a basis for Kn and there is aunique linear transformation A : Kn → Kn with Aei = P1−P0. The affine transformation[A,P0] given by the linear transformation A followed by translation along P0 is clearlythe unique affine tranformation with the desired properties.

Theorem 1.4.5. A function f : R2 → R2 is an affine transformation iff f is a continuousfunction with the following property (*): For any P,Q,R ∈ R2, P,Q,R are collinear ifff(P ), f(Q), f(R) are collinear.

Proof. It is clear that any affine transformation of R2 is a continuous function satisfying(*) and that a composition of continuous functions satisfying (*) is also a continuousfunction satisfying (*). Suppose now that h is a continuous function satisfying (*). Thenh(0), h(e1) and h(e2) are non-collinear, so by the previous proposition there is an affinetransformation g : R2 → R2 with g(0) = h(0), g(e1) = h(e1) and g(e2) = h(e2). We claimthat h = g—equivalently, g−1h = Id. Since g−1 is an affine transformation, our earlierobservations show that f := g−1h is a continuous function satisfying (*) and fixing 0, e1,and e2. We thus reduce to proving that such a function f must be the identity.

In the rest of the proof we assume that f : R2 → R2 satisfies (*). We first note thatf must be injective: Indeed, suppose f(P ) = f(Q), but P 6= Q. Pick any point R ∈ R2

so that P,Q,R are not collinear. But f(R), f(P ), and f(Q) are certainly collinear,contradicting (*). We next note that if P,Q are distinct points of R2, then the image

1.4 Affine transformations 15

f(PQ) of the line PQ must be contained in the line f(P )f(Q). Finally, we note that ifPQ, RS are parallel lines, then the lines f(P )f(Q) and f(R)f(S) must be parallel.

From these simple remarks, we next deduce the key observation: If f fixes threevertices P,Q,R of a parallelogram with vertices P,Q,R, S, then it must also fix the forthvertex S. To see this, we can assume the vertices are labelled so that the lines PQ andRS are parallel and the lines PS and QR are parallel. Since f fixes P and Q and satisfies(*), it takes the line PQ into itself, so, since f also fixes R, it takes RS into Rf(S),hence, since f takes parallel lines to parallel lines, f(S) must lie on the line RS. But wesee similarly that f(S) must also lie on the line PS, so f must fix S.

Now, if f fixes 0, e1, e2, then we can repeatedly apply the above observation toparallelograms of one of the following five forms

1. P , P + e1, P + e2, P + e1 + e2

2. P , P + e1, P + e1 + e2, P + 2e1 + e2

3. P , P + e1, P + e2, P − e1 − e2

4. P , P + e1, P + e2, P + e1 − e2

5. P , P + e1, P − e2, P + e1 + e2

with P ∈ Z2 ⊆ R2 chosen appropriately to see that f must fix each point of Z2 ⊆ R2. Forexample: We first apply the observation to the parallelogram of the first type with P = 0to see that f fixes e1 + e2. Next we repeatedly apply the observation to parallelogramsof the second and third types to see that f fixes all points of the form (a, b) with a ∈ Z,b ∈ {0, 1}. Finally we apply the observation repeatedly to parallelograms of the fourthand fifth types to conclude that f fixes every point of Z2.

Next we argue that if f fixes 0, e1, and e2, then it must fix all points with rationalcoordinates. Consider a point S = (x, y) ∈ Q2 ⊆ R2. Take any point P ∈ Z2 not equalto S. Since the line PS has rational slope and contains P ∈ Z2, it will contain somepoint R ∈ Z2 distinct from P . Since f fixes P and R it takes PR into itself, hencef(S) ∈ PR = PS. Now just take another point P ′ ∈ Z2 not lying on the line PR = PSand repeat the same argument to show that f(S) ∈ P ′S, hence f(S) = S as desired.

Now we have shown that a function f : R2 → R2 satisfying (*) and fixing 0, e1, e2fixes all points with rational coordinates, hence, if f is also continuous, it must be theidentity.

Remark 1.4.6. For a general field K, a function f : K2 → K2 satisfying (*) in Theo-rem 1.4.5 is called a collineation. (One defines a line in K2 to be a subset of K2 arisingas a translation of a one-dimensional linear subspace of the vector space K2.) Here isone way to construct collineations: Let G be the automorphism group of the field K(the “Galois group” of K). Each σ ∈ G gives a collineation fσ : K2 → K2 by applyingσ to the coordinates; let us call such a collineation a Galois collineation. Notice thateach Galois collineation fixes (0, 0), (1, 0), and (0, 1). In fact, it can be shown that anycollineation of K2 fixing (0, 0), (1, 0), and (0, 1) is a Galois collineation. (The argumentfor this is elementary—I have omitted it only because it is a bit long and tedious.) It


follows from the argument in the above proof that any collineation of K2 can be writtenas a composition of an affine transformation of K2 and a Galois collineation. It can alsobe shown that the automorphism group of the field R is trivial (the idea here is that theordering of R can be interpreted in terms of the field structure because, in R, being ≥ 0 isthe same thing as being a square, which is a field-theoretic notion). It follows that everycollineation of R2 is, in fact, an affine transformation. In other words, Theorem 1.4.5remains true even if the word “continuous” is deleted. See Propositions 3.11 and 3.12 in[H2].

Remark 1.4.7. The proof of Theorem 1.4.5 shows that any collineation f : Q2 → Q2 isan affine transformation.

1.5 Classification of isometries of the plane

In this section we will classify all isometries of R2.

Lemma 1.5.1. Let C1, C2 be circles in R2 with centers P1, P2, respectively. Assumethat C1 and C2 intersect in precisely two points: P and Q. (In particular this implies thatP1 6= P2.) Then the line segment PQ is orthogonal to the line P1P2 and PQ intersectsP1P2 at the midpoint of PQ.

Proof. Let f : R2 → R2 be the isometry of R2 given by reflection (Example 1.3.6) acrossthe line L = P1P2. Since P1, P2 ∈ L, we have f(Pi) = Pi (i = 1, 2). Since f is an isometrywith f(Pi) = Pi, f takes Ci into itself, hence f(P ), f(Q) are in C1 ∩ C2 = {P,Q}. Wecannot have f(P ) = P and f(Q) = Q, for then P , Q, and Pi would all lie on L (the fixedlocus of f), and P,Q would be equidistant from both P1 and P2, hence we would haveP = Q. So, since f is bijective, we must have f(P ) = Q and hence P = f(f(P )) = f(Q).The conclusion follows easily.

Lemma 1.5.2. Suppose f : R2 → R2 is a distance preserving function such thatf(P1) = P , f(P2) = P2 and f(P3) = P3 for three non-collinear points P1, P2, P3 ∈ R2.Then f is the identity.

Proof. We need to show that f(Q) = Q for an arbitrary Q ∈ R2. Suppose—toward acontradiction—that f(Q) 6= Q for some Q ∈ R2. Then Q cannot be one of the Pi, sothe distances di := d(Pi, Q) are positive. Let Ci be the circle of radius di centered atPi. Then Q ∈ C1 ∩ C2 ∩ C3. Since f preserves distance and fixes Pi it takes Ci intoCi, hence f(Q) ∈ C1 ∩ C2 ∩ C3. Suppose i, j ∈ {1, 2, 3} are distinct. Then the circlesCi ∩Cj cannot intersect in more than two points, for then they’d be equal and we’d havePi = Pj , contradicting the assumption that the Pi aren’t collinear. We conclude thatCi ∩ Cj = {Q, f(Q)}. Let Lij be the line through Pi and Pj . By the previous lemma,

the lines L12, L13, and L23 are all orthogonal to Qf(Q) and each intersects Qf(Q) atthe midpoint of the line segment Qf(Q), so we have L12 = L13 = L23, contradicting theassumption that the Pi are not collinear.

Lemma 1.5.3. A function f : R2 → R2 is distance preserving and satisfies f(0) = 0 ifff is an orthogonal linear transformation.

1.5 Classification of isometries of the plane 17

Proof. Since any linear transformation fixes zero (takes 0 to 0), Proposition 1.3.1 impliesthat any orthogonal linear transformation is an isometry fixing zero. Now supposef : R2 → R2 is distance preserving and satisfies f(0) = 0. Then, since f preservesdistance, f(e1) and f(e2) have to be on the unit circle (i.e. at distance one from theorigin) and

d(f(e1), f(e2)) = d(e1, e2) =√

2.

As in the proof of Proposition 1.3.1, the Law Of Cosines then implies that f(e1) and f(e2)are orthogonal unit vectors, so the matrix A with columns f(e1), f(e2) is an orthogonalmatrix and the corresponding linear transformation fA : R2 → R2 is an orthogonal lineartransformation with fA(e1) = f(e1), fA(e2) = f(e2), and fA(0) = 0 = f(A). Then thecomposition f−1A f is an isometry of R2 fixing the three non-collinear points 0, e1, e2, hencef−1A f = Id by the previous lemma, hence f = fA.

Theorem 1.5.4. Every distance preserving function f : R2 → R2 can be written uniquelyas the composition of an orthogonal linear transformation followed by a translation. Inparticular, every such f is an isometry.

Proof. Set t := −f(0) ∈ R2, and let ft : R2 → R2 be the corresponding translation. Thenthe composition ftf : R2 → R2 is a distance preserving function satisfying f(0) = 0,so by the previous lemma we have ftf = g for some orthogonal linear transformationg : R2 → R2, and hence f = f−tg displays f as a composition of an orthogonal lineartransformation followed by a translation. The uniqueness of this expression is establishedin the same manner one establishes uniqueness of the corresponding expression for affinetransformations in §1.4.

Corollary 1.5.5. Every distance preserving function f : R2 → R2 is an affine transfor-mation.

The following corollary could also be proved directly:

Corollary 1.5.6. Every isometry f : R2 → R2 can be expressed as a composition ofreflections (Example 1.3.6), hence the group Isom(R2) can be generated by elements oforder 2.

Proof. In light of the theorem it suffices to show that every translation and everyorthogonal transformation can be expressed as a composition of reflections. Given twoparallel lines L,L′, the composition rL′rL of the corresponding reflections will be thetranslation f2t, where t ∈ R2 is the vector such that L′ = ft(L) (exercise!), so by choosingL and L′ appropriately, we can realize every translation as a composition of (at most)two reflections.

In Example 1.3.6 we noted that every orthogonal translation of R2 is either a rotationaround the origin, or a reflection over the x-axis followed by such a rotation. It thusremains only to prove that the counter-clockwise rotation fθ around the origin throughan angle θ can be expressed as a composition of reflections. Let L (resp. L′) be the linethrough the origin obtained by rotating the x-axis counter-clockwise around the originthrough the angle θ/2 (resp. θ).


I claim that the composition rL′rL is fθ. As discussed in Example 1.3.6, one cansee by linear algebraic methods that both rL and rL′ are orthogonal transformationsof determinant −1, hence rL′rL is an orthogonal transformation of determinant 1, soit must be a counter-clockwise rotation around the origin through some angle by theclassification of orthogonal transformations in Proposition 1.3.3. This reduces us toshowing that (rL′rL)(e1) is equal to fθ(e1), which is (cos θ, sin θ). This is exactly whatwe arranged by our choice of L and L′: We have rL(e1) = fθ(e1) by the choice of L, andthis point is fixed by rL′ since it lies on L′, by our choice of L′.

Remark 1.5.7. The proof of the above corollary shows that every isometry f : R2 → R2

is the composition of at most five reflections. In fact at most three are needed.

In light of the classification of isometries of R2 in Theorem 1.5.4, we can make thefollowing definition, which will be useful later:

Definition 1.5.8. An isometry f : R2 → R2 is called orientation preserving iff, in theunique expression f = ftg of f as a composition of an orthogonal transformation gfollowed by a transflation ft, the orthogonal transformation g is a special orthogonallinear transformation (i.e. a rotation—see Example 1.3.5).

Remark 1.5.9. It is possible to define the notion of “orientation preserving” for anyhomeomorphism f : Rn → Rn in a manner restricting to the notion in Definition 1.5.8 forisometries of R2 and satisfying the expected properties: The identity map is orientationpreserving and a composition fg of homeomorphisms is orientation preserving iff eitherboth f and g are orientation preserving, or neither f nor g is orientation preserving.

1.6 Algebraic automorphisms

Let K be a field. Let us agree that a function f : Kn → Km is algebraic iff there arepolynomials

f1(x1, . . . , xn), . . . , fm(x1, . . . , xn) ∈ K[x1, . . . , xm]

such that

f(x1, . . . , xn) = (f1(x1, . . . , xn), . . . , fm(x1, . . . , xn))

for every x = (x1, . . . , xn) ∈ Kn. A function f : Kn → Kn is called an algebraicautomorphism of Kn iff f is bijective and both f and f−1 are algebraic.

For example, every affine transformation is an algebraic automorphism. One cancheck that every algebraic automorphism of K1 is of the form x 7→ λx+a for some λ ∈ K∗,a ∈ K—in other words, every algebraic automorphism of K1 is an affine transformation.For K2, however, there are algebraic automorphisms K2 → K2 which are not affinetransformations:

Example 1.6.1. The function f : K2 → K2 defined by f(x, y) := (x, x2 + y) is analgebraic automorphism of K2 with inverse g given by g(x, y) = (x, x2 − y). The readercan check as an exercise that f is not an affine transformation (unless K = F2 is “the”field with two elements, for then x2 = x for “every” (i.e. “both”) x ∈ F2).

1.7 Exercises 19

The Jacobian Conjecture is a famous unsolved conjecture about algebraic automor-phisms of Cn. It says that if f1, . . . , fn ∈ C[x1, . . . , xn] are polynomials such that thepolynomial

det

∂f1∂x1· · · ∂f1∂xn

......

∂fn∂x1· · · ∂fn∂xn

(the Jacobian determinant) is a non-zero constant, then the corresponding algebraicfunction f : Cn → Cn is an algebraic automorphism. As mentioned above, this is nothard to see when n = 1, though it is unknown even when n = 2.

1.7 Exercises

Exercise 1.1. Let f : R2 → R2 be the linear transformation corresponding to the(invertible) matrix

A =

(1 1

0 1

).

Let S1 := {x ∈ R2 : d(x, 0) = 1} be the unit circle. Draw a picture of the image f(S1) ofS1 under f and a picture of f(f(S1)).

Exercise 1.2. Let f : R2 → R2 be the linear transformation corresponding to the matrix

Aθ :=

(cos θ − sin θ

sin θ cos θ

)

(for some θ ∈ R/2πZ). Show, by explicit calculation with the distance formula, that f isan isometry.

Exercise 1.3. Prove that the special orthogonal group SO2 is a normal subgroup of theorthogonal group O2.

Exercise 1.4. For points x, y ∈ R2, the line segment xy is the set of points z in R2 thatcan be written in the form z = tx+ (1− t)y for some t ∈ [0, 1]. Prove that a point z ∈ R2

lies on the line segment xy iff the triangle inequality d(x, y) ≤ d(x, z) + d(z, y) is actuallyan equality.

Exercise 1.5. Suppose f : R2 → R2 is an isometry and let L ⊆ R2 be a line. Prove“directly” (without making use of the classification of isometries of R2) that f(L) ⊆ R2 isa line. For clarity, let us take as the definition of a line: a subset L ⊆ R2 for which thereexist real numbers a, b, c (with a, b not both equal to zero) such that

L = {x = (x1, x2) ∈ R2 : ax1 + bx2 + c = 0}.

(The previous exercise might be helpful.)

Chapter 2

Parametric Plane Curves

There are many possible things one can mean by a “curve” in the plane R2. In thischapter we will return to the study of parametrized plane curves, which should be familiarfrom calculus classes. Here are the main definitions:

Definition 2.0.1. A parametrized plane curve is a function γ : [a, b]→ R2, defined onsome closed interval [a, b] ⊆ R. We use t to denote a point of [a, b] and x, y : [a, b]→ R todenote the coordinates of γ, so that γ(t) = (x(t), y(t)). We assume that the derivativesx′, y′, x′′, y′′ with respect to t exist and are continuous. We call γ′ := (x′, y′) the tangentvector to γ and we say that γ is smooth iff γ′(t) 6= (0, 0) for all t ∈ [a, b]. A smoothparametrized plane curve γ is said to be parameterized by arc length iff a = 0 and|γ′(t)| = 1 for all t ∈ [0, b]—in this case we generally use s instead of t for the parameterand we write ` instead of b, for reasons that will become clear momentarily.

Example 2.0.2. Fix a positive real number r. The function

γ : [0, 2π] → R2

γ(t) := (r cos t, r sin t)

is a smooth parametrized curve whose image γ([0, 2π]) is a circle of radius r centered atthe origin. The function

η : [0, 2πr] → R2

η(s) := (r cos(s/r), r sin(s/r))

is a smooth parametrized plane curve parametrized by arc length with the same imageas γ.

In this chapter, we will use the word curve (resp. curve parametrized by arc length)to mean smooth parameterized plane curve (resp. smooth parametrized plane curveparametrized by arc length). We will study various geometric quantities associated withcurves—for example, the length. Our assumption that the derivatives x′ and y′ exist andare continuous ensures that we can define the length in a fairly simple manner. Whilethere are much more general definitions of length, they are less amenable to calculation

20

2.1 Reparametrization 21

and, in any case, are beyond the scope of these notes. The other important invariantattached to a curve is its curvature. Since our curves are plane curves, one can actuallyattach a sign to the curvature (this is not usually done in calculus classes) to define thesigned curvature of a curve γ—this is a continuous function k : [a, b]→ R. This is ensuredby our assumption that the second derivatives x′′ and y′′ exist and are continuous.

Here I want to focus on the interaction between the group of isometries of the planeand the properties of plane curves. The upshot is that the length and curvature areisometry invariant and they are, in some sense we will make precise, the “only” isometryinvariants of a curve.

2.1 Reparametrization

The reader will probably agree that the two curves γ, η in Example 2.0.2 are in somesense “the same.” After all, the function γ is just the composition of multiplication byr, viewed as a function [0, 2π]→ [0, 2πr], followed by η. In this section we are going tomake this notion of “the same” precise. This point is implicit in the calculus treatmentof parametrized curves, but it never really made explicit, probably because it is a bittechnical.

Consider two closed intervals [a, b], [c, d] ⊆ R. Let us use s for the coordinate on[c, d] and t for the coordinate on [a, b] for clarity. Let us agree that a nice bijectiong : [c, d]→ [a, b] is a bijection g : [c, d]→ [a, b] which is twice continuously differentiablewith g′(s) > 0 for all s ∈ [c, d].

It follows from the Implicit Function Theorem (see Theorem 6.3.1 in the appendixfor a precise statement) that the inverse of g is also a nice bijection. Let us discuss thispoint a bit further, at it is rarely explained well in an introductory calculus sequence.For clarity, let us denote the inverse of g by f = f(t) : [a, b] → [c, d], rather than byg−1. Then we have fg = Id, or, in other words, f(g(s)) = s for every s ∈ [c, d]. If wedifferentiate both sides of this equality with respect to s by using the Chain Rule, wefind that

f ′(g(s))g′(s) = 1. (2.1)

Then, using the fact that g′(s) > 0, we solve (2.1) for f ′(g(s)) to find f ′(g(s)) = 1/g′(s).Since every t ∈ [a, b] is g(s) for some (in fact a unique) s ∈ [c, d], we conclude thatf ′(t) > 0 for all t ∈ [a, b]. Notice that we have to use the Implicit Function Theoremhere to know a priori that f is differentiable, so that the Chain Rule can be appliedto calculate the derivative of the composition f(g(s)). We can use the same method tocalculate the second derivative of g, since we know from the Implicit Function Theoremthat it exists (and is continuous): We just differentiate (2.1) again with respect to s tofind

f ′(g(s))g′′(s) + g′(s)2f ′′(g(s)) = 0

and then we use that g′(s) > 0 to solve for f ′′(g(s)). This gives

f ′′(g(s)) = −f′(g(s))g′′(s)

g′(s)2.

22 Chapter 2 Parametric Plane Curves

The Implicit Function Theorem says that this formula is indeed valid, so f ′′ exists and iscontinuous because g′′ exists and is continuous.

It is customary in the situation described above to get rid of the notation for g andits inverse, and just view t = t(s) as a function of s, and s = s(t) as a function of t. Withthis understanding, formula (2.1) is rewritten

ds

dt

dt

ds= 1. (2.2)

Definition 2.1.1. If γ : [a, b] → R2 is a (smooth parametrized plane) curve andt(s) : [c, d] → [a, b] is a nice bijection, then the composition γt : [c, d] → R2 is alsoa curve (by the Chain Rule). A curve obtained from γ in this manner is called areparametrization of γ. Two curves γ : [a, b]→ R2, η : [c, d]→ R2 are called equivalent iffη is a reparametrization of γ. A parametric plane curve is an equivalence class of smoothparametrized plane curves.

The notion of “equivalence” defined above is indeed an equivalence relation on the setof curves. The observation at the beginning of this section implies that the two curves inExample 2.0.2 are equivalent.

2.2 Length and the arc length parametrization

Among all reparametrizations of a given curve, there is one particularly nice choice, calledthe arc length parametrization, which we will now describe.

Given a curve γ = (x, y) : [a, b]→ R2, recall that we call γ′ = γ′(t) = (x′(t), y′(t)) thetangent vector to γ at γ(t). The smoothness assumption on γ says that γ′ is never thezero vector, hence its magnitude

|γ′| =√

(x′)2 + (y′)2

is a function |γ′| : [a, b]→ R>0.We define the length ` = `(γ) of γ to be

`(γ) :=

∫ b

a|γ′(t)|dt. (2.3)

More generally, we define the arc length function s : [a, b]→ [0, `] by

s(t) :=

∫ t

a|γ′(τ)|dτ.

By the Fundamental Theorem of Calculus we have

ds

dt= |γ′(t)|, (2.4)

which is > 0 by the assumption that γ is smooth. Our assumption that x′′ and y′′ existand are continuous also shows that

d2s

dt2=x′x′′ + y′y′′

|γ′(t)|

2.3 Curvature and the Frenet formulas 23

exists and is continuous, hence our arc length function s is a nice bijection, and hence sois its inverse

t = t(s) : [0, `] → [a, b].

As discussed in §2.1, we have

dt

ds=

1

ds/dt(2.5)

=1

|γ′(s(t))|.

The reparametrization η(s) := γ(t(s)) : [0, `] → R2 of γ is called the arc lengthparametrization of γ. As the terminology suggests, the curve η is parameterized by arclength because we compute ∣∣∣∣dηds

∣∣∣∣ =

∣∣∣∣dγdt dtds∣∣∣∣

=

∣∣∣∣dγdt∣∣∣∣ ∣∣∣∣ dtds

∣∣∣∣= 1

using the Chain Rule and (2.5).

The upshot of this section is that if one considers curves up to reparametrization,then nothing is lost (in theory) by considering only curves parametrized by arc length.However, it should be mentioned that the integrals defining `(γ) and s(t) can rarely becarried out in terms of elementary functions, so if one ever needs to explicitly compute aquantity associated to the arc length parametrization of a curve γ, one usually tries toexpress that quantity in terms of the original parametrization of γ by using the ChainRule.

2.3 Curvature and the Frenet formulas

Consider a curve γ = γ(s) : [0, `]→ R2 parametrized by arc length. We call T(s) := γ′(s)the unit tangent vector to γ at γ(s). Since γ is parametrized by arc length, we have |T| = 1,independent of s. By linear algebra, there is a unique unit vector N = N(s) orthogonal toT such that the matrix (T,N) with columns given by T and N has determinant 1. Thevector N(s) is called the principal unit normal to γ at γ(s). Since |T| = 1 is constant, ifwe differentiate both sides with respect to s we find that T · T′ = 0. So T′ is orthogonalto T, hence it must be some scalar multiple of N:

T′ = kN. (2.6)

The scalar multiple k = k(s) is called the signed curvature of γ at γ(s). The formula fork in Exercise 2.3 makes it clear that k is a continuous function k : [0, `]→ R.


One can now consider the derivative N′. By the same arguments used for T, wesee that N′ will be some scalar multiple of T: N′ = αT. In fact this α doesn’t give usanything “new”: By differentiating 〈N,T〉 = 0, we find that

〈N′,T〉+ 〈N,T′〉 = 0.

Using the definitions of the signed curvature k and α we can rewrite this

〈αT,T〉+ 〈N, kN〉 = 0.

Using bilinearity of the inner product and the fact that T and N are unit vectors, thisgives α+ k = 0, or α = −k. The discussion thus far establishes the Frenet formulas (forcurves in the plane parametrized by arc length):

T′ = kN (2.7)

N′ = −kT.

Example 2.3.1. Consider the curve parametrized by arc length η : [0, 2πr]→ R2 fromExample 2.0.2 whose image is the circle of radius r centered at the origin in R2. Notethat η traces out its image in the counter-clockwise direction. We have

T = (r cos(s/r), r sin(s/r))′

= (− sin(s/r), cos(s/r)).

Since (− sin(s/r) − cos(s/r)

cos(s/r) − sin(s/r)

)is an orthogonal matrix, we have

N = (− cos(s/r),− sin(s/r)).

We compute

T′ = (−(1/r) cos(s/r),−(1/r) sin(s/r))

= (1/r)N,

so the signed curvature k of η is constant, equal to 1/r.

If there is any chance of confusion (for example, if we consider two curves parametrizedby arc length at the same time) then we will write Tγ , Nγ , and kγ instead of T, N, and k.

As mentioned in the previous section, the arc length parametrization of a curveγ = γ(t) is generally difficult, or—in some sense—impossible, to write down, so it isworth having formulas for T, N, and k in terms of an arbitrary parametrization. This isdone by using formula (2.5), together with the Chain Rule, to re-express the derivatives

2.4 Plane curves up to isometry 25

with respect to arc length s used to define T, N, and k in terms of differentiation withrespect to t. For example,

T(t) =dγ(t(s))

ds

=dγ

dt

dt

ds

=1

|γ′(t)|γ′(t),

where the primes denote differentiation with respect to t. Then N(t) is defined from T(t)in the same way that N(s) is defined from T(s). Similarly we compute

dTds

=dTdt

dt

ds

=1

|γ′|T′,

where the primes denote derivatives with respect to t. Then k(t) is defined by

dTds

= k(t)N(t).

2.4 Plane curves up to isometry

If f : R2 → R2 is any nice enough function (a smooth bijection with smooth inverse, say),and γ : [a, b]→ R2 is a smooth parametrized curve, then the composition fγ : [a, b]→ R2

is also a smooth parametrized curve. We will be interested in this construction onlywhen f is an orientation preserving isometry (Definition 1.5.8) and γ : [0, `]→ R2 is acurve parametrized by arc length. In this situation we can write f = ftg, where g ∈ SO2

and ft is a translation. Let us write A = Ag for the orthonormal matrix correspondingto g. One checks (Exercise 2.4) that:

1. (fγ)′ = Aγ′, hence fγ is also parameterized by arc length and Tfγ = ATγ .

2. `(fγ) = `(γ).

3. The principal unit normal vectors of γ and fγ are related by Nfγ = ANγ .

4. The curvatures of γ and fγ are equal: kγ = kfγ .

Theorem 2.4.1. Suppose γ : [0, `]→ R2 and γ : [0, `]→ R2 are two curves parametrizedby arc length. Then the following are equivalent:

1. γ = fγ for some orientation preserving isometry f : R2 → R2.

2. The curves γ and γ have the same length and curvature: That is ` = ` and k = k.


Proof. The fact that (1) implies (2) is discussed above and left for the reader as Exer-cise 2.4. To see that (2) implies (1), suppose γ and γ are curves parametrized by arclength with the same length and curvature. Let t := γ(0)− γ(0) ∈ R2 and let A ∈ SO2

be the rotation taking the unit vector γ′(0) to the unit vector γ′(0). Then f := ftfA isan isometry of R2 such that the curves γ and fγ have the same initial point and initialtangent vector. By the implication already “proved,” these two curves also have thesame length and curvature. By replacing γ with fγ, we thus reduce to proving that if γand γ are curves parametrized by arc length with the same initial point, the same initialtangent vector, and the same length ` and curvature k, then they are equal. Let T andN (resp. T and N) be, respectively, the unit tangent vector and principal unit normal forthe curve γ (resp. γ). Using the Frenet Formulas (2.7) we compute(

〈T− T,T− T〉+ 〈N− N,N− N〉)′

= 2〈T′ − T′,T− T〉+ 2〈N′ − N′,N− N〉= 2〈kN− kN,T− T〉 − 2〈kT− kT,N− N〉= −2k〈N,T〉 − 2k〈N,T〉+ 2k〈N,T〉+ 2k〈N,T〉= 0.

This shows that the function in parentheses is constant—but it is zero at s = 0 sinceT(0) = T(0) (hence N(0) = N(0) as well), so it is zero. But this function is the sum ofthe magnitude of T− T and the magnitude of N−N, both of which are non-negative, sowe must have T = T. Since T = γ′ and T = γ′, this proves that the derivatives of thecoordinate functions of γ and γ are equal. But we also have γ(0) = γ(0), hence γ = γ bythe Fundamental Theorem of Calculus.

Remark 2.4.2. There is an analogue of Theorem 2.4.1 for curves in R3 parametrizedby arc length. In this case, the curvature cannot be given a sign, and one must alsoconsider another invariant called the torsion. (The proof is essentially the same, thoughthe Frenet formulas become a bit more complicated in three dimensions.) For this, seethe first chapter of the book Differential Geometry of Curves and Surfaces by ManfredoP. Do Carmo, from which the material in this section has been “borrowed.”

2.5 Exercises

Exercise 2.1. Suppose f : R2 → R2 is an isometry of R2 which is orientation reversing(not orientation preserving) and γ is a curve parametrized by arc length in R2. Showthat fγ is a curve parametrized by arc length in R2 with signed curvature kfγ = −kγ .

Exercise 2.2. Let P and Q be distinct points of R2. Let v := (Q− P )/|Q− P | be theunit vector pointing from P to Q and let ` := |Q−P | be the distance from P to Q. Showthat the curve γ : [0, `]→ R2 defined by γ(s) := P + sv is a smooth curve parametrizedby arc length whose signed curvature k is identically zero.

Exercise 2.3. Consider a smooth plane curve γ = γ(s) = (x(s), y(s)) : [0, `] → R2

parametrized by arc length. Show that the signed curvature k of γ is given by

k = x′y′′ − x′′y′.

2.5 Exercises 27

Hint: First figure out an explicit formula for the principal unit normal N in terms of x, yand their derivatives. You also have an obvious such formula for T′, so you just have tocheck that the proposed formula for k actually satisfies T′ = kN. For this, it might behelpful to have the identity obtained by writing out the condition |T| = 1 in terms ofx, y and their derivatives then differentiating both sides.

Exercise 2.4. Check the four listed assertions at the beginning of §2.4.

Chapter 3

Affine Algebraic Plane Curves

Generally speaking, a curve should be a subset of the plane R2 obtained by imposingsome “reasonable” relationship between the two coordinates x, y of a point (x, y) ∈ R2.For example, the graph Γf ⊆ R2 of a function f : R→ R is a curve since it is obtainedby requiring that the second coordinate be the value of f on the first coordinate:

Γf := {(x, y) ∈ R2 : y = f(x)}.

It seems reasonable then, to say that an algebraic curve (more precisely, an affine algebraicplane curve) is a subset of R2 obtained by imposing an “algebraic” (i.e. polynomial)relationship between the coordinates. For example, if

f(x) = a0 + a1x+ · · ·+ anxn

is a polynomial function of x, then the graph Γf ⊆ R2 is an algebraic curve. Here aresome other algebraic curves that you should have encountered at some point:

C1 := {(x, y) ∈ R2 : x2 + y2 = 1}C2 := {(x, y) ∈ R2 : x = 0}C3 := {(x, y) ∈ R2 : x2 − y2 = 1}

The curve C1 is a circle of radius one centered at the origin, C2 is the y-axis, and C3

is a hyperbola. Of course, not all plane curves are graphs of functions (e.g. C1, C2, C3

above).The notion of an algebraic curve makes sense over any field K. For example, replacing

R everywhere with the complex numbers C, one can also consider complex algebraiccurves C ⊆ C2 in the complex plane C2. It may seem at first that this might complicatematters: For one thing, it becomes difficult to visualize a complex plane curve, sincethe complex plane C2 itself has “four real dimensions.”On the other hand, the field ofcomplex numbers C has certain technical advantages over R, particularly the fact that itis algebraically closed : Any polynomial (of positive degree in one variabel) with complexcoefficients can be factored as a product of linear (degree one) polynomials. In general,the theory of algebraic curves, as we have defined them, is most satisfactory over analgebraically closed field, where the relationship between a polynomial and the algebraiccurve it defines can be most satisfactorily understood.

28

3.1 Polynomials and zero loci 29

As this course is meant to be a course on projective geometry, we will not dwell overlyon the subject of affine curves. Our purpose in this section is mainly just to give thereader a taste for the theory of algebraic plane curves, without with our later study ofprojective plane curves would seem rather unmotivated. In order for this chapter to havesome actual content, I have chosen to discuss the problem of classifying affine conics(over R and C) up to an appropriate notion of affine equivalence. The classification willbe given without proof, though we shall give a complete treatment of the analogousclassification problem in the projective setting. Already in this classification problem wewill see that C has an important “advantage” over R: Every complex number is a square,whereas real numbers less than zero are not squares.

3.1 Polynomials and zero loci

Recall that a polynomial in a single variable x with coefficients in a field K is a formalexpression f(x) of the form

f(x) = a0 + a1x+ a2x2 + · · ·+ anx

n,

where the coefficients ai are elements of K. The set of polynomials with coefficients in Kis denoted K[x]. This set carries various additional structure: It is an K-algebra, whichbasically means that one can add, subtract, and multiply polynomials in a meaningfulway, and that one can regard an element of K as a constant polynomial.

Though it is perhaps a stretch to emphasize this point now, it might be worthexplaining—for the more mathematically sophisticated student—what is meant by “formalexpression.” We really mean here that a polynomial is just a way of writing (and thinkingabout) a sequence

(a0, a1, a2, . . . )

of elements of K which is eventually zero. We can add, subtract, and multiply suchsequences (by thinking of them as coefficients of polynomials). Now, of course, we canalso view the polynomial as a function f : K → K, as we often do in calculus whenK = R. It might not seem like such a big deal to distinguish between the formal list ofcoefficients and the corresponding function, and, indead, it is not in many senses: youcan recover the polynomial (i.e. its list of coefficients) from the corresponding functionf : R→ R by the rule

an =f (n)(0)

n!

(as you learn in the study of Taylor series). On the other hand, if you work with coefficientsin a finite field K, say, then of course we can still view a polynomial f(x) ∈ K[x] as afunction f : K→ K, but obviously there can be no hope of recovering the polynomial fromthe corresponding function because K[x] is infinite, while there are only finitely manyfunctions K→ K)! For example, if K = F2 = Z/2Z = {0, 1} is the two-element field, thenx = x2 as functions K→ K, but we do not think of x = 0 + 1x and x2 = 0 + 0x+ 1x2 asbeing the same polynomial. This distinction is largely a pedagogical matter, but it mightbe wise for the reader to forget everything he/she knows about functions R→ R for awhile and just think about polynomials in a completely formal way.

30 Chapter 3 Affine Algebraic Plane Curves

Definition 3.1.1. The degree of a polynomial f as above is the largest n for which thecoefficient of xn in f is non-zero:

deg(f) := max{i : ai 6= 0}.

Thus a non-zero constant polynomial f(x) = a0 6= 0 has degree zero. It is often convenientto define the degree of the zero polynomial to be −∞. The terms “non-constant” and “ofpositive degree” are equivalent; similarly “linear” and “degree one” are interchangeable.

We can make the same definitions for polynomials in two (or more) variables x, y.Now a polynomial is a formal expression of the form

f(x, y) =

∞∑i,j=0

ai,jxiyj ,

where all but finitely many of the real numbers aij are zero. The degree of such apolynomial is defined by

deg(f) := max{i+ j : aij 6= 0}.

Evidently, any polynomial f(x, y) of degree d can be written

f =∑i+j≤d

ai,jxiyj

with at least one ofad,0, ad−1,1, . . . , a1,d−1, a0,d

non-zero. The set K[x, y] of polynomials in variables x, y forms an K-algebra.

Definition 3.1.2. Given a polynomial f(x, y) ∈ K[x, y], we call the subset

Z(f) := {(x, y) ∈ K2 : f(x, y) = 0}

of K2 the zero locus of f(x, y). A subset of K2 of the form Z(f), for some non-constantf(x, y) ∈ R[x, y] will be called an affine algebraic plane curve over K, or simply, analgebraic curve. If C ⊆ K2 is an algebraic curve, then any f ∈ K[x, y] for which C = Z(f)will be called a defining equation for C.

When discussing algebraic curves, we often call K2 the affine plane over K. Usuallywe refer to R2 simply as the plane and to C2 as the complex plane.

Remark 3.1.3. If the field K is contained in a larger field K, then of course we canalso consider the zero locus of a polynomial f(x, y) ∈ K[x, y] in the larger affine plane

K2. If there is any chance of confusion, we will add some notation to Z(f) to emphasize

which field we are working with. We will write Z(f)(K) for the zero locus of f in K2, to

distinguish it from the zero locus of f in K2, which we will denote Z(f)(K). The interested

student may note that the Galois group G = Gal(K/K) (the group of automorphisms ofthe field K which restrict to the identity on the subfield K ⊆ K) acts on Z(f)(K). The

3.1 Polynomials and zero loci 31

student who wants to think about this at all should think about the case where the fieldextension is R ⊆ C, in which case the Galois group is the group with two elements becausecomplex conjugation z 7→ z is the unique non-identity automorphism of C restricting tothe identity on R. If f ∈ R[x, y] and (a, b) ∈ Z(f)(C) ⊆ C2, then (a, b) ∈ Z(f)(C) ⊆ C2

as well. This is because, for any a, b ∈ C, we have f(a, b) = f(a, b) since the coefficients off are real. It was realized early on in the development of algebraic geometry that the best“geometric” picture of the polynomial f is obtained by considering all the sets Z(f)(K),for all extension fields K ⊆ K, together with the actions of the relevant Galois groups. Inmodern algebraic geometry this information is all packaged together by considering thering K[x, y]/(f) and its associated “prime spectrum.” The historical development of thisperspective is described in [D].

For example, the subsets C1, C2, C3 of R2 from the introduction to this chapter can bewritten Z(f1), Z(f2), Z(f3) for f1(x, y) = x2 + y2− 1, f2(x, y) = x, f3(x, y) = x2− y2− 1.

Since a product of two elements of K is zero iff at least one of them is zero, we have

Z(fg) = Z(f) ∪ Z(g).

For example, Z(xy) is the union of the y-axis and the x-axis.

The subset Z(f) does not determine the polynomial f : It is certainly possible thatZ(f) = Z(g) for different polynomials f, g. For example, if λ ∈ K∗ = K \ {0}, then clearlyZ(f) = Z(λf) since f(x, y) = 0 iff λf(x, y) = 0. For similar reasons,

Z(f) = Z(f2) = Z(f3) = · · · .

“Even worse,” there are lots of polynomials f ∈ R[x, y], such as f(x, y) = x2 + y2 + 1,for which Z(f) ⊆ R2 is empty, and lots of polynomials such as f(x, y) = x2 + y2, orf(x, y) = x4 + y4, for which Z(f) consists of a single point.

At this point one might wonder whether “algebraic curve” is really a good name forthe sets Z(f) which can be finite (or empty), despite being called “curves,” which suggeststhey are one dimensional in some sense. One such sense is that the Krull dimension of thering K[x, y]/(f) is 1 for every non-constant f(x, y) ∈ K[x, y] (by Krull’s Hauptidealsatz ).Furthermore, the examples where Z(f) is empty or a finite set of points occur only whenworking over the real numbers R. Over C, it turns out that, after possibly removingfinitely many points from Z(f), the set Z(f) carries, in some reasonably natural way, thestructure of a (non-empty!) complex one dimensional manifold. We will not use thesefacts in anything that follows; they are mentioned only to justify our terminology.

Over the complex numbers, one can, roughly speaking, recover f from the subsetZ(f) ⊆ C2 up to the sort of ambiguities discussed two paragraphs above. The precisestatement is Hilbert’s Nullstellensatz which says that the subset (ideal, in fact)

{g(x, y) ∈ C[x, y] : g(x, y) = 0 for all (x, y) ∈ Z(f) ⊆ C2}

of C[x, y] coincides with the subset (ideal)

{g(x, y) : gn = hf for some h ∈ C[x, y] and some n ∈ {1, 2, . . . }}.


We mention this only for the sake of completeness; we will not use this fact elsewhere.Since the subset Z(f) is determined from f , but not vice-versa, we often think of

the polynomial f as being the more fundamental object. This point of view will beespecially important in §3.3 where we discuss the problem of “classifying” the algebraicplane curves (of, say, a given degree).

In particular, we would like to say that the degree of an algebraic plane curve C ⊆ K2

is the degree of “the” polynomial f for which C = Z(f). Of course this makes no sensebecause there are many such polynomials f ! There are various ways of resolving this—wecould, for example, consider the minimum degree of all those f for which C = Z(f).Rather than worry about this technicality, let us just agree to the following:

Definition 3.1.4. A line (resp. conic, cubic, quartic, . . . ) is an algebraic curve C ⊆ K2

which can be written C = Z(f) for some polynomial f ∈ K[x, y] of degree one (resp. two,three, four, . . . ).

Notice that, according to this definition, every “line” is also a conic, because ifC = Z(f) with f linear, then we also have C = Z(f2), and f2 will have degree two.

3.2 Smooth and singular points

Just as there is a notion of “smoothness” for parametric curves (Chapter 2), there is avery closely related notion of smoothness for algebraic curves.

Notice that the partial derivatives fx and fy of a polynomial f ∈ K[x, y] makesense over any field K (not just over the real numbers, where they have a “calculus”interpretation in terms of limits). Explicitly, if

f =∑i,j

ai,jxiyj

then we define fx and fy to be

fx :=∑i,j

iai,jxi−1yj

fy :=∑i,j

jai,jxiyj−1.

It is perhaps not so obvious that these “definitions” of the derivative should have anygeometric significance over other fields, though we shall see that this is the case.

Definition 3.2.1. For f ∈ K[x, y], a point P = (x0, y0) ∈ Z(f) ⊆ K2 is called a non-singular (or smooth) point of Z(f) iff at least one of fx(P ), fy(P ) is non-zero. A pointP ∈ Z(f) for which fx(P ) = fy(P ) = 0 is called a singular point of Z(f). The subset ofZ(f) consisting of smooth (resp. singular) points is denoted Z(f)sm (resp. Z(f)sing). Wesay that f is non-singular or smooth iff Z(f)sing(K) is empty for every field extensionK ⊆ K.1

1It equivalent to ask that Z(f)sing(K) = ∅ for some algebraically closed field K containing K.

3.3 Change of variables and affine equivalence 33

It is worth emphasizing that the question of whether P ∈ Z(f) is a smooth pointor a singular point depends highly on f , not just on the curve C = Z(f). For example,for any (non-constant) f ∈ K[x, y], we have Z(f) = Z(f2), but every point of Z(f2) is asingular point (Exercise 3.2).

In defining smoothness, we require that Z(f)sing(K) be empty for all field extensionsK ⊇ K (rather than for just K itself) for various reasons. For one thing, if we have apolynomial f ∈ K[x, y] and we regard it as a polynomial in f ∈ K[x, y], we don’t wantthe meaning of “smooth” to change! A good example to keep in mind is the polynomialf = (1 + x2)2 ∈ R[x, y], which has empty zero locus over R, even though every point ofZ(f)(C) is singular (c.f. Exercise 3.2). This phenomenon illustrates the general philosophyof Remark 3.1.3 that one should consider all zero loci of f over all field extensions of Ktogether.

Definition 3.2.2. If P = (x0, y0) is a non-singular point of Z(f), we define the tangentline to Z(f) at P to be the line

TP Z(f) := Z(xfx(P ) + yfy(P )− x0fx(P )− y0fy(P )).

Notice that the assumption that P is non-singular is needed to ensure that the linearpolynomial whose zero locus defines TP Z(f) is non-constant, so that TP Z(f) is actuallya line. Note also that the constant term of the linear polynomial is chosen so thatP ∈ TP Z(f).

Proposition 3.2.3. Let P = (a, b) be a smooth point of a curve Z(f) ⊆ R2. Then thereare open subsets U , V of R, containing a and b, respectively, such that Z(f)∩ (U × V ) iseither the graph of a smooth (infinitely differentiable) function g : U → V (this will bethe case if fx(P ) 6= 0) or the graph of a smooth function h : V → U (if fy(P ) 6= 0). Inother words, at least one of the projections

π1 : Z(f) ∩ (U × V ) → U

π2 : Z(f) ∩ (U × V ) → V

will be a continuous bijection with smooth inverse.

Proof. This is a special case of the Implicit Function Theorem (Theorem 6.3.1 in theappendix).

3.3 Change of variables and affine equivalence

Recall the group Aff(K2) of affine transformations of the plane from §1.4. Given [A, t] ∈Aff(K2) and f ∈ K[x, y], we define a new polynomial f · [A, t] ∈ K[x, y] by setting

(f · [A, t])(x, y) := f([A, t](x, y)).

To be careful here, we should stick more precisely to the conventions of linear algebraand write every element of K2, as well as the argument of f , as a column vector. Then


we would write:

(f · [A, t])

(x

y

):= f(A

(x

y

)+ t).

To be completely explicit, if

A =

(a b

c d

)and t =

(t1

t2

),

then

f(x, y) = f(ax+ by + t1, cx+ dy + t2).

Usually I will be more sloppy with the notation since it is painful to write column vectorsin the inline text. A good compromise is to write x (or x if you prefer) for the vectorof variables (x, y) (thought of as a column vector, but often abusively written as a rowvector). Then f · [A, t] is defined by

(f · [A, t])(x) := f(Ax+ t), (3.1)

where the argument of f is thought of as a column vector.The above construction of f · [A, t] defines an action (more precisely, a right action)

of Aff(K2) on K[x, y]. This means that

1. f · Id = f , where Id = [Id, 0] is the identity element of Aff(K2), and

2. (f · [A, t]) · [A′, t′] = f · ([A, t][A′, t′])

for every f ∈ K[x, y], [A, t], [A′, t′] ∈ Aff(K2).The first equality is obvious and the second is a simple calculation using definition

(3.1) and the formula (1.3) for products in Aff(K2):

((f · [A, t]) · [A′, t′])(x) = (f · [A, t])(A′x+ t′)

= f(A(A′x+ t′) + t)

= f(AA′x+At′ + t)

= (f · [AA′, At′ + t])(x)

= (f · ([A, t][A′, t′]))(x).

One can also think of this action by noting that any [A, t] ∈ Aff(K2) gives rise to anautomorphism of the K-algebra K[x, y] given by

x 7→ ax+ by + t1

y 7→ cx+ dy + t2,

where we have written out A and t as above.More, general, the same formula (3.1) may be viewed as defining an action of Aff(Kn)

on K[x1, . . . , xn] (think of x as the column vector of variables (x1, . . . , xn)).The actions of Aff(K2) on K2 and K[x, y] are “compatible” in various ways. For now,

let us just note:

3.3 Change of variables and affine equivalence 35

Lemma 3.3.1. For f ∈ K[x, y], we have

Z(f · [A, t]) = [A, t]−1 · Z(f)

= {[A−1,−A−1t](x) : x ∈ Z(f)}.

Proof. A point x ∈ K2 is in Z(f · [A, t]) iff f(Ax+t) = 0, which is true iff Ax+t = [A, t](x)is in Z(f), which is true iff

x = [A, t]−1 · ([A, t]x) ∈ [A, t]−1 Z(f).

In other words, the zero locus of f · [A, t] is obtained from the zero locus of f byapplying the inverse of the affine transformation [A, t] to K2. The reader should befamiliar with the appearance of the “inverse” here, as this also occurs (for the samereason) when one thinks about, say, the relationship between the group of f(x) and thegraph of f(ax+ b), with a ∈ R∗, b ∈ R.

Proposition 3.3.2. The degree of a polynomial is invariant under the action of affinetransformations. That is,

deg f = deg(f · [A, t])

for every f ∈ K[x, y] and every [A, t] ∈ Aff(K2).

Proof. Consider a degree d polynomial f(x, y) =∑

i+j≤d aijxiyj . Then if we write out

A and t as above, we have

f · [A, t] =∑i+j≤d

aij(ax+ by + t1)i(cx+ dy + t2)

j .

From the rule for Multinomial Expansion and the fact that the degree of a product isthe sum of the degrees of the factors, it is clear that (when i+ j ≤ d) the polynomial(ax + by + t1)

i(cx + dy + t2)j clearly has degree at most d. Since f · [A, t] is a sum of

such things, we see that the degree of f · [A, s] cannot be any bigger than d = deg f :

deg(f · [A, t]) ≤ deg f.

But now we apply this same observation with the polynomial f replaced by the polynomialf · [A, t], and [A, t] replaced by its inverse to find the opposite inequality (here we usethe fact that we have an action to know that f = (f · [A, t]) · [A, t]−1).

Definition 3.3.3. Two polynomials f, g ∈ K[x, y] are called affine equivalent iff thereis an affine transformation [A, t] ∈ Aff(K2) and a non-zero scalar λ ∈ K∗ such thatλg = f · [A, t]. When this is the case, one says that g is obtained from f by affine linearchange of variables (and rescaling). Similarly, we say that two subsets S, T ⊆ K2 areaffine equivalent iff S = [A, t](T ) for some affine transformation [A, t] ∈ K2.


One can check, using the fact that our construction of f · [A, t] is an action of Aff(K2)on K[x, y], that “being affine equivalent” is indeed an equivalence relation (Exercise 3.4).Notice that Lemma 3.3.1 (plus the fact that rescaling a polynomial doesn’t change itszero locus) implies that if f, g ∈ K[x, y] are affine equivalent, then their zero loci Z(f),Z(g) are affine equivalent as subsets of K2. The converse is certainly not true: we willsoon see that there are inequivalent degree two polynomials f, g ∈ R[x, y] such that bothZ(f) and Z(g) are empty!

A very ambitious goal would now be to classify all polynomials and all plane curvesup to affine equivalence. (That is, to “explicitly” describe all affine equivalence classes.)To attempt this, one has to think of many “affine invariants,” which is to say, “quantitiesassociated to a polynomial f in K[x, y] (or a subset C ⊆ K2) that depend only on theaffine equivalence class of f (or C).” Proposition 3.3.2 gives us one such quantity, namelythe degree of a polynomial. This lets us break up our classification problem degree bydegree.

As a warmup, let us classify linear (that is, degree one) polynomials up to affineequivalence. The corresponding classification of plane curves will be the classificationof lines, in the sense of Definition 3.1.4. We will again take up the subject of lines, inthe projective setting, in §5.4. We claim that in fact every degree one polynomial isequivalent to the polynomial x. A degree one polynomial is a polynomial of the formf(x, y) = Ax + By + C for some real numbers A,B,C with (A,B) 6= (0, 0). Afterapplying the (invertible linear) transformation (x, y) 7→ (y, x) if necessary, we see thatf is equivalent to a polynomial g of the same form where in fact A 6= 0. By applyingthe invertible linear transformation (x, y) 7→ (A−1x, y), we see that g is equivalent to apolynomial h of the form x+By+C. After applying the invertible linear transformation(x, y) 7→ (x − By, y), we see that h is equivalent to a polynomial j of the form x + C.Finally, applying the translation (x, y) 7→ (x−C, y), we see that j is equivalent to x. Weleave it as an easy exercise for the reader to see that “a linear polynomial f(x, y) can berecovered (up to multiplication by a non-zero real number) from the corresponding planecurve (line) Z(f)”.

We should also emphasize at this point that our notion of affine equivalence is notthe only “reasonable” one: There are very legitimate reasons for rejecting the ideathat plane curves differing by affine transformation are “equivalent.” For one thing,affine transformations of R2 do not always distances, or even angles. It would be quitereasonable to try to classify conics in R2 up to isometry (§1.2). While this can be donein an elementary manner, it is rather tedious and we shall not pursue this here. Wewill see this issue (the difference between affine equivalence and isometry-equivalence)manifested in our classification of conics in various ways: all ellipses are equivalent tothe unit circle, all pairs of intersecting lines are equivalent (regardless of the angle ofintersection), and so forth.

3.4 Five points determine a conic

If P and Q are distinct points of K2, then it is easy to see that there is a unique lineL ⊆ K2 for which P,Q ∈ L. Here we prove an analogous statement for conics. The

3.4 Five points determine a conic 37

technique used in the proof will be recycled many times later.

Proposition 3.4.1. Suppose P1, . . . , P5 ∈ K2 are any five points, no three of which arecollinear. Then, up to multiplying by a non-zero real number, there is a unique degreetwo polynomial f(x, y) ∈ K[x, y] such that P1, . . . , P5 ∈ Z(f).

Proof. Let us first treat the special case where P1 = (0, 0), P2 = (1, 0), and P3 = (0, 1).The general degree two polynomial takes the form

f(x, y) = Ax2 +Bxy + Cy2 +Dx+ Ey + F,

where (A,B,C) 6= (0, 0, 0). The condition that (0, 0) ∈ Z(f) is equivalent to F = 0, andthe conditions that (1, 0) ∈ Z(f) and (0, 1) ∈ Z(f) are then equivalent to D = −A andE = −C, so our f must take the form

f(x, y) = Ax2 +Bxy + Cy2 −Ax− Cy.

If we write P4 = (x1, y1), P5 = (x2, y2), then the conditions that P4, P5 ∈ Z(f) areequivalent to the system of linear equations

Ax1(x1 − 1) +Bx1y1 + Cy1(y1 − 1) = 0

Ax2(x2 − 1) +Bx2y2 + Cy2(y2 − 1) = 0.

By basic linear algebra, we can conclude that this system has a unique non-zero solution(A,B,C), up to scaling, provided that we can show the coefficient vectors

(x1(x1 − 1), x1y1, y1(y1 − 1)) and (x2(x2 − 2), x2y2, y2(y2 − 1))

are linearly independent.Suppose not. Then there is some λ ∈ K so that

λ(x1(x1 − 1), x1y1, y1(y1 − 1)) = (x2(x2 − 2), x2y2, y2(y2 − 1)).

Now, since P4 isn’t on the line joining (0, 0) and (1, 0) (the x-axis), we have y1 6= 0. SinceP4 isn’t collinear with (0, 0) and (1, 0), we have x1 6= 0. For similar reasons, we havex2, y2 6= 0. We can then solve for λ using equality of the second coordinates above tofind λ = (x2y2)/(x1y1). Plugging in for λ and cancelling the non-zero real number x2,equality of the first coordinates above then yields:

y2(x1 − 1) = y1(x2 − 1).

But y(x1 − 1) = y1(x− 1) is the equation defining the line containing P4 = (x1, y1) andP2 = (1, 0), so this latter equality shows that P2, P4, and P5 are collinear—a contradiction.

We deduce the general case from the special case as follows: By Proposition 1.4.3 wecan find an affine transformation [A, t] : K2 → K2 taking P1, P2, and P3 to (0, 0), (1, 0),and (0, 1), respectively. Since an affine transformation takes lines to lines, the five points

(0, 0) = [A, t](P1), (1, 0) = [A, t](P2), (0, 1) = [A, t](P3), [A, t](P4), [A, t](P5)


are in general position (no three collinear), so by the special case treated above, thereis a unique (up to rescaling) degree two f ∈ K[x, y] such that Z(f) contains these fivepoints. By Proposition 3.3.2 and Lemma 3.3.1, the polynomial f · [A, t] is of degree twoand its zero locus contains P1, . . . , P5. If g is any other degree two polynomial whosezero locus contains P1, . . . , P5, then that proposition and lemma imply that g · [A, t]−1 isa degree two polynomial whose zero locus contains the five points listed above, henceg · [A, t]−1 must be a rescaling of f by the special case treated above, and hence g mustbe a rescaling of f · [A, t].

3.5 Plane conics up to affine equivalence

Having dispensed with the classification of degree one polynomials up to affine equivalencein §3.3, we now turn to the case of degree two polynomials. Unlike the classification oflinear polynomials, this classification is quite sensitive to the field K over which we work.The classification over the real numbers R is as follows:

Theorem 3.5.1. Every polynomial f(x, y) ∈ R[x, y] of degree two is affine equivalent(Definition 3.3.3) to exactly one of the following:

1. x2 + y2 − 1

2. x2 − y2 − 1

3. y − x2

4. xy

5. x2

6. x(x− 1)

7. x2 + y2

8. x2 + y2 + 1

9. x2 + 1

Over the complex numbers C, the classification is simpler:

Theorem 3.5.2. Every polynomial f(x, y) ∈ C[x, y] of degree two is affine equivalent toexactly one of the following:

1. x2 + y2 − 1

2. y − x2

3. xy

4. x2

5. x(x− 1)

3.5 Plane conics up to affine equivalence 39

We are not going to prove these theorems, though we will prove their “projectiveanalogues” in §5.6. The proofs are elementary: the basic technique, which will also beused in the proofs our projective analogues of these theorems, amounts to little more thanthe idea of “completing the square” familiar from calculus (or from the derivation of theQuadratic Formula, say). However, the proofs in the affine case are slightly more tedious,as there is more case-by-case analysis. The projective analogues of this theorem are alsoin many ways more important and fundamental, as they are related to the problem ofclassifying symmetric bilinear forms, which is important in linear algebra.

From the classification of degree two polynomials, we can easily obtain the classificationof the conics given by their zero loci:

Corollary 3.5.3. For every polynomial f(x, y) ∈ R[x, y] of degree two, the correspondingplane curve Z(f) ⊆ R2 is either empty, or affine equivalent to exactly one of the following:

1. the circle Z(x2 + y2 − 1)

2. the hyperbola Z(x2 − y2 − 1)

3. the parabola Z(y − x2)

4. the pair of intersecting lines Z(xy)

5. the “doubled” line Z(x2)

6. the pair of parallel lines Z(x(x− 1))

7. the single point Z(x2 + y2)

Proof. This will follows from Theorem 3.5.1 and Lemma 3.3.1 once we prove that notwo of the non-empty plane curves on the list are affine equivalent. (A priori, it is quiteconceivable that two inequivalent polynomials determine equivalent plane curves, and,indeed, this is the case—the two inequivalent polynomials x2 + y2 + 1 and x2 + 1 bothdetermine the empty curve, but, what we are about to argue is that, in fact, this doesnot happen, except in the aforementioned “empty curves” case.)

This can be done in various ways. Perhaps the most direct way is to rule out theequivalence of different curves on the list on elementary topological grounds: affinetransformations are, in particular, homeomorphisms, and the point is that very fewpairs of curves on the list are even homeomorphic. (The reader who doesn’t knowwhat “homeomorphism” means should just take this as a precise way of making precisethe obvious “differences” we observe such as the fact that the parabola and circleare “connected,” whereas the hyperbola “has two components.” Similarly, the circle is“bounded” (compact), while the parabola and hyperbola are not.) The exceptions are:the hyperbola is homeomorphic to the two parallel lines (both are just disjoint unions oftwo copies of the real line), and the parabola is homeomorphic to the “double” line (bothare homeomorphic to the real line). But these pairs of plane curves are easily seen to bedistinct (inequivalent) on the grounds that affine transformations take lines to lines. Forexample, one need only write down three non-colinear points on the parabola to knowthat it cannot be equivalent to the “double” line.


3.6 Invariants of affine conics

The most difficult thing to do in a classification problem is often to prove that variousthings are not equivalent. To do this, one typically tries to attach some kind of invariant—a number, say—to the objects being classified in such a way that this invariant dependsonly on the equivalence class of the object.

The relevant invariants for plane conics are the discriminant and the degeneracy.Consider a typical polynomial

f(x, y) = Ax2 +Bxy + Cy2 +Dx+ Ey + F (3.2)

of degree at most two with coefficients in a field K where 2 ∈ K∗ (for example: K = Q,R, or C). A key point is to note that f(x, y) can de described in terms of matrixmultiplication, as follows. We let M(f) be the symmetric 3× 3 matrix with entries in Kdefined by

M(f) :=

A B/2 D/2

B/2 C E/2

D/2 E/2 F

. (3.3)

(Here we need to assume 2 ∈ K∗.) Then f(x, y) and the matrix M(f) are related via theformula:

f(x, y) = (x, y, 1)

A B/2 D/2

B/2 C E/2

D/2 E/2 F

xy

1

. (3.4)

We call the matrix M(f) the matrix associated to f . Given any symmetric 3× 3 matrixM with entries in K, we can define a polynomial f(M) ∈ K[x, y] of degree at most twoby setting

f(M) = (x, y, 1)M(x, y, 1)tr.

It is clear from formula (3.4) that f 7→M(f) establishes a bijection from polynomials inx, y of degree at most two to symmetric 3× 3 matrices with inverse M 7→ f(M).

Definition 3.6.1. For f as above, we define the discriminant D(f) ∈ K and thedegeneracy ∆(f) ∈ K of f by

D(f) := B2 − 4AC

∆(f) := detM(f).

We say that f(x, y) is degenerate if ∆(f) = 0 and non-degenerate otherwise.

The behaviour of D(f) and ∆(f) under affine transformations is as follows.

3.6 Invariants of affine conics 41

Lemma 3.6.2. For a polynomial f(x, y) ∈ K[x, y] of degree at most two, an affinetransformation [A, t] ∈ Aff(K2), and λ ∈ K, we have

D(λf · [A, t]) = λ2(detA)2D(f)

∆(λf · [A, t]) = λ3(detA)2∆(f).

Proof. The most direct proof is to just compute everything explicitly. The behaviourof D and ∆ under multiplying by λ is easy to understand (multiplying every entry of a3× 3 matrix by λ changes its determinant by λ3), so it is enough to deal with the caseλ = 1. Suppose

A =

(a b

c d

)t =

(s

t

)

and f is given by (3.2). Then

f · [A, t] = f(ax+ by + s, cx+ dy + t)

= A′x2 +B′xy + C ′y2 +D′x+ E′y + F ′,

where

A′ = Aa2 +Bac+ Cc2

B′ = 2Aab+B(ad+ bc) + 2Ccd

C ′ = Ab2 +Bbd+ Cd2

D′ = 2Aas+B(at+ sc) + 2Cct+Da+ Ec

E′ = 2Abs+B(bt+ ds) + 2Cdt+Db+ Ed

F ′ = As2 +Bst+ Ct2 +Ds+ Et+ F.

Now a certain amount of laborious (but elementary) calculation yields

D(f · [A, t]) = (B′)2 − 4A′C ′

= (ad− bc)2(B2 − 4AC)

= det(A)2D(f)

∆(f · [A, s]) = det

A′ B′/2 D′/2

B′/2 C ′ E′/2

D′/2 E′/2 F ′

= (ad− bc)2 det

A B/2 D/2

B/2 C E/2

D/2 E/2 F

= (detA)2∆(f).


A slightly more sophisticated treatment of ∆(f) using the matrix M(f) proceeds asfollows. We calculate

f · [A, t] = (ax+ by + s, cx+ dx+ t, 1)M(f)

ax+ by + s

cx+ dy + t

1

= (x, y, 1)

a c 0

b d 0

s t 1

M(f)

a b sc d t

0 0 1

xy

1

so we see that the matrix

M(f · [A, t]) =

A′ B′/2 D′/2

B′/2 C ′ E′/2

D′/2 E′/2 F ′

corresponding to f · [A, s] is given by

M(f · [A, t]) =

a b sc d t

0 0 1

tr

M(f)

a b sc d t

0 0 1

.

The second formula of the lemma now follows from standard properties of determinants(invariance under transpose, determinant of a product is the product of the determinants)using the trivial calculation

det

a b sc d t

0 0 1

= detA.

Corollary 3.6.3. Let K be a field with 2 ∈ K∗ and let f ∈ K[x, y] be a polynomial ofdegree 2. Then ∆(f) = 0 iff ∆(g) = 0 for some (equivalently any) g affine equivalentto f . Similarly, when K = R, the “sign” of Df in {+,−, 0} depends only on the affineequivalence class of f .

We can use D and ∆ to distinguish between most of the pairs of plane curves on thelist in Theorem 3.5.1. The values of D and ∆ for the curves on that list are tabulated

3.7 Exercises 43

below:f(x, y) D(f) ∆(f)

x2 + y2 − 1 (−) 6= 0

x2 − y2 − 1 (+) 6= 0

y − x2 (0) 6= 0

xy (+) 0

x2 (0) 0

x(x− 1) (0) 0

x2 + y2 (−) 0

x2 + y2 + 1 (−) 6= 0

x2 + 1 (0) 0

At this point, we shouldn’t be shy about admitting that the definitions of the discrim-inant and the ∆-invariant seem to “come out of the blue.” Later, in Proposition 5.5.3,we will give a completely “geometric” meaning of the discriminant.

3.7 Exercises

Exercise 3.1. Suppose K is an algebraically closed field (take K = C if you want) andf ∈ K[x, y] is a non-constant polynomial. Prove that Z(f) 6= ∅.

Exercise 3.2. Suppose f ∈ K[x, y] is a non-constant polynomial and n is an integergreater than one. Prove that every point of Z(fn) is a singular point. More generally,prove that if f = gh with g and h non-constant, then any point of Z(g) ∩ Z(h) is asingular point of Z(f).

Exercise 3.3. Let f ∈ K[x, y] be a polynomial, [A, t] ∈ Aff(K2) an affine transformation.Prove that a point P ∈ Z(f) is a singular point of Z(f) iff [A, t]−1(P ) is a singular pointof Z(f · [A, t]).

Exercise 3.4. Prove that being affine equivalent (Definition 3.3.3) is an equivalencerelation on K[x, y].

Exercise 3.5. Consider the degree two polynomials f := x2 + y2, g := −x2 − y2 inR[x, y]. Show that there is no A ∈ Aff(R2) such that g = f · A. (Nevertheless, we doconsider f and g to be affine equivalent because (−1)f = g.)

Exercise 3.6. For each of the following polynomials f ∈ R[x, y] decide which of the fivepolynomials in the list of Theorem 3.5.1 is affine equivalent to f :

1. zy − x2

2. xy

3. x2 + xy + z2

4. 2x2 + 2xz + z2 + 2xy + y2


Exercise 3.7. For f ∈ R[x, y], let Stab f be the set of A ∈ Aff(R2) such that f ·A = f .

1. Show that Stab f is a subgroup of Aff(R2)—i.e. show that Id ∈ Stab f , that A−1 ∈Stab f whenever A ∈ Stab f , and that AA′ ∈ Stab f whenever A,A′ ∈ Stab f .

2. Note that Stab f = Stab(λf) for any non-zero λ ∈ R. Show that if f and g areAff-equivalent, then Stab f is conjugate to Stab g in Aff(R2)—i.e. there is someA ∈ Aff(R2) such that

Stab g = {A−1A′A : A′ ∈ Stab f}.

3. Show that Stab(x2 + y2) = Stab(x2 + y2 − 1) is equal to the orthogonal groupO2 < Aff(R2).

4. Calculate Stab f for f = y − x2. Do you recognize this group? (Forget about thedescription of it as a subgroup of Aff(R2).)

Chapter 4

Projective Spaces

In this brief chapter we shall introduce and study the projective spaces KPn for a fieldK. The idea of the construction of KPn is to enlarge Kn by adding some additional“points at infinity,” one for each line through the origin in Kn. One surprising feature ofthe construction is that the newly added points “at infinity” are actually “on the samefooting” as the old “finite” points. We shall make this precise in §4.4 by introducing andstudying the projective general linear groups PGLn(K). The group PGLn(K) acts onKPn through projective transformations in a manner similar to the way Aff(Kn) acts onKn. Indeed, we shall see that Aff(Kn) can be identified with the subgroup of PGLn(K)consisting of those A ∈ PGLn(K) such that the associated map A : KPn → KPn takesthe locus of “finite points” Kn ⊆ KPn into itself.

In §4.2 we discuss linear subspaces of projective spaces—this concept leads to thedefinition of a line and to notions of “general position” which we be important later.

Special attention will be given to the real projective spaces RPn and the complexprojective spaces CPn in §4.3. A full appreciation of this section requires some backgroundin topology—this material is not strictly necessary elsewhere in the text.

4.1 Motivation and construction

Before we give the general definition of the projective spaces KPn, let us motivate theconstruction by discussing RP1.

One “problem” with the real line R is that it is not compact. The general notionof compactness is a concept of topology with which we shall not particularly concernourselves here, but in this case, let us rephrase the issue a bit by noting that thereare sequences of points a1, a2, . . . in R which do not have a convergent subsequence.For example, 1, 2, 3, . . . is such a sequence. Now, the astute reader with some calculusbackground would probably suggest that the limit of this sequence should be “∞,” so letus try to make some sense of this. We are led to attempt to enlarge the real line R byadding one point ∞, thus obtaining a (slightly) larger space RP1 = R

∐{∞}. But now

we have to decide how to “topologize” this larger set RP1. In other words, we have togive some precise definition of when a sequence of points a1, a2, . . . in RP1 “converges to∞.” A very reasonable way to do this is to agree that a1, a2, . . . converges to ∞ iff the

45

46 Chapter 4 Projective Spaces

sequence 1/a1, 1/a2, . . . converges to 0 (in the usual epsilon-delta sense), where we agreethat 1/∞ := 0.

With this understanding, let us prove that any sequence a1, a2, . . . in RP1 has aconvergent subsequence. First of all, if infinitely many of the ai are equal to ∞, thenthose ai of course give a subsequence converging to∞. On the other hand, if only finitelymany of the ai are equal to ∞, then we discard them to obtain a subsequence containedin R ⊆ RP1. We thus reduce to the case where our sequence a1, a2, . . . is contained inR ⊆ RP1. Now, if the magnitudes |ai| of the ai are bounded, say by M , then we can finda convergent subsequence by basic results from calculus or topology (because the closedinterval [−M,M ] is compact). On the other hand, if the magnitudes |ai| are unbounded,then we can find an increasing sequence of integers

0 < i1 < i2 < · · ·

so that |aij | > j for each j ∈ {1, 2, . . . }. The subsequence ai1 , ai2 , . . . of a1, a2, . . . thenconverges to ∞ because |1/aij | < 1/j for each j ∈ {1, 2, . . . }.

There are many other ways to think about this construction of RP1. For example, wecan view RP1 as being obtained from two copies R1, R2 of the real line R by identifyinga non-zero point x ∈ R1 with the (non-zero) point 1/x ∈ R2. Every point in the resulting“glued up” space will have a unique representative in R1, with the exception of the origin0 ∈ R2, which plays the role of ∞ in the previous construction.

Another way to think about the construction of RP1 more in line with what we willdo in general, is to notice that any point x ∈ R determines a line Lx passing through theorigin in R2—namely, the line with slope x. Notice that every line through the originin R2 will be equal to Lx for a unique real number x with one exception: the verticalline L∞ through the origin, which we might reasonably describe as the “line throughthe origin with slope ∞.” The reader with some background in linear algebra will nodoubt realize that a line through the origin in R2 is the same thing as a one dimensionalsubspace of the vector space R2. This is the point of view that will generalize well.

Yet another way to think about RP1, also in line with what we will do later, is tothink of a real number x as a ratio of two real numbers r = x/1 = 2x/2 = · · · wherethe denominator is non-zero. With this understanding, we might reasonable agree that∞ can also be viewed as a ratio of two real numbers: ∞ = 1/0 = 5/0 = · · · , wherethe numerator is non-zero. To make this a bit more formal, we can consider the setR2 \ {(0, 0)} of pairs of real numbers (x, y), not both zero, and impose the equivalencerelation ∼ where we declare (x1, y1) ∼ (x2, y2) iff x1y2 = y1x2. Notice that, if y1 and y2are non-zero, we have (x1, y1) ∼ (x2, y2) iff the real numbers x1/y1 and x2/y2 are equal.Denote the equivalence class of (x, y) by [x : y]. Notice that every such equivalence classis of the form [x : 1] for a unique real number x with one exception: the equivalence class[1 : 0] (which is equal to the equivalence class [x : 0] for any non-zero real number x).

With all this motivation, we are led to the general definition of the projective spacesKPn:

Definition 4.1.1. For a field K, we let KPn denote the set of one dimensional subspacesof the K-vector space Kn+1. We call KPn the n-dimensional projective space over K.For a non-zero vector x ∈ Kn+1 \ {0}, we let [x] ∈ KPn denote the span of x. If

4.1 Motivation and construction 47

x = (x0, . . . , xn) ∈ Kn+1, we write [x0 : · · · : xn] as an alternative to [x]. More abstractly,if V is a K-vector space, then we let P(V ) denote the set of one dimensional subspaces ofV . We call P(V ) the projectivization of V .

Notice that every point of KPn is of the form [x] for some x ∈ Kn+1 \ {0} (becauseany one dimensional subspace of a vector space is spanned by any non-zero element ofit) and we have [x] = [y] iff x = λy for some λ ∈ K∗. We can thus view KPn as the setof equivalence classes in Kn+1 \ {0} where we declare x ∼ y iff x = λy for some λ ∈ K∗.With this understanding, we can also view [x] as denoting the ∼-equivalence class ofx ∈ Kn+1 \ {0}.

If (x0, . . . , xn) = λ(y0, . . . , yn) for λ ∈ K∗, then clearly xi = 0 iff yi = 0, so thequestion of whether xi = 0 for a point x = (x0, . . . , xn) ∈ Kn+1 \ {0} depends only onthe equivalence class [x] of x. Hence we can make the following:

Definition 4.1.2. Let U0, . . . , Un ⊆ KPn be the subsets defined by

Ui := {[x0 : · · · : xn] ∈ KPn : xi 6= 0}.

The subset Un of KPn will be called the set of finite points of KPn, and its complement

KPn \ Un = {[x0 : · · · : xn] ∈ KPn : xi = 0}

will be called the set of infinite points of KPn.

Note that KPn = U0 ∪ · · · ∪ Un.

Definition 4.1.3. Define a bijection φi : Kn → Ui ⊆ KPn by

φi : Kn → KPn

(a1, . . . , an) 7→ [a1 : · · · : ai−1 : 1 : ai : · · · : an].

The inverse φ−1i of φi is given by

φ−1i [x0 : · · · : xn] = (x0/xi, . . . , xi−1/xi, xi+1/xi, . . . , xn/xi).

Similarly, we let ψi : KPn−1 → KPn \ Ui be the bijection defined by

ψi : KPn−1 → KPn \ Ui[x0 : · · · : xn−1] 7→ [x0 : · · · : xi−1 : 0 : xi : · · · : xn−1].

In particular, φn gives a bijection between Kn and the set Un of finite points of KPn.We often suppress notation for this bijection and simply regard Kn as a subset of KPnby identifying Kn with Un via φn. In other words, we don’t make much distinctionbetween a point (x1, . . . , xn) ∈ Kn and the “corresponding” point [x1 : · · · : xn : 1] ofUn ⊆ KPn. Similarly, we identify KPn−1 with the set of infinite points in KPn via thebijection ψi, so if there is no chance of confusion we do not make any distinction between[x0 : · · · : xn−1] ∈ KPn−1 and [x0 : · · · : xn−1 : 0] ∈ KPn. In particular, notice that wehave identified the set of points at infinity in KPn with the set of lines through the originin (one dimensional linear subspaces of) Kn = Un ⊆ KPn.

For example, in the case n = 1, the “finite points” of KP1 are those of the form [x : y]with y 6= 0, and there is a single infinite point [1 : 0], often denoted ∞, corresponding tothe single point of KP0.


4.2 Linear subspaces of projective space

The definition of P(V ) for an abstract K-vector space V is convenient, even if one isonly interested in KPn, for essentially the same reasons that it is convenient to have anabstract notion of “vector space” even if one is only interested in the vector space Kn—forexample, because “abstract” vector spaces are going to arise naturally as subspaces ofKn. By definition, we have KPn = P(Kn+1). (The notation is unfortunate.) Notice alsothat if V is a subspace of a vector space W , then P(V ) is a subset of P(W ) because anyone dimensional subspace of V is also a one dimensional subspace of W . In particular, asubspace V of the vector space Kn+1 (a “linear subspace of Kn+1”) gives rise to a subsetP(V ) ⊆ KPn.

Definition 4.2.1. A linear subspace of KPn of dimension d is a subsset of KPn+1 ofthe form P(V ), for V a linear subspace of Kn+1 of dimension d + 1 (as a K vectorspace). A linear subspace of dimension one (resp. two) is often called a line (resp. plane),while a linear subspace of KPn of dimension n− 1 (“codimension one”) is often called ahyperplane.

For example, the set of “infinite” points of KPn is a hyperplane—it is P(V ) where Vis the subspace of Kn+1 spanned by the first n basis vectors e0, . . . , en−1.

Definition 4.2.2. Points P1, . . . , Pm ∈ KPn are said to be in general position iff, for allk ∈ {1, . . . ,maxm,n+ 1}, no k of the Pi are contained in a linear subspace of KPn ofdimension less than k − 1.

For example, when n ≥ 2, three points P1, P2, P3 ∈ KPn are in general position iffthey are not contained in any line. Three points P1, P2, P3 ∈ KP1 are in general positioniff they are distinct.

4.3 Real and complex projective spaces

In order to get a feeling for the general projective spaces KPn introduced in §4.1 it helpsto think about the cases where K = R and K = C.

By definition, RPn is the set of line through the origin in Rn+1. Each such line L willintersect the n-dimensional sphere

Sn := {x ∈ Rn+1 : |x| = 1}

in precisely two points, which are “antipodal” in the sense that one is obtained from theother by multiplying the coordinates by −1. Indeed, if we choose any non-zero pointx ∈ L, then the two points in question are x/|x| and −x/|x|. “Conversely,” if we have apoint x ∈ Sn ⊆ Rn+1, then we get a line through the origin Lx by considering the spanof x. The line Lx intersects Sn at the points x and the antipodal point −x. We thus seethat RPn can be obtained from the n-dimensional sphere Sn by identifying antipodalpoints.

Alternatively, we can consider the upper hemisphere

Sn+ := {x = (x0, . . . , xn) ∈ Sn : xn > 0}

4.3 Real and complex projective spaces 49

of Sn. Notice that projecting onto the first n coordinates defines a (continuous) bijectionfrom Sn+ to the n-dimensional disc

Dn := {x = (x0, . . . , xn−1) ∈ Rn : |x| ≤ 1}.

The inverse of this bijection is given by

x = (x0, . . . , xn−1) 7→ (x0, . . . , xn−1,√

1− |x|2).

By reasoning as above, we see that each line through the origin in Rn+1 will intersectthe upper hemisphere Sn+ in either one point or two points; this intersection will consistof two antipodal points iff the line is contained in the hyperplane Rn ⊆ Rn+1 where thelast coordinate xn is zero. From this perspective we see that RPn can be obtained fromthe n-dimensional disc Dn by identifying antipodal points of its boundary sphere Sn−1.

For example, RP1 is the quotient of S1 by the antipodal map x 7→ −x, but itcan also be described by identifying the two boundary points of the upper half circle(which is homeomorphic to a closed interval), so it is also a circle. The quotient mapS1 → RP1 = S1 can be viewed as the map z 7→ z2 if we view the circle S1 as the set ofcomplex numbers z of magnitude one.

The set RP2 can similarly be viewed as the quotient of the usual sphere S2 ⊆ R3

obtained by identifying antipodal points, or as the quotient of the unit disc D2 ⊆ R2

obtained by identifying antipodal points on the boundary circle S1 ⊆ D2.The story for complex projective space CPn is similar. For any point [z] ∈ CPn, we

can find a representative z ∈ [z] lying in the sphere

S2n+1 := {z ∈ Cn+1 : |z| = 1}.

Two points z, z′ ∈ S2n+1 determine the same point [z] = [z′] of CPn iff z = λz′ for someλ ∈ C—in fact, taking | |, we see that |λ| = 1, so λ will lie in the unit circle

S1 = {λ ∈ C : |λ| = 1}.

This shows that CPn can be described as the quotient of S2n+1 by the free action of S1

given by (λ, z) 7→ λz.For the student with some background in topology, we remark that all of the descrip-

tions of RPn and CPn given in these notes can actually be viewed as descriptions of atopological space (not just a set). In fact all of these descriptions are describing the sametopology:

Definition 4.3.1. We view RPn as a topological space by declaring a subset U ⊆ RPnto be open iff the following equivalent conditions are satisfied:

1. The preimage of U under the map

Rn+1 \ {0} → RPn

given by x 7→ [x] is open in Rn+1.

2. The preimage of U under the map Sn → RPn given by x 7→ [x] is open in Sn.


3. The preimage of U under the map Sn+ → RPn given by x 7→ [x] = Spanx is openin Sn+.

4. The preimage of U ∩ Ui under the bijection φi : Rn → Ui ⊆ RPn (Definition 4.1.3)is open for each i ∈ {0, . . . , n}.

(The equivalence of these conditions is Exercise 4.2.)

Definition 4.3.2. We topologize CPn by declaring a subset U ⊆ CPn to be open iff thefollowing equivalent conditions are satisfied:

1. The preimage of U under the map

Cn+1 \ {0} → CPn

given by z 7→ [z] is open in Cn+1.

2. The preimage of U under the map S2n+1 → CPn given by z 7→ [z] is open in S2n+1.

3. The preimage of U ∩ Ui under the bijection φi : Cn → Ui ⊆ CPn is open for eachi ∈ {0, . . . , n}.

In particular, CPn and RPn are both quotients of spheres and spheres are compact,so we have:

Proposition 4.3.3. For each n, real projective space RPn and compact projective spaceCPn are compact topological spaces.

In Exercise 4.4 you will show that CP1 is homeomorphic to the 2-sphere S2, so thatthe map S3 → CP1 given by z 7→ [z] discussed above (the quotient of the circle action onS3) can be viewed as a map f : S3 → S2 for which f−1(x) is a circle for each x ∈ S2.The map f is called the Hopf fibration—it plays a role in various branches of topology.It is the most basic example of an “interesting” map from a sphere to a sphere of lowerdimension.

4.4 Projective transformations

Now we will define the projective analogue of the affine transformations studied in§1.4. First, notice that an invertible linear transformation A ∈ GLn+1(K) gives rise to abijection

A : KPn → KPn (4.1)

[x] 7→ [Ax].

This is well-defined because if x = λy for λ ∈ K∗, then λAx = Aλx = Ay, so [Ay] = [Ax].We also need to know that A is invertible to ensure that Ax 6= 0 when x 6= 0, so that [Ax]makes sense as a point of KPn. The bijections (4.1) are called projective transformations.Clearly the composition A ◦B of the projective transformations associated to invertiblematrices A,B ∈ GLn+1(K) is the projective transformation AB associated to their

4.4 Projective transformations 51

product. In particular, the inverse of the projective transformation A associated toA ∈ GLn+1(K) is the projective transformation A−1 associated to its inverse. Noticethat for λ ∈ K∗ and A ∈ GLn+1(K), we have A = λA (for the same reason that A iswell-defined). We are thus led to consider the following group:

Definition 4.4.1. The projective general linear group PGLn(K) is the quotient ofGLn+1(K) by the (normal—in fact central) subgroup {λI : λ ∈ K∗}. We denote theimage of A ∈ GLn+1(K) in PGLn(K) by A.

Explicitly, two invertible matrices A,B ∈ GLn+1(K) have the same image in PGLn(K)iff A = λB for some λ ∈ K∗.

Remark 4.4.2. The indexing of PGLn(K) varies. Some people might write PGLn+1(K)for what I have called PGLn(K).

The next result justifies our use of the notation A for both the image of a matrixA ∈ GLn+1(K) in PGLn(K) and for the associated projective transformation A : KPn →KPn.

Proposition 4.4.3. Two invertible matrices A,B ∈ GLn+1(K) have the same image inPGLn(K) iff the projective transformations A,B : KPn → KPn are equal.

Proof. By definition of PGLn(K), if A and B have the same image in PGLn(K), thenwe can write A = λB for some λ ∈ K∗ and we already noted above that this implies thatthe projective transformations A and B are equal.

Conversely, suppose the projective transformations A,B : KPn → KPn are equal.Then, since A[ei] = B[ei] for i = 0, . . . , n, we can write Aei = λiBei for some λi ∈ K∗.Also, since A[1 : · · · : 1] = B[1 : · · · : 1], we can write

Ae0 + · · ·+Aen = λ(Be0 + · · ·+Ben)

for some λ ∈ K∗. Putting these together we see that

λ0Be0 + · · ·+ λnBen = λBe0 + · · ·+ λBen.

But B is invertible, so Be0, . . . , Ben is a basis for Kn+1, hence the above equality impliesthat λ = λ1 = · · · = λn and we conclude that A = λB, so A and B have the same imagein PGLn(K).

Theorem 4.4.4. The subgroup

G := {A ∈ PGLn+1(K) : A(Kn) = Kn}

of PGLn+1(K) consisting of projective transformations that “preserve the locus of finitepoints Kn ⊆ KPn” is isomorphic to the group Aff(Kn) of affine transformations of Kn

via the restriction map A 7→ A|Kn.


Proof. Suppose A ∈ G, so the projective transformation A : KPn → KPn takes the locusKn ⊆ KPn of finite points into itself. Equivalently, since A : KPn → KPn is bijective,this means that A takes the locus KPn−1 ⊆ KPn of infinite points into itself. PickA ∈ GLn+1(K) mapping to A in PGLn(K). Since [e0], . . . , [en−1] are infinite points ofKPn, the points A[ei] = [Aei] must also be infinite points for i = 0, . . . , n − 1, whichmeans that the last entry of each vector Aei (i = 0, . . . , n− 1) must be a zero. On theother hand, A must be invertible, so the last entry λ of Aen therefore cannot be zero.Then L(A) := λ−1A ∈ GLn+1(K) also maps to A in PGLn(K) and is the unique suchmatrix with lower right entry equal to 1. Note that the (n+ 1)× (n+ 1) matrix L(A)takes the “block upper triangular” form

L(A) =

(B t

0 1

), (4.2)

where B is an n × n matrix and t is a column vector with n rows. A matrix in thisblock upper triangular form is invertible iff B is invertible (for example, because one cancheck that detL(A) = detB for such a “block upper triangular” matrix L(A)). From thedefinition of matrix multiplication we see that two such block upper triangular matricesmultiply according to the rule(

B t

0 1

)(B′ t′

0 1

)=

(BB′ Bt′ + t

0 1

). (4.3)

In particular, this shows that L(A1A2) = L(A1)L(A2), so that L may be viewed as anisomorphism of groups from G to the subgroup G′ of GLn+1(K) consisting of invertiblematrices of the “block upper triangular form” described above. Furthermore, if wecompare the multiplication formula (4.3) to the formula (1.3) for composition of affinetransformations, we see that (

B t

0 1

)7→ [B, t]

is an isomorphism from G′ to Aff(Kn). To see that the isomorphism A 7→ [B, t] from Gto Aff(Kn) just described is given by the indicated restriction map, just notice that if wefix a point x ∈ Kn, then the corresponding point of KPn is [x : 1] and we have

A[x : 1] = [L(A)(x, 1)]

=

[(B t

0 1

)(x

1

)]

=

[(Bx+ t

1

)],

which is nothing but [B, t](x) = Bx + t ∈ Kn viewed as a point of KPn in the usualmanner.

4.5 Exercises 53

We now prove the analogue of Proposition 1.4.3 for projective transformations.

Proposition 4.4.5. If P0, . . . , Pn+1 are n+ 2 points of KPn in general position (Defini-tion 4.2.2), then there is a unique A ∈ PGLn(K) satisfying A[ei] = P for i = 0, . . . , nand A[1 : · · · : 1] = Pn+1.

Proof. Choose representatives x0, . . . , xn+1 ∈ Kn+1 \ {0} for the Pi. Since the Pi are ingeneral position, x0, . . . , xn must be a basis for Kn+1, so we can write

xn+1 =

n∑i=0

λixi

for unique λi ∈ K. Each λi must be non-zero, otherwise xn+1 would be in the span of aproper subset of {x0, . . . , xn}, in violation of the general position assumption. The imageA ∈ PGLn+1(K) of the matrix A ∈ GLn+1(K) whose ith column is λixi is as desired.

4.5 Exercises

Exercise 4.1. In Definition 4.1.2 we defined subsets U0, U1, U2 of KP2, and, moregenerally, U0, . . . , Un of KPn. In Definition 4.1.3 we defined bijections φi : K2 → Ui ⊆ KP2,and, more generally, φi : Kn → Ui ⊆ KPn.

a) Describe the subsets φ−11 (U1 ∩ U2), φ−11 (U0 ∩ U1 ∩ U2), and φ−12 (U1 ∩ U2) of K2.

b) We obtain a bijection φ12 from φ−11 (U1 ∩ U2) to φ−12 (U1 ∩ U2) by mapping a pointin φ−11 (U1 ∩ U2) into U1 ∩ U2 using the map φ1, then taking the inverse of theresulting point under φ2. Give an explicit formula for φ12.

c) More generally, for an arbitrary projective space KPn, describe the subsets φ−1i (Ui∩Uj) and the bijections φij : φ−1i (Ui ∩ Uj)→ φ−1j (Ui ∩ Uj) defined in the analogousmanner.

The next three exercises probably require some background in topology.

Exercise 4.2. Show that, for a subset U ⊆ RPn, the conditions of Definition 4.3.1 areequivalent. Also show that the conditions of Definition 4.3.2 are equivalent for a subsetU ⊆ CPn.

Exercise 4.3. For K = R or C, prove that the subsets Ui ⊆ KPn defined in Defini-tion 4.1.2 are open in the topology on KPn defined in Definitions 4.3.1 and 4.3.2. (Ifyou haven’t done the previous exercise, then just use, say, the first condition in eachof those definitions.) Prove that the bijections φi : Kn → Ui of Definition 4.1.3 arehomeomorphisms when Ui is given the subspace topology from the inclusion Ui ⊆ KPn.

Exercise 4.4. Show that CP1 is homeomorphic to the 2-sphere S2.

Exercise 4.5. Let A ∈ GLn+1(K), x ∈ Kn+1 \ {0}. In terms of basic concepts fromlinear algebra, what does it mean to say that [x] ∈ KPn is a fixed point of the projectivetransformation A : KPn → RPn associated to A?


Exercise 4.6. Let L1, L2, L3 be three lines in R2 with empty intersection: L1∩L2∩L3 = ∅.How many connected components does R2 \ (L1 ∪ L2 ∪ L3) have? What happens if R2 isreplaced by RP2? Hint: Move one of the lines to infinity by a projective transformation.(In fact you can move them even “more freely” as described in Exercise 4.10.)

There are an infinite number of possible variations on Proposition 4.4.5. Here aresome:

Exercise 4.7. Suppose L and M are lines in KP3 and P is a point of KP3. Prove thatthere is a line in KP3 containing P and intersecting both L and M .

Exercise 4.8. Suppose P(V ) ⊆ KPn is a linear subspace. Prove that every projectivetransformation of P(V ) (invertible linear transformation V → V , up to rescaling) extendsto a projective transformation of KPn.

Exercise 4.9. Let L be a line in KP2, P a point of L, M a line not containing P . Provethat there is a projective transformation A : KP2 → KP2 taking L to the line at infinity,P to the point [0 : 1 : 0], and M to the line Z(y) = P(Span(e0, e1)).

Exercise 4.10. Suppose L1, . . . L4 are four lines in KP2 such that the intersection ofany three is empty. Suppose L′1, . . . , L

′4 are four lines in KP2 with the same property.

Prove that there is a unique projective transformation A : KP2 → KP2 taking Li onto L′ifor i = 1, . . . , 4. Hint: This statement is “dual” to the statement of Proposition 4.4.5 inthe case n = 2.

Chapter 5

Projective Plane Curves

In this chapter, we will specialize our study of projective spaces in Chapter 4 to the caseof the projective plane KP2 over a field K. Recall that we identify the affine plane K2

over K with the subset

U2 = {[x : y : z] ∈ KP2 : z 6= 0}

of the projective plane KP2 via the map (x, y) 7→ [x : y : 1], thus we view K2 as a subsetof KP2, which we call the set of finite points. The complementary subset of infinite pointsis called the line at infinity ; it is identified with KP1 via [x : y : 0]↔ [x : y]. We will seethat the projective plane has many nice features that are lacking in the affine plane K2.For example, in K2, two distinct lines may or may not intersect, whereas in KP2, anytwo distinct lines intersect in precisely one point (§5.4).

Just as a polynomial in two variables f ∈ K[x, y] gives rise to an affine algebraicplane curve Z(f) ⊆ K2, we shall see in §5.1 that a homogeneous polynomial (also called aform) in three variables f ∈ K[x, y, z] gives rise to a projective plane curve Z(f) ⊆ KP2.Using this construction, we shall revisit some of the geometry we studied in Chapter 3.

In §5.8 we give a brief treatment of intersection multiplicity and Bezout’s Theorem,to the extent that we shall need it elsewhere. This section is rather tedious and not aparticularly satisfactory treatment of intersection multiplicity, as my hands were tied bymy refusal to make use of any concepts from commutative algebra (besides the notionof fields and polynomials). In §5.5 we see that the discriminant of a real affine conic,introduced in §3.6 has a natural “geometric” interpretation in projective geometry. In§5.6 we carry out the classification of projective conics (over R and C) promised in §3.5.

The remaining sections of this chapter are largely independent of one another andconsist of a hodge-podge of topics in the geometry of projective plane curves. In §5.7, wediscuss the cross ratio, which may be thought of as a classification of ordered lists of fourpoints in KP1, up to projective equivalence. In §5.9 we prove some “classical” theoremsof projective geometry. The theorem of Desargues proved there plays an important rolein axiomatic projective geometry. In §5.10 we show that the set of non-singular pointsof (certain) smooth cubic curves can be given the structure of an abelian group. Wedescribe this group explicitly in the case of the “cuspital cubic” in §5.11.

55

56 Chapter 5 Projective Plane Curves

5.1 Homogeneous polynomials and zero loci

A polynomial f(x, y, z) ∈ K[x, y, z] is called homogeneous of degree d (or a form of degreed) iff it can be written in the form

f(x, y, z) =∑

i+j+k=d

Aijkxiyjzk

for coefficients Aijk ∈ K. Usually we assume that at least one of the Aijk is non-zero, sothat f 6= 0, though in some situations it is convenient to have the convention that thezero polynomial is homogeneous of degree d for every d. A form of degree one (resp. two,three, . . . ) is often called a linear (resp. quadratic, cubic, . . . ) form.

If f is a form of degree d then for any λ ∈ K we have

f(λx, λy, λz) =∑

i+j+k=d

(λx)i(λy)j(λz)k (5.1)

=∑

i+j+k=d

λi+j+kxiyjzk

= λdf(x, y, z).

To any degree d polynomial

f(x, y) =∑i+j≤d

Aijxiyj

in two variables we associate a homogeneous polynomial f in three variables, called thehomogenization of f(x, y), by setting

f(x, y, z) :=∑i+j≤d

Aijxiyjzd−i−j .

This association f 7→ f yields a bijection between degree d polynomials in x, y andhomogeneous degree d polynomials in x, y, z whose inverse is given by f(x, y, z) 7→f(x, y, 1).

For any homogeneous polynomial f(x, y, z), we can consider the subset

Z(f) := {[x : y : z] ∈ KP2 : f(x, y, z) = 0} (5.2)

of the projective plane KP2. This makes sense because the question of whether f(x, y, z) =0 depends only on the equivalence class [x : y : z] of (x, y, z) ∈ K3 \ {0}, in light of theformula (5.1) (in the case of a non-zero λ).

Definition 5.1.1. For a homogeneous polynomial (form) f ∈ K[x, y, z] (usually assumedto be of positive degree), the subset Z(f) ⊆ KP2 defined by (5.2) is called the zerolocus of f . A subset S ⊆ KP2 equal to Z(f) for some homogeneous polynomial f (ofpositive degree) is called a projective plane curve and f is called a defining equationfor S. If f ∈ K[x, y] is any non-constant polynomial, the zero locus Z(f) ⊆ KP2 of thehomogenization of f is called the projective completion of Z(f) ⊆ K2.

5.2 Smooth and singular points 57

The usual remarks are in order: What we call the “projective completion of Z(f)”really depends highly on the polynomial f defining Z(f)—it is not really intrinsic to thesubset Z(f) ⊆ K2. Notice that, for a polynomial f ∈ K[x, y], we have

f(x, y, 1) = f(x, y),

so (x, y) ∈ K2 is in Z(f) ⊆ K2 iff [x : y : 1] ∈ KP2 is in Z(f) ⊆ KP2. In other words, thefinite points of the projective completion Z(f) ⊆ KP2 of Z(f) ⊆ K2 are just the points ofZ(f):

Z(f) ∩K2 = Z(f).

If f ∈ K[x, y, z] is homogeneous, notice that f(x, y, 0) ∈ K[x, y] is homogeneous (thoughpossibly zero even if f is non-zero)—the intersection Z(f) ∩KP1 of Z(f) ⊆ KP2 with theline at infinity KP1 ⊆ KP2 is given by

Z(f) ∩KP1 = {[x : y] ∈ KP1 : f(x, y, 0) = 0}.

This intersection with the line at infinity is often of interest even in the case where f = gis the homogenization of a polynomial g ∈ K[x, y]—it encodes, in an appealing geometricway, the “asymptotic” behaviour of g(x, y) for “large” (x, y) ∈ K2. We shall return tothis point in §5.5.

Remark 5.1.2. Given a homogeneous polynomial f ∈ K[x, y, z], the notation Z(f), usedin isolation could possibly be confusing, because it might be understood as the projectiveplane curve Z(f) ⊆ KP2, or just as the “affine” zero locus

Z(f) = {(x, y, z) ∈ K3 : f(x, y, z) = 0}

of f . In these notes, we always have the first meaning of Z(f) in mind. These two “zeroloci” are not unrelated: The homogeneity of f and (5.1) show that the “affine” zero locusof f—let us call it C(Z(f)) for the moment, for clarity—is “invariant under rescaling,”meaning: If (x, y, z) ∈ C(Z(f)), then λ(x, y, z) ∈ C(Z(f)) for any λ ∈ K (and converselyif λ ∈ K∗). Note also that the projection map K3 \ {0} → KP2 given by x 7→ [x] takesC(Z(f)) (minus the origin) onto Z(f) ⊆ KP2. For these reasons, the affine zero locusC(Z(f)) is often called the cone on the projective zero locus Z(f) ⊆ KP2.

5.2 Smooth and singular points

In this section we give the projective analogues of the definitions from §3.2.

Definition 5.2.1. For a homogeneous polynomial f ∈ K[x, y, z], a point P ∈ Z(f) ⊆ KP2

is called a non-singular (or smooth) point of Z(f) iff at least one of fx(P ), fy(P ), fz(P )is non-zero. A point P ∈ Z(f) for which fx(P ) = fy(P ) = fz(P ) = 0 is called a singularpoint of Z(f). The subset of Z(f) consisting of smooth (resp. singular) points is denotedZ(f)sm (resp. Z(f)sing). We say that f is non-singular or smooth iff Z(f)sing(K) is emptyfor every field extension K ⊆ K.1

1It equivalent to ask that Z(f)sing(K) = ∅ for some algebraically closed field K containing K.


As in the affine case, the question of whether P ∈ Z(f) is a smooth point or a singularpoint depends highly on f , not just on the curve C = Z(f). The following lemma is oftenuseful when trying to determine the singular points of a projective curve:

Lemma 5.2.2. For a homogeneous polynomial f ∈ K[x, y, z] of degree d, we have

df = xfx + yfy + zfz.

Consequently, if d is not zero in K (which holds if d > 0 and K = Q, R, C, or anyother field of characteristic zero), then a point P ∈ KP2 is a singular point of Z(f) ifffx(P ) = fy(P ) = fz(P ) = 0.

Proof. Exercise 5.11.

Definition 5.2.3. If P is a non-singular point of Z(f), we define the tangent line toZ(f) at P to be the line

TP Z(f) := Z(xfx(P ) + yfy(P ) + zfz(P )).

The previous lemma implies that P ∈ TP Z(f). The projective notions of smoothness,tangent lines, etc. are compatible in every imaginable sense with those from the affinesetting of §3.2. Some precise statements along these lines are given in Exercise 5.12.

5.3 Linear change of variables and projective equivalence

The notion of affine equivalence introduced in §3.3 also has an analogue in the projectivesetting.

Given a form (homogeneous polynomial) f ∈ K[x, y, z] = K[x] of degree d and aninvertible matrix A ∈ GL3(K), we obtain a form f ·A ∈ K[x] of degree d by setting

(f ·A)(x) := f(Ax).

As in §3.3, we think of the vector x = (x, y, z) of variables as a column vector, and thearguments of our polynomials as column vectors.

Definition 5.3.1. Two homogeneous polynomials f, g ∈ K[x, y, z] = K[x] are calledprojectively equivalent iff there is an invertible matrix A ∈ GL3(K) and a λ ∈ K∗ so thatf(x) = λg(Ax). Two subsets S, T ⊆ KP2 are called projectively equivalent iff there is anA ∈ GL3(K) such that A(S) = T .

When g = f ·A for an invertible matrix A, we often say that g is obtained from f bya linear change of variables. Similarly, if g = λf for λ ∈ K∗, we say that g is a rescalingof f .

The formula (5.1) shows that if A ∈ GL3(K) and f ∈ K[x, y, z] is homogeneous ofdegree d, then f · (λA) = λdf ·A. So as long as we consider our homogeneous polynomialsonly up to rescaling (as is reasonable if we are concerned with their zero loci), f · Adepends only on the image A ∈ PGL2(K) of A in the projective general linear group.

5.4 Lines and linear functionals 59

As in the affine setting, it is straightforward to check that projective equivalence is infact an equivalence relation (Exercise 5.1). As in the affine case (Lemma 3.3.1), one seesthat when f(x) = λg(Ax), the corresponding projective plane curves are related by

Z(f) = A−1(Z(g)). (5.3)

In particular, projectively equivalent homogeneous polynomials have projectively equiva-lent zero loci. As in the affine case, it is easy to classify homogeneous linear polynomialsup to projective equivalence—we shall see in the next section that there is only oneequivalence class, regardless of the field K over which we work.

5.4 Lines and linear functionals

Recall that, in Definition 4.2.1, we defined a line to be a subset L of KP2 equal to P(V ) forsome two dimensional subspace V of the K vector space K3. By linear algebra, every suchV arises as the kernel of some surjective (equivalently: non-zero) linear transformation(A,B,C) : K3 → K, hence a line may equivalently be defined as a subset L ⊆ KP2 suchthat

L = {[x : y : z] ∈ KP2 : Ax+By + Cz = 0}

for some (A,B,C) 6= (0, 0, 0) in K3. In other words, a line is the zero locus in KP2 of alinear form. As in Definition 5.1.1, we call “Ax+By + Cz” a defining equation for L.(Compare the affine version of a line, discussed in §3.3.)

Recall from linear algebra that if V is a K-vector space, then a linear transformationV → K is called a linear functional on V . The set

V ∗ := HomK(V,K)

of linear functionals on V is itself a K-vector space (under “pointwise addition andscalar multiplication”), called the dual vector space of V . The next lemma says that[f ] 7→ P(Ker f) establishes a bijection from KP2∗ = P((K3)∗) to the set of lines in KP2.

Lemma 5.4.1. The line with defining equation Ax+ By + Cz coincides with the linewith defining equation A′x+B′y +C ′z iff (A,B,C) = λ(A′, B′, C ′) for some λ ∈ K∗. Inother words, the kernel of a linear functional f : K3 → K coincides with the kernel of thelinear functional f ′ : K3 → K iff f = λf ′ for some λ ∈ K∗.

Proof. If (A′, B′, C ′) = λ(A,B,C) for some λ ∈ R∗, then Z(Ax+By + Cz) = Z(A′x+B′y + C ′z) because A′x + B′y + C ′z = λ(Ax + By + Cz), so A′x + B′y + C ′z = 0 iffAx+By+Cz = 0. On the other hand, suppose that Z(Ax+By+Cz) = Z(A′x+B′y+C ′z)and let us show that (A′, B′, C ′) = λ(A,B,C) for some λ ∈ R∗. After possibly permutingthe three coordinates everywhere, we can assume that A 6= 0. Then [−B : A : 0] ∈Z(Ax+ By + Cz) ⊆ RP2, so, since Z(Ax+ By + Cz) = Z(A′x+ B′y + C ′z), we have[−B : A : 0] ∈ Z(A′x+B′y + C ′z), hence AB′ = BA′, or, equivalently, B′ = λB, wherewe set λ := A′/A. Similarly, since [−C : 0 : A] ∈ Z(Ax + By + Cz), we find thatAC ′ = CA′, hence C ′ = λC, and then (A′, B′, C ′) = λ(A,B,C) as desired.


In other words, lines in KP2 are in bijective correspondence with points of “another”projective plane KP2∗ (whose coordinates we will call A,B,C, rather than x, y, z to avoidconfusion) via the map taking the line with defining equation Ax+By+Cz to the point[A : B : C]. The following result sums up basically everything there is to know aboutlines:

Proposition 5.4.2. Let K be a field, KP2 the projective plane over K.

1. For any line L = Z(Ax + By + Cz), any point P of L is a smooth point andL = TPL is the tangent line to L at P .

2. The “line at infinity” KP1 ⊆ KP2 is a line.

3. Any line L, other than the line at infinity, is the projective completion of a uniqueline in K2.

4. Any two distinct lines in KP2 intersect in exactly one point.

5. If L ⊆ KP2 is a line and A ∈ GL3(K), then the image A(L) ⊆ KP2 of L under theprojective transformation A : KP2 → KP2 is also a line.

6. For any line L, there is some A ∈ GL3(K) such that A(L) is the line at infinity.

Proof. (1): The partial derivations of f = Ax+By+Cz are fx = A, fy = B, and fz = C,at least one of which is a non-zero element of K by definition of a line. The equalityL = TPL is immediate from the definition of TPL (Definition 3.2.2).

(2): Note that z = 0x + 0y + 1z is a defining equation for the line at infinity.Alternatively, note that the line at infinity is the projectivization of the subspace of K3

spanned by the first two standard basis vectors e0, e1 (cf. §4.2).(3): If L = Z(Ax+By+Cz) is not the line at infinity, then by Lemma 5.4.1 and the

description of the line at infinity as Z(z), we see that either A or B is non-zero—let usassume A 6= 0 as the case B 6= 0 is similar. Then Z(x+ (B/A)y + (C/A)) is a line in K2

whose projective completion is

Z(x+ (B/A)y + (C/A)z) = Z(A(x+ (B/A)y + (C/A)z)) = L.

If Z(A′x + B′y + C ′) (with (A′, B′) 6= (0, 0)) is another line in K2 whose projectivecompletion is L, then we have L = Z(A′x+B′y + C ′z), hence (A′, B′, C ′) = λ(A,B,C)for some λ ∈ K∗ by Lemma 5.4.1. Then we have

(λA)(x+ (B/A)y + (C/A)z) = A′x+B′y + C ′,

and hence

Z(x+ (B/A)y + (C/A)z) = Z(A′x+B′y + C ′).

(4): Suppose L = Z(Ax+ By + Cz) and L′ = Z(A′x+ B′y + C ′z) are two distinctlines. Then (A,B,C), (A′, B′, C ′) ∈ K3 are linearly independent by Lemma 5.4.1, so, bylinear algebra, the matrix A whose rows are (A,B,C) and (A′, B′, C ′) gives a surjective

5.5 Invariants of affine conics revisited 61

(i.e. rank two) linear map A : K3 → K2; the kernel of such a map is a one dimensionalsubspace of K3 which, viewed as a point of KP2, is the intersection of L and L′.

(5): This follows from (5.3).(6): Write L = P(V ) for a two dimensional linear subspace V ⊆ K3. By linear

algebra, we can find a basis v0, v1, v2 for K3 such that v0, v1 is a basis for V . By linearalgebra there is a unique A ∈ GL3 with Avi = ei, i = 0, 1, 2. This A takes the twodimensional linear subspace V bijectively onto Span(e0, e1) ⊆ K3, hence A : KP2 → KP2

takes L = P(V ) bijectively onto the line at infinity KP1 = P(Span(e0, e1)).

5.5 Invariants of affine conics revisited

The discriminant (Definition 3.6.1 in §3.6) of a degree two polynomial f(x, y) ∈ R[x, y]has an elegant geometric interpretation in terms of projective plane curves, which wewill give momentarily after a brief lemma.

Lemma 5.5.1. For A,B,C ∈ R (not all zero), the subset

S := {[x : y] ∈ RP1 : Ax2 +Bxy + Cy2 = 0}

of the real projective line is empty (resp. consists of a single one point, consists of twodistinct points) if B2 − 4AC < 0 (resp. B2 − 4AC = 0, B2 − 4AC > 0).

Proof. First of all, the point ∞ = [1 : 0] ∈ RP1 is in S iff A = 0. Every point other than∞ in RP1 can be written as [x : 1] for a unique real numer x, and such a point is in S iffAx2 +Bx+ C = 0.

Now we divide into cases. If A = 0, then ∞ ∈ S and [x : 1] is in S iff Bx+ C = 0.The equation Bx+ C = 0 has exactly one solution when B 6= 0 and no solutions whenB = 0 (because then C 6= 0 since we assume A,B,C aren’t all zero). In the latter case,we have B2 − 4AC = B2 > 0, and S has two points (namely ∞, [−C/B : 1]), while inthe former case, we have B2 − 4AC = 0 and S consists only of the point ∞. The lemmais hence correct in the case A = 0.

If A 6= 0, then ∞ /∈ S, and hence the points of S are just the real numbers x forwhich Ax2 +Bx+ C = 0. Since A 6= 0, the number of such real numbers x is as statedin the lemma by a an exercise with the Quadratic Formula that should be familiar fromhigh-school algebra.

Remark 5.5.2. Over the complex numbers, the corresponding result says that S consistsof a single point if B2 − 4AC = 0 and two points otherwise. The “single” point shouldbe thought of as occurring with “multiplicity two.” Cf. Bezout’s Theorem (§5.8).

Now we can see that the discriminant of a degree two polynomial f(x, y) ∈ R[x, y] isjust a way of encoding how the projective completion Z(f) ⊆ RP2 of Z(f) ⊆ R2 intersectsthe line at infinity RP1 ⊆ RP2:

Proposition 5.5.3. Let f = Ax2 +Bxy+Cy2 +Dx+Ey+F be a polynomial of degreeat most two with real coefficients A, . . . , F and discriminant D(f) = B2 − 4AC. Then:

1. D(f) < 0 iff Z(f) is disjoint from the line at infinity,


2. D(f) > 0 iff Z(f) ∩ RP1 consists of two points, and

3. D(f) = 0 iff Z(f) ∩ RP1 consists of a single point.

Proof. Note that the homogenization of f is

f = Ax2 +Bxy + Cy2 +Dxz + Eyz + Fz2,

so a typical point [x : y : 0] of the line at infinity is on ZP(f) iff f(x, y, 0) = 0 iffAx2 +Bxy + Cy2 = 0. The result now follows from the previous lemma.

For a generalization of these ideas, see Exercise 5.2.

5.6 Classification of projective conics

Just as we studied the problem of classifying affine plane curves and polynomials in twovariables up to affine equivalence (§3.3), we can also study the problem of classifyingprojective plane curves and homogeneous polynomials in three variables up to projectiveequivalence (Definition 5.3.1). In this section, we will classify homogeneous degreetwo polynomials (with real or complex coefficients) and the corresponding projectiveplane curves (also called conics, as in the affine case) up to projective equivalence. Theprojective classification is simpler than the affine one, so we can give complete proofs inthis section. As in the affine case, the classification of conics is highly sensitive to thefield over which we work. The classification is “easier” over C than it is over R.

Many steps in the classification make sense over an arbitrary field K where 2 = 1+1 ∈K∗ (i.e. a field whose characteristic is not equal to 2). Consider a quadratic form

f = Ax2 +Bxy + Cy2 +Dxz + Eyz + Fz2 (5.4)

with coefficients A, . . . , F in such a field K. As in §3.5, we consider the symmetric 3× 3matrix

M(f) :=

A B/2 D/2

B/2 C E/2

D/2 E/2 F

(5.5)

with entries in K—notice that we have to assume 2 ∈ K∗ to be able to divide by 2 indefining the off-diagonal entries of our matrix M(f). We have

f(x) = xtrM(f)x, (5.6)

where x is our vector of variables (x, y, z), viewed as a column vector as usual. Notice that(5.6) is a little more natural looking than the analogous formula (3.4) we encountered inthe affine case.

In particular, notice that for A ∈ GL3(K), the formula (5.6) gives

f(Ax) = (Ax)trM(f)Ax

= xtrAtrM(f)Ax,

5.6 Classification of projective conics 63

thus we see that M(f) and M(f ·A) are related by

M(f ·A) = AtrM(f)A. (5.7)

Proposition 5.6.1. Let K be a field. For a quadratic form f ∈ K[x, y, z], the followingare equivalent:

1. We can write f = gh for linear forms g, h ∈ K[x, y, z], hence Z(f) = Z(g) ∪ Z(h) isa union of “two” (possibly equal) lines.

2. Z(f) contains a line.

These equivalent conditions imply condition (3) below. If 2 ∈ K∗, then the following twoconditions are equivalent:

3. There is a singular point in Z(f).

4. The matrix M(f) is singular (not invertible).

Proof. Obviously (1) implies (2). Conversely, suppose Z(f) contains a line L. By

Proposition 5.4.2(6) there is an invertible matrix A ∈ GL3(K) such that A−1

(L) is theline at infinity, hence Z(f ·A) contains the line at infinity by Formula 5.3. If f ·A = ghfactors as a product of two homogeneous linear polynomials, then f = (g ·A−1)(h ·A−1)also has such a factorization. We thus reduce to proving that if Z(f) contains the line atinfinity, then f factors as a product of two homogeneous linear polynomials. SupposeZ(f) contains the line at infinity. Then Z(f) contains the three points [0 : 1 : 0], [1 : 0 : 0],and [1 : 1 : 1], so we have

f(0, 1, 0) = f(1, 0, 0) = f(1, 1, 0) = 0.

If we write f as in (5.4), then these equalities are equivalent to A = B = C = 0, in whichcase we have the desired factorization:

f = z(Dx+ Ey + Fz).

(1) implies (3): Since the intersection of two lines in KP2 is non-empty by Proposi-tion 5.4.2(4), this follows from Exercise 5.3.

(3) iff (4): Notice that for x ∈ K3, we have

2M(f)x = (fx(x), fy(x), fz(x)) (5.8)

(up to a transpose). By linear algebra, the matrix M(f) is singular iff M(f)x = 0 forsome x ∈ K3 \ {0}. From (5.8), we see that this is equivalent to saying there is some[x] ∈ KP2 for which

fx(x) = fy(x) = fz(x) = 0.

Since 2 ∈ K∗ this is equivalent to saying there is a singular point in Z(f) by Lemma 5.2.2.


Definition 5.6.2. A homogeneous degree two polynomial f ∈ K[x, y, z] is called degen-erate iff Z(f) contains a singular point; otherwise f is called non-degenerate.

Theorem 5.6.3. Suppose K is a field with 2 ∈ K∗ and f ∈ K[x, y, z] is a homogeneouspolynomial of degree two. Then there is an invertible matrix M ∈ GL3(K) such that

f ·M = Ax2 +By2 + Cz2

for some A,B,C ∈ K.

Proof. Start with an arbitrary such f as in (5.4). Our goal is to show that by making afinite sequence of linear changes of variables, we can eliminate the “cross terms” xy, xz,and yz. Our procedure for doing this is a careful version of completing the square whichsometimes goes by the name of Lagrange’s reduction.Step 1. Our first step is to eliminate the cross terms Bxy and Dxz involving the firstvariable x. If we already have B = D = 0, then we proceed immediately to Step 3 below.If A 6= 0, then proceed to Step 2. If A = 0, but C 6= 0, just exchange the variables x andy, then proceed to Step 2. If A = C = 0, but B 6= 0, then our polynomial

Bxy +Dxz + Eyz + Fz2

is taken via the linear change of variables (x, y, z) 7→ (x, x+ y, z) to the polynomial

Bx2 + (B + E)xy + (D + E)xz + Eyz + Fz2,

which has non-zero x2 coefficient, so we can proceed to Step 2. In the remaining case whereA = C = 0, but D 6= 0, we can make a similar change of variables (x, y, z) 7→ (x, y, x+ z),then proceed to Step 2.Step 2. At this step, we are given a polynomial

Ax2 +Bxy + Cy2 +Dxz + Eyz + Fz2

with A 6= 0. By making the linear change of variables

(x, y, z) 7→ (x− B

2Ay − C

2Az, y, z)

(we need 2 ∈ K∗ here) our polynomial becomes

Ax2 +

(C +

B2

4A

)y2 +

(BD

2A2− BD

A+ E

)yz +

(D2

4A2− D2

2A+ F

)z2.

The exact coefficients are not important, but note that there are no cross terms involvingx, so we can proceed to Step 3.Step 3. At this step we are given a polynomial

Ax2 + Cy2 + Eyz + Fz2

with no cross terms involving x, and we wish to eliminate the one remaining cross termEyz (without creating any new cross terms). Of course, if E = 0, then we are done. If

5.6 Classification of projective conics 65

C 6= 0, proceed to Step 4. If C = 0, but F 6= 0, then interchange y and z and proceed toStep 4. The remaining case is where C = F = 0, but E 6= 0. In this case, our polynomial

Ax2 + Eyz

is taken by the invertible change of variables (x, y, z) 7→ (x, y, y + z) to the polynomial

Ax2 + Ey2 + Eyz,

which has non-zero y2 coefficient, hence we can proceed to Step 4.Step 4. At this step we are given a polynomial

Ax2 + Cy2 + Eyz + Fz2

with C 6= 0. By making the invertible change of variables

y 7→ y − E

2Cz

fixing x and z, our polynomial becomes

Ax2 + Cy2 +

(−E

2

4C+ F

)z2,

which is of the desired form.

Corollary 5.6.4. Every (non-zero) homogeneous degree two polynomial f ∈ R[x, y, z] isprojectively equivalent (Definition 5.3.1) to exactly one of the following:

1. x2 + y2 + z2

2. x2 + y2 − z2

3. x2 + y2

4. x2 − y2

5. x2.

Proof. By Theorem 5.6.3, f is equivalent to a polynomial of the form

Ax2 +By2 + Cz2,

for some A,B,C ∈ R. For each of the coefficients A,B,C which is non-zero, we canrescale the corresponding variable by the inverse of the square root of the absolute valueof its coefficient, thus we see that f will be equivalent to a polynomial of the same formwhere, now, A,B,C ∈ A,B,C ∈ {0, 1,−1}. Multiplying everything through by (−1)if necessary, we can assume there are at least as many +1’s as −1’s. Permuting thevariables if necessary, we can assume the +1’s come first, then the −1’s, then the zeros.Since we assume that the polynomial we started with was non-zero, the polynomial weget at the end is non-zero. The resulting list of possibilities is tabulated in the statementof the corollary.

Now, it still remains to show that no two polynomials on the list are equivalent. Buteven the corresponding projective plane curves


1. Empty

2. Circle

3. Point ([0 : 0 : 1])

4. Two intersecting lines

5. One “doubled” “line”

can be distinguished by elementary topological concerns, with the exception that the “line”is a copy of RP1, which is also homeomorphic to the circle; but this can be distinguishedfrom the other “circle” on the grounds that projective transformations take lines to linesand the circle contains three non-collinear points.

Corollary 5.6.5. Every (non-zero) homogeneous degree two polynomial f ∈ C[x, y, z] isprojectively equivalent (Definition 5.3.1) to exactly one of the following:

1. x2 + y2 + z2

2. x2 + y2

3. x2.

Proof. This is proved in exactly the same was as the previous corollary, except, in C,we can also make a change of variables like x 7→

√−1x to ensure that every such f is

projectively equivalent to one of the form Ax2 +By2 + Cz2 with A,B,C ∈ {0, 1}.

There is a lot more to say about the material in this section, but we will contentourselves with the following remarks:

Remark 5.6.6. Corollary 5.6.5 is valid over any algebraically closed field K with 2 ∈ K∗,by essentially the same proof—the point is that, in such a field K, every element a is asquare since the polynomial x2 − a ∈ K[x] must have a root.

Remark 5.6.7. Theorem 5.6.3 also makes sense in an arbitrary number of variables.The proof is basically the same: Suppose

n∑i=1

Aix2i +

∑i<j

Bijxixj

is a typical degree two homogeneous polynomial over a field K with 2 ∈ K∗. Suppose wehave eliminated all the cross terms involving x1, . . . , xk−1 (i.e. the second sum is actuallyover k ≥ i < j) and we want to further eliminate the ones involving xk. To do this,proceed as follows:Step 1. We first ask whether Al 6= 0 for some l ∈ {k, . . . , n}. If so, then we exchangexk and xl if necessary (obviously this preserves the property that cross terms involvingx1, . . . , xk−1 are zero) to assume that, in fact, Ak 6= 0, then we proceed to Step 2. Inthe other case where Ak = Ak+1 = · · · = An = 0, we find some l ∈ {k + 1, . . . , n} with

5.7 The cross ratio 67

Bkl 6= 0 (if there is no such l, then all the cross terms involving xk are already zero,so there is nothing to do). We then make the linear change of variables xl 7→ xk + xl(fixing the other xi). It is easy to see that this doesn’t create any cross terms involvingx1, . . . , xk−1 and that the new x2k coefficient is Bkl 6= 0, hence we can proceed to Step 2.

Step 2. At this step, we have a homogeneous degree two polynomial as above, with nocross terms involving x1, . . . , xk−1, with the further property that Ak 6= 0. The changeof variables

xk 7→ xk −Bk,k+1

2Ak−Bk,k+2

2Ak− · · · −

Bk,n2Ak

(fixing the other xi) then yields a polynomial with no cross terms involving x1, . . . , xk.

Obviously repeating Steps 1 and 2 successively for k = 0, 1, . . . , n− 1 eliminates allthe cross terms.

Remark 5.6.8. If K is of characteristic 2, one cannot eliminate the cross terms, even inthe 2 variable case (Exercise 5.6).

5.7 The cross ratio

Recall from Proposition 4.4.5 that any three distinct points of KP1 can be moved to0 = [0 : 1], 1 = [1 : 1], ∞ = [1 : 0] (respectively) by a unique projective transformationA ∈ PGL2(K). Now consider four distinct points P1, P2, P3, P4 ∈ KP1. By the sameproposition, there is a unique A ∈ PGL2(K) so that

(AP1, AP2, AP3) = (0, 1,∞),

thus we can associate the element

AP4 ∈ KP1 \ {0, 1,∞} = K \ {0, 1}

of K to our configuration (P1, . . . , P4) of four distinct points. The question arises: Can wegive an explicit formula for AP4 in terms of the homogeneous coordinates of P1, . . . , P4?

Indeed, if Pi = [xi : yi], then we claim that the cross ratio

χ(P1, P2, P3, P4) := [(x4y1 − x1y4)(x3y2 − x2y3) : (x2y1 − x1y2)(x3y4 − x4y3)] (5.9)

is the desired formula. The key calculation we need is:

Lemma 5.7.1. For A ∈ GL2(K), we have

χ(AP1, AP2, AP3, AP4) = χ(P1, P2, P3, P4).

Proof. If we write

A =

(a b

c d

),


then APi = [axi + byi : cxi + dyi], and we make the following calculation:

(axi + byi)(cxj + dyj)− (axj + byj)(cxi + dyi)

= (ad− bc)(xiyj − xjyi).

Thus we see that, replacing xi by axi + byi and yi by cxi + dyi everywhere in (5.9) willonly rescale both coordinates by (ad− bc)2, which is non-zero since detA = (ad− bc) 6= 0(as A is invertible) hence the corresponding point of KP1 is unchanged.

If we now specialize the formula (5.9) to the case P1 = 0 = [0 : 1], P2 = 1 = [1 : 1],P3 =∞ = [1 : 0], P4 = [x : 1], we find

χ(0, 1,∞, x) = [(x · 1− 0 · 1)(1 · 1− 1 · 0) : (1 · 1− 0 · 1)(1 · 1− x · 0)]

= [x : 1]

= x,

which shows that our formula has the desired property.Notice that the formula (5.9) makes sense as long as no three of the Pi coincide, so

we can in fact define the cross ratio

χ(P1, P2, P3, P4) ∈ KP1

for any four points P1, . . . , P4 ∈ KP1, as long as no three of them are the same. In fact,we have:

1. χ(P1, P2, P3, P4) = 0 iff P1 = P4 or P2 = P3

2. χ(P1, P2, P3, P4) = 1 iff P1 = P3 or P2 = P4

3. χ(P1, P2, P3, P4) =∞ iff P1 = P2 or P3 = P4

Now suppose we have a line L ⊆ KP2. Write L = P(V ) for a two-dimensionalsubspace V ⊆ K3. Choose an isomorphism of vector spaces f : V → K2. Given fourpoints P1, . . . , P4 ∈ L, no three the same, we can define their cross ratio, using the chosenisomorphism f , by setting

χf (P1, P2, P3, P4) := χ(f(P1), f(P2), f(P3), f(P4)) ∈ RP1.

We claim that this is independent of the choice of f . Indeed, suppose g : V → K2 isanother vector space isomorphism. Then by linear algebra, we can write g = Af for avector space isomorphism (invertible 2× 2 matrix) A : K2 → K2. We then compute

χg(P1, . . . , P4) = χ(g(P1), g(P2), g(P3), g(P4))

= χ(Af(P1), Af(P2), Af(P3), Af(P4))

= χ(f(P1), f(P2), f(P3), f(P4))

= χf (P1, P2, P3, P4),

using Lemma 5.7.1 for the second equality. Having established that our notion of crossratio is independent of the choice of f , we can unambiguously denote it χ(P1, P2, P3, P4).

5.8 Intersection multiplicity 69

5.8 Intersection multiplicity

As it will be necessary at many points later in the notes, we are now forced to discuss theidea of intersection multiplicity. Suppose that f1 and f2 are (non-constant) polynomialsin K[x, y] (for some field K), or (non-constant) homogeneous polynomials in K[x, y, z].Let C1 := Z(f1), C2 := Z(f2) be the corresponding affine algebraic curves in K2, as in§3.1, or projective curves in KP2 (§5.1), as appropriate. Assume that there are only

finitely many points in C1 ∩ C2 in K2(or KP2, as appropriate) for some (equivalently

any) algebraic closure K of K. (For simplicity, the reader may wish to keep in mind thecase where K = K = C.) Under these assumptions, one can assign a positive integerm(f1, f2, P ) to each point P ∈ C1 ∩ C2, called the intersection multiplicity. (One has toassume the finiteness of C1∩C2 over K, rather than just over K, because it could happen,for example, that f1 = f2 and the common curve C = Z(fi) has only one point P overK, and in this situation there is no “reasonable” way to define m(f1, f2, P ).) Althoughm(f1, f2, P ) certainly depends on the defining equations fi for the curves Ci, one oftenabusively writes m(C1, C2, P ) instead of m(f1, f2, P ). One can then prove the followinggeneral result:

Theorem 5.8.1. (Bezout’s Theorem) Suppose K is an algebraically closed field andf1, f2 ∈ K[x, y, z] are homogeneous polynomials of degrees d1, d2 > 0. Let Ci := Z(fi) ⊆KP2. Assume that C1 ∩ C2 is finite. Then∑

P∈C1∩C2

m(f1, f2, P ) = d1d2.

The general definition of m(f1, f2, P ) involves more commutative algebra than I wishto assume for the purposes of these notes, so it will not be given here. We will contentourselves with defining m(f1, f2, P ) only in the case where one of the fi, say f1, is ofdegree one. We will also prove Bezout’s Theorem in this case. This special case will besufficient for our later purposes.

Since we force ourselves to avoid any commutative algebra and to avoid the generallanguage of algebraic geometry, our definition of m(f1, f2, P ) will be rather tedious,though elementary. I will relegate almost everything to the exercises. Our constructionis based on the idea of multiplicity of roots for a polynomial in a single variable, whichwe shall now briefly review.

Let f ∈ K[x] be a polynomial of degree d > 0. It is a general fact of commutativealgebra that one can find a field K containing K as a subfield such that, in K[x], one canfactor f as a product of linear polynomials—that is, one can write

f(x) = Cd∏i=1

(x− αi) (5.10)

with C ∈ K∗, α1, . . . , αd ∈ K. (For example, if K is Q, R, or C, then the FundamentalTheorem of Algebra says one can take K = C. There is, in some sense, a “smallest” suchK, called a splitting field for f .) It is clear that α ∈ K is a root of f (i.e. f(α) = 0) iff α


is one of the αi. If α is a root of f , we define its multiplicity m = m(α, f) to be

m(α, f) := |{i ∈ {1, . . . , d} : α = αi}|. (5.11)

By construction, m ∈ {1, . . . , d}. Of course, the formula (5.11) also makes sense whenα is not a root of f , in which case m(α, f) = 0. If the factorization (5.10) of f canbe done in K (which holds, for example, if K is algebraically closed), then we clearlyhave

∑αm(f, α) = d, where the sum runs over the roots of f in K. This simple fact

will ultimately yield our special case of Bezout’s Theorem. The reader is asked toestablish some basic properties of this notion of “multiplicity” for roots of polynomialsin Exercise 5.7.

Definition 5.8.2. Let f ∈ K[x, y, z] be a form. A smooth point P of Z(f) is called aninflection point of Z(f) iff the intersection multiplicity of the tangent line TP Z(f) andZ(f) at P is greater than 2. (Cf. Exercise 5.14.)

5.9 The theorems of Pascal, Pappus, and Desargues

In this section we shall prove some classic theorems of projective geometry. The firstof these is a famous theorem of Blaise Pascal (1623-1662), which he apparently provedat the age of 16. Unfortunately his own proof of the theorem was lost, though severaldifferent proofs have accumulated since 1639. H. S. M. Coxeter includes a proof ofPascal’s Theorem in at least three of his books [C1, 3.35], [C3, 7.21], [C2, 9.23]. In fact,[C3, 1.7] also gives the “simple proof” due to van Yzeren’s proof [vY]. Pascal’s Theoremis also mentioned in [R, p. 140], and appears as an exercise in [H]. There are several“proofs” of Pascal’s Theorem on the Wikipedia, none of which is particularly clear in myopinion. Many “modern” proofs of Pascal’s Theorem use Bezout’s Theorem, which isridiculous overkill. It is much harder to formulate and prove Bezout’s Theorem than tojust prove Pascal’s Theorem from first principles. The proof we give here is the simplestI could come up with, relying only on a tiny bit of linear algebra and high-school levelalgebraic manipulation.

Theorem 5.9.1. (Pascal’s Theorem) Suppose P1, . . . , P6 are six distinct points con-tained in a non-degenerate conic C ⊆ RP2. Then the points X := P1P2 ∩ P4P5,Y := P2P3 ∩ P5P6, and Z := P3P4 ∩ P1P6 are collinear.

Proof. It follows from properties of intersection multiplicity that no three points ofa non-degenerate conic can be collinear, hence P1, . . . , P6 are in general position. Aprojective transformation takes lines to lines and conics to conics, so, after applying aprojective transformation, we can assume (by Proposition 4.4.5) that

P1 = [1 : 0 : 0]

P2 = [0 : 1 : 0]

P3 = [0 : 0 : 1]

P4 = [1 : 1 : 1]

P5 = [a : b : c]

P6 = [u : v : w].

5.9 The theorems of Pascal, Pappus, and Desargues 71

The lines P1P2, P1P3, P1P4, P2P4, P3P4 are described, respectively, by the equations

z, y, y − z, x, x− z, x− y,

so the fact that P5 and P6 are not on any of these lines is equivalent to the fact that thefollowing numbers are non-zero:

c, b, b− c, a, a− c, a− b, w, v, v − w, u, u− w, u− v.

We will divide by these non-zero numbers freely in what follows.Since P1 ∈ C (resp. P2 ∈ C, P3 ∈ C), the equation

Ax2 +Bxy + Cy2 +Dxz + Eyz + Fz2

defining C has A = 0 (resp. C = 0, F = 0). Then P4 ∈ C implies E = −B −D. Wecan’t have B = 0 (because then our conic would be degenerate), so we can assume, afterrescaling that B = 1, that this equation takes the form y(x− z) +Dz(x− y). Solving forD using the fact that P5 ∈ C yields

D =b(c− a)

c(a− b).

The fact that P6 is on C is then equivalent to:

v(u− w) +b(c− a)

c(a− b)w(u− v) = 0

Clearing the (non-zero!) denominator and expanding out, the bcvw term cancels and wefind:

abuw − bcuw + acvw = acuv − bcuv + abvw. (5.12)

The line P1P2 is given by z = 0, so X = P1P2∩P4P5 is given by the linear combinationof P4 = [1 : 1 : 1] and P5 = [a : b : c] with vanishing third coordinate, namely

X = [a− c : b− c : 0].

By similar reasoning, we find

Y = [0 : bu− av : cu− aw]

Z = [v : v : w].

To show that X, Y , and Z are colinear, we need to show that the coordinates of Y are alinear combination of those of X and Z, so we need to show that

Y = X − a− cv

Z.

Expanding this out and clearing (non-zero!) denominators, we see that this is equivalentto:

[0 : bu− av : cu− aw] = [0 : v(a− b) : w(a− c)],


which in turn is equivalent to

w(a− c)(bu− av) = (cu− aw)v(a− b). (5.13)

But if we expand out and cancel the term −a2vw common to both sides of (5.13), wesee that (5.13) is the same as (5.12).

There is also a variant of Pascal’s theorem for certain configurations of six pointson a degenerate conic—it can be proved by the same technique of using a projectivetransformation to move the points into some convenient locations, then doing a littleelementary algebra:

Theorem 5.9.2. (Pappus’ Theorem) Let L and L′ be distinct lines, A,B,C threedistinct points of L not equal to P := L ∩ L′. Let A′, B′, C ′ be three points of L′ distinctfrom P . Then the points X := AB′ ∩ A′B, Y := AC ′ ∩ A′C, and Z := BC ′ ∩ B′C arecollinear.

Proof. Exercise 5.16

The same general technique can also be used to prove:

Theorem 5.9.3. (Desargues’ Theorem) For points A,B,C,A′, B′, C ′ ∈ KP2 in gen-eral position, the following are equivalent:

1. The three lines AA′, BB′, and CC ′ have non-empty intersection.

2. The three points X := AB ∩ A′B′, Y := AC ∩ A′C ′, and Z := BC ∩ B′C ′ arecollinear.


Remark 5.9.4. Desargues’ Theorem does not hold without the general position assump-tion, even if the six points are in “general enough” position that the statement of thetheorem makes sense. See Exercise 5.18.

5.10 Cubic curves and the group law

Up to this point, our discussion of projective plane curves has been limited to the casesof lines and conics—the zero loci of linear and quadratic forms. Now it is time to look atthe zero locus of a homogeneous degree three polynomial—a cubic form. For technicalreasons, and to ease some calculations and exposition, we will often assume in thissection that 2 ∈ K∗, and occasionally, that 3 ∈ K∗. For reasons that we will discussmomentarily, we will also restrict our attention to a certain class of homogeneous degreethree polynomials:

Definition 5.10.1. Let K be a field with 2 ∈ K∗. A degree three polynomial g(x, y) ∈K[x, y] is said to be in Weierstrass form iff

g(x, y) = y2 − f(x) (5.14)

5.10 Cubic curves and the group law 73

for some monic (leading coefficient one) degree three polynomial f(x) ∈ K[x]. A cubicform g(x, y, z) ∈ K[x, y, z] is called a Weierstrass cubic (or is said to be in Weierstrassform) iff it is equal to the homogenization of a degree three polynomial g(x, y) ∈ K[x, y]in Weierstrass form. In other words, a Weierstrass cubic is a cibic form g that can bewritten

g(x, y, z) = y2z − f(x, z), (5.15)

where f(x, z) ∈ K[x, z] is the homogeneous degree three polynomial obtained by homoge-nizing a monic degree three polynomial f(x) ∈ K[x].

Here are the basic features of Weierstrass cubics:

Proposition 5.10.2. Assume 2 ∈ K∗. Let g(x, y) = y2 − f(x) ∈ K[x, y] be a degreethree polynomial in Weierstrass form, with homogenization g(x, y, z) ∈ K[x, y, z].

1. The singular points of Z(g) are the points of the form (α, 0) = [α : 0 : 1], where αis a repeated root of f . In particular, Z(g) has at most one singular point and if fhas no repeated roots in an algebraic closure of K, then Z(g) is smooth.

2. The point ∞ := [0 : 1 : 0] is the unique point of Z(g) on the line at infinity. It isan inflection point (Definition 5.8.2) of Z(g) with tangent line T∞ Z(g) equal to theline at infinity.

3. The polynomial g(x, y, z) is invariant under the linear change of variables (x, y, z) 7→(x,−y, z), hence Z(g) is invariant under the involution [x : y : z] 7→ [x : −y : z] ofKP2.

4. The points of Z(g) fixed by the involution from the previous part are the point ∞,together with the points of the form (0, α) = [α : 0 : 1] where α is a root of f .

5. g cannot be factored as a product of a linear form g1 and a quadratic form g2, evenif the coefficients of g1 and g2 are allowed to be in a field extension K ⊇ K.

Proof. (1) and (2): The partial derivatives of g are gx = −f ′(x) and gy = 2y. Since2 ∈ K∗, a point (a, b) ∈ K2 satisfies

g(a, b) = gx(a, b) = gy(a, b) = 0

iff b = 0 and f(a) = f ′(a) = 0—these latter equalities are equivalent to saying that a is arepeated root of f (Exercise 5.7(1)). This shows that the only “finite” singular points ofZ(g) (cf. Exercise 5.12) are those described in (1). Since f(x) is monic of degree three, wehave f(x, 0) = x3 and hence g(x, y, 0) = x3, hence ∞ = [0 : 1 : 0] is the unique point ofZ(g) on the line at infinity and the line at infinity intersects Z(g) at ∞ with multiplicitythree. Next notice that fz(x, z) is homogeneous of degree two, so certainly fz(0, 0) = 0.Then, since gz = y2 − fz, we see that gz(0, 1, 0) = 1 6= 0, so ∞ is a smooth point of Z(g).

(3): The invariance of g under the indicated change of variables is clear since g isactually a polynomial in x, y2, and z, so the invariance of Z(g) under the indicatedprojective transformation is just a special case of the formula (5.3).


(4): The involution in question takes the locus K2 ⊆ KP2 of finite points into itselfvia the involution (x, y) 7→ (x,−y). Clearly then, a finite point is fixed by this involutioniff it is of the form (α, 0). Such a point is in Z(g) iff f(α) = 0. On the line at infinity,the involution is given by [x : y : 0] 7→ [x : −y : 0]. This has exactly two fixed points:[1 : 0 : 0] and [0 : 1 : 0], only the second of which is actually in Z(g).

(5): Suppose we could write g = g1g2 in K[x, y, z] for some field K ⊇ K for a linearform g1 ∈ K[x, y, z] and a quadratic form g2 ∈ K[x, y, z]. Since the coefficient of x3 is 1in g, we could rescale g1 and g2 to arrange that they take the form

g1 = x+Gy +Hz

g2 = x2 +Bxy + Cy2 +Dxy + Eyz + Fz2

for some B, . . . ,H ∈ K. Since g(x, y, 0) = x3, we find that B, . . . ,H must satisfy

G+B = 0 (5.16)

CG = 0 (5.17)

C +BG = 0. (5.18)

First suppose that G = 0. Then these equations imply that B = C = 0. But then g1g2looks like

(x2 +Dxz + Eyz + Fz2)(x+Hz),

which has no y2z term despite the fact that the coefficient of y2z in g is 1. Next supposethat G 6= 0. Then (5.17) implies that C = 0 and (5.18) implies that B = 0, but then(5.16) doesn’t hold. We conclude that there can be no such factorization.

Now we prove that any cubic polynomial with these same geometric features isprojectively equivalent to a Weierstrass cubic:

Theorem 5.10.3. Suppose h ∈ K[x, y, z] is a cubic form for which there exist aninflection point P ∈ Z(h) and a projective transformation M satisfying:

1. M is an involution, meaning: M2

= Id,

2. the fixed locus of M consists of the point P and some line L not containing P , and

3. h ·M is a non-zero multiple of h for some (equivalently any) M ∈ GL3(K) mappingto M in PGL2(K).

Then 2 ∈ K∗ and h is projectively equivalent to a cubic g = y2z − f(x, z) in Weierstrassform. If, furthermore, 3 ∈ K∗, then we can also arrange that the coefficient of x2 in f iszero.

Proof. By Exercise 4.9, we can find A ∈ GL3(K) such that the projective transformationA takes P to ∞ = [0 : 1 : 0], the tangent line TP Z(h) to the line at infinity, and the line

L to the line Z(y). Replacing h, P , and M with h ·A−1, ∞, and AMA−1

, respectively,we can assume that P =∞, TP Z(h) = Z(z), and L = Z(y).

5.10 Cubic curves and the group law 75

Since ∞ is an inflection point for Z(h) and T∞ Z(h) = Z(z), Z(h) must intersect Z(z)only at ∞ with multiplicity three, hence h(x, y, 0) must be a non-zero multiple of x3, so,after rescaling h, we can assume h(x, y, 0) = −x3. This means

h = By2z − x3 −Ax2z − Cxz2 +Dyz2 − Ez3 + Fxyz.

Pick a lift M ∈ GL3(K) of M . Since M fixes ∞ = [e1] and Z(y) = P(Span(e0, e2)),e1 must be an eigenvector for M , say with eigenvalue λ, and Span(e0, e2) must be aneigenspace for M , say with eigenvalue µ. We can’t have λ = µ, for then M would be

λ = µ times the identity and M = Id would fix all of KP2. Since M2

= Id, we must haveλ2 = µ2, hence λ = −µ. (Since λ 6= µ this implies 2 ∈ K∗.) Replacing M with µ−1M ,we can assume µ = 1, λ = −1, hence M : K3 → K3 is given by (x, y, z) 7→ (x,−y, z) andM is given by [x : y : z] 7→ [x : −y : z].

Since h ·M must be a non-zero scalar multiple of h, we find that D = F = 0. Wecan’t have B = 0, for then

h = −x3 −Ax2z − Cxz2 − Ez3

would be singular at∞. So, after doing the change of variables z 7→ B−1z, we can assumeB = 1, which puts h in Weierstrass form:

h = y2z − x3 −A′x2z −B′xz2 − C ′z3.

If 3 ∈ K∗ we can do the linear change of variables x 7→ (x−A′/3z) to get rid of the x2zterm.

Remark 5.10.4. There are “better” reasons than Theorem 5.10.3 for restricting ourattention to Weierstrass cubics. It turns out that any smooth cubic C with a pointP ∈ C is “abstractly isomorphic” to a Weierstrass cubic by an “isomorphism” taking Pto ∞. In these notes, we have not developed the notion of an “abstract curve” or an“isomorphism” between such things—we have only the notion of projective equivalence.Now, it is not true that any smooth cubic C with a point P ∈ C is projectively equivalentto a Weierstrass cubic by a projective equivalence taking P to ∞. This is simply becauseP might not be an inflection point of C. Even “worse,” there are smooth cubics C (overK = Q, say) with non-empty zero locus (over K), but which have no inflection points(over K). The upshot of this discussion is supposed to be that not much generality is lostin restricting our attention to cubics in Weierstrass form.

The essential geometric features of Weierstrass cubics, that we have made precisein Theorem 5.10.3, allow one to endow the set E := Z(g)ns of non-singular points ofZ(g) ⊆ KP2 with the structure of an abelian group! For a Weierstrass cubic g, we letP 7→ P denote the involution [x : y : z] 7→ [x : −y : z] of Z(g). Since this involution is aprojective equivalence, it takes smooth points to smooth points, so it can also be viewedas an involution of E. The group law � on E is defined as follows:

Definition 5.10.5. For distinct points P,Q ∈ E, let L(P,Q) ⊆ KP2 be the unique linecontaining P and Q. If P = Q, then L(P, P ) := TPE is defined to be the tangent line to


E at P (here we need P to be a non-singular point). By “Bezout’s Theorem” (really:Exercise 5.13, plus Exercise 5.14 in the P = Q case), L(P,Q) ∩ Z(g) consists of threepoints when counted with multiplicity, so we can write L(P,Q) ∩ Z(g) = {P,Q,R} withthe understanding that R might be equal to P , Q, or both. In fact the point R mustbe a smooth point of Z(g), otherwise, using Exercise 5.15, we would find at least fourpoints, counted with multiplicity, in L(P,Q) ∩ Z(g). We define P �Q := R.

It is clear from the definition of � that � is commutative. In the interest of brevity ofthese notes, we shall not include here a proof that � is associative—we will now establishthe other group axioms.

Proposition 5.10.6. Let g be a Weierstrass cubic, � the binary operation on E = Z(g)ns

defined in Definition 5.10.5.

1. The point ∞ ∈ E is the identity element for �. That is, ∞�P = P for all P ∈ E.

2. For all P ∈ E, we have P ⊕ P =∞, so P is the inverse of P .

Proof. We saw in Proposition 5.10.2(2) that ∞ is an inflection point of Z(g), henceL(∞,∞) = Z(z) is the line at infinity and L(∞,∞) ∩ E = {∞,∞,∞}. We also sawin Proposition 5.10.2(4) that ∞ = ∞. From the definition of �, it is now clear thatboth parts of the proposition hold when P =∞. Since ∞ is the unique point of E onthe line at infinity (Proposition 5.10.2(2)), we can assume in the rest of the proof thatP = (s, t) = [s : t : 1] is a finite point of E. Then L(P,∞) = Z(x − sz) intersects Eat the point P = [s : −t : 1] (with multiplicity two if t = 0, in which case P = P andZ(x− sz) is the tangent line to E at P ) so we find that

L(P,∞) ∩ E = {P,∞, P}= L(P, P ) ∩ E.

The two parts of the proposition now follow immediately from the definition of �.

There is a tremendous amount to say about the group (E,�), most of which is beyondthe scope of these notes. We just mention the following celebrated theorem:

Theorem 5.10.7. (Mordell-Weil) If K = Q (or, more generally, any number field)and g ∈ K[x, y, z] is a Weierstrass cubic such that Z(g) has no singular points, then thegroup (E = Z(g),�) is finitely generated.

Some standard references for this material are [Sil] and [Tat].

5.11 Nodal and cuspital cubics

In this section we will explicitly describe the abelian group (E,�) for two specialWeierstrass cubics. We continue to assume that K is a field with 2 ∈ K∗.

Definition 5.11.1. The Weierstrass cubic forms y2z − x2(x+ z) and y2z − x3 will becalled the nodal cubic and the cuspital cubic, respectively.

5.11 Nodal and cuspital cubics 77

Note that the point [0 : 0 : 1] is the unique singular point of both the nodal cubicand the cuspital cubic, corresponding to the fact that 0 is the unique repeated root ofthe polynomials x2(x+ 1) and x3 (cf. Proposition 5.10.2(1)). We leave it as an exerciseto prove that any Weierstrass cubic form with a singular point is projectively equivalentto either the nodal cubic or the cuspital cubic (Exercise 5.21).

Theorem 5.11.2. The map φ(s) := [s : 1 : s3] defines an isomorphism of groupsφ from K (under addition) to E = Z(y2z − x3)ns (under the group operation � ofDefinition 5.10.5).

Proof. Note that [0 : 0 : 1] is the only singular point of Z(y2z − x3) and ∞ = [0 : 1 : 0] isthe only infinite point of E; any other point of E is of the form (x, y) = [x : y : 1] wherey2 = x3 6= 0—such a point can be written in the form (s−2, s−3) for a unique s ∈ K∗. Forsuch an s, we have

(s−2, s−3) = [s−2 : s−3 : 1] = [s : 1 : s3].

Since φ takes s = 0 to ∞, this discussion proves that φ is bijective and takes the identityelement 0 of (K,+) to the identity element ∞ of (E,�). It remains to check that φpreserves addition—that is, that

[s+ t : 1 : (s+ t)3] = [s : 1 : s3] � [t : 1 : t3] (5.19)

for all s, t ∈ K. This is clear if s or t is zero: If, say, s = 0, then s = 0 and φ(s) =∞ arethe identity elements and both sides of (5.19) are [t : 1 : t3]. It is also clear if s = −t, forthen φ(s) = φ(t) and both sides of (5.19) are ∞. The case where s = t 6= 0 is specialbecause we need to calculate the tangent line to E at (s−2, s−3) in order to calculatethe right hand side of (5.19)—we will leave this case as Exercise 5.22 and just treat theremaining “generic” case where s and t are non-zero, s 6= t, and s 6= −t.2 Looking at thedefinition of �, what needs to be shown is that the line L(s, t) ⊆ K2 containing (s−2, s−3)and (t−2, t−3) also contains the point ((s + t)−2,−(s + t)−3). As one probably knowsfrom high school, the equation for the line containing (x0, y0) and (x1, y1) is

y(x0 − x1)− x(y0 − y1) + y0x1 − x0y1,

so the equation for L(s, t) is

y(s−2 − t−2)− x(s−3 − t−3) + s−2t−2(s−1 − t−1). (5.20)

Indeed, one checks easily that (s−2, s−3) and (t−2, t−3) are in the zero locus of this linearpolynomial—notice that we need to know s 6= t to know that this polynomial is actuallynon-constant. Finally, we need to show that

−(s+ t)−3(s−2 − t−2)− (s+ t)−2(s−3 − t−3) + s−2t−2(s−1 − t−1) (5.21)

2When K = R or C one could argue on continuity grounds that this “generic” case implies all theother cases. In fact, one can make similar arguments over an arbitrary field K—this is one reason why Idon’t feel so bad about relegating the special case s = t 6= 0 to the exercises.


is zero. Since s 6= −t, s + t 6= 0, so we can check that (5.21) is zero after multiplyingthrough by

(s+ t)3 = s3 + 3s2t+ 3st2 + t3,

in which case (5.21) becomes

−s−2 + t−2 − (s+ t)(s−3 − t−3) + (st−2 + 3t−1 + 3s−1 + s−2t)(s−1 − t−1).

Expanding this out, one sees that it is zero.

Theorem 5.11.3. The map φ([x : y : z]) := (y + x)/(y − x) defines an isomorphism ofgroups from E = Z(y2z − x2(x+ z))ns to K∗ (under muliplication).


5.12 Exercises

Exercise 5.1. Show that “projective equivalence” (Definition 5.3.1) defines an equiv-alence relation on the set of homogeneous polynomials and on the set of subsets ofKP2.

Exercise 5.2. Let f ∈ K[x, y] be a polynomial of degree d. Write f =∑

i+j≤d ai,jxiyj .

Let f =∑

i+j≤d ai,jxiyjzd−i−j be the homogenization of f and let f :=

∑i+j=d ai,jx

iyj

be the “leading order part of f”.

1. Show that the intersection Z(f) ∩ KP1 of Z(f) ⊆ KP2 with the line at infinityKP1 ⊆ KP2 is equal to Z(f), so that the intersection of the projective completionof Z(f) and the line at infinity depends only on the leading order part of f . This isreasonable because the set of points “at infinity” in the projective completion ofZ(f) should have to do only with the asymptotic behaviour of f .

2. Calculate Z(f) ∩ RP1 when

f = x3 − y3 + 5x2 − 7xy + 5x+ 13 ∈ R[x, y

What is the cardinality of Z(f) ∩ RP1? What would change if we replaced the realnumbers with the complex numbers everywhere and asked for the cardinality ofZ(f) ∩ CP1?

Exercise 5.3. State and prove the analogue of Exercise 3.2 for a homogeneous polynomialf ∈ K[x, y, z].

Exercise 5.4. Let f ∈ K[x, y, z] be a homogeneous polynomial, A ∈ GL3(K) an invertible

matrix. Prove that a point P ∈ Z(f) ⊆ KP2 is a singular point of Z(f) iff A−1

(P ) is asingular point of Z(f ·A). (Cf. Exercise 3.3.)

Exercise 5.5. Prove Lemma 5.4.1.

5.12 Exercises 79

Exercise 5.6. Let K be a field of characteristic 2. Show that the homogeneous degreetwo polynomial xy ∈ K[x, y] is not projectively equivalent to one of the form Ax2 +By2.

Exercise 5.7. Let f(x) ∈ K[x] be a polynomial in one variable over a field K. Themultiplicity of a root α of f is defined in §5.8.

1. If α is a root of f with multiplicity m, show that

f(α) = f ′(α) = · · · = f (m−1)(α) = 0,

but f (m)(α) 6= 0. (This gives an alternative definition of m.)

2. Show that if f has at least d− 1 roots in K, counted with multiplicity (that is, atleast d− 1 of the αi are in K for a factorization of f in K[x] as in (5.10)), then infact f has d roots in K, counted with multiplicity (that is, all the αi are in K, sothe factorization of f above can actually be done in K[x]). The point here is that,when we write f = C(xd + ad−1x

d−1 + · · ·+ a0), the coefficients ad−1, . . . , a0 are inK, but they are also expressible somehow (how?) in terms of the αi.

3. Clearly m(λf, α) = m(f, α) for any λ ∈ K∗. Show also that m(f(λx), λ−1α) =m(f, α) for λ ∈ K∗ and m(f(x + a), α − a) = m(f, α) for any a ∈ K. That is,“multiplicity is invariant under Aff(K).”

Exercise 5.8. Suppose f ∈ K[u, v] is a homogeneous polynomial of degree d > 0 invariables u, v and P = [s : t] ∈ Z(f) ⊆ KP1. If P ∈ U0 ⊆ KP1 (resp. P ∈ U1), we definethe multiplicity m = m(f, P ) of P (as a point of Z(f)) to be the multiplicity of t/s(resp. s/t) as a root of g(x) := f(1, x) ∈ K[x] (resp. h(x) := f(x, 1) ∈ K[x]). Show that ifP ∈ U0 ∩ U1, then both ways of defining m give the same result, so this actually makessense!

Exercise 5.9. Let f and P be as in the previous exercise and let

A =

(a b

c d

)∈ GL2(K)

be an invertible 2 × 2 matrix. Let A : KP1 → KP1 be the associated projective

transformation defined by A([x]) := [Ax]. Show that m(f, P ) = m(f ·A,A−1(P )). Recallthat

(f ·A)(x, y) = f(au+ bv, cu+ dv).

In other words, multiplicity is invariant under projective transformations of KP1.

Exercise 5.10. Now let g ∈ K[x, y, z] be a homogeneous polynomial of degree d > 0, letL ⊆ KP2 be a line, and let P be a point of L ∩ Z(g). We want to define the multiplicitym = m(P, g, L) of the intersection L∩Z(g) at P when it makes sense to do so. As we sawon a previous homework, L = PL for a unique two dimensional linear subspace L ⊆ K3.Choose an ordered basis (q, r) for L and set

f(u, v) := g(uq + vr).


More precisely, if q = (q1, q2, q3) and r = (r1, r2, r3), then

f(u, v) := g(uq1 + vr1, uq2 + vr2, uq3 + vr3).

We also have a bijection

KP1 → L

[u, v] 7→ [uq + vr],

so we can write P = [sq+tr] for a unique [s, t] ∈ KP1. The polynomial f(u, v) ∈ K[u, v] isclearly “of degree d,” except that it might be zero. (Note that if f is zero, then L ⊆ Z(g).)Assuming that L is not contained in Z(g), so f is not zero, we define m(P, g, L) to be themultiplicity of [s, t] as a point of Z(f) in the sense of Exercise 5.8. Show that m(P, g, L)is independent of the choice of ordered basis (b, c) for V . Hint: If (q, r) is another choiceof ordered basis, with f ∈ K[u, v] the corresponding polynomial, then consider the changeof basis matrix in GL2(K) relating (q, r) and (q, r). Show that f and f are related as inthe previous exercise. To do this without making a mess, I strongly advise you to strictlyfollow the conventions of linear algebra and write elements of K3 as column vectors, alsowriting arguments of polynomials as column vectors. Notice that I didn’t follow my ownadvice above... it would probably be better to write

f

(u

v

):= g

((u

v

)tr(q

r

)).

Exercise 5.11. Prove Lemma 5.2.2.

Exercise 5.12. Let f ∈ K[x, y, z] be a homogeneous polynomial of degree d, so thatf(x, y, 1) ∈ K[x, y] is a polynomial of degree d, though not necessarily homogeneous.Show that a point P = (x0, y0) of the affine algebraic curve Z(f(x, y, 1)) ⊆ K2 is asmooth point in the sense of Definition 3.2.1 iff the corresponding point [x0 : y0 : 1](also abusively denoted P ) is a smooth point of the projective curve Z(f) ⊆ KP2 in thesense of Definition 5.2.1. Furthermore, if these equivalent smoothness notions hold, showthat the tangent line to Z(f) ⊆ KP2 at P defined in Definition 5.2.3 is the projectivecompletion of the tangent line to Z(f(x, y, 1)) ⊆ K2 at P defined in Definition 3.2.2.

Exercise 5.13. Let g ∈ K[x, y, z] be a (non-zero) homogeneous polynomial of degree 3,L ⊆ KP2 a line. Assume that L is not contained in Z(g). Show that if L∩Z(g) containsat least two points (counted with multiplicity), then it contains precisely three points(counted with multiplicity).

Exercise 5.14. Let f ∈ K[x, y, z] be a form, P a smooth point of Z(f). Show that theintersection multiplicity m(TP Z(f), f, P ) of the tangent line TP Z(f) and Z(f) at P is atleast two.

Exercise 5.15. Let f ∈ K[x, y, z] be a form, P a singular point of Z(f), L a line in KP2

containing P . Show that the intersection multiplicity m(L, f, P ) of L and Z(f) at P isat least two.

5.12 Exercises 81

Exercise 5.16. Prove Pappus’ Theorem (Theorem 5.9.2).

Exercise 5.17. Prove Desargues’ Theorem (Theorem 5.9.3).

Exercise 5.18. Suppose A,B,C,A′, B′, C ′ ∈ R2 ⊆ RP2 are six distinct points in generalposition, except that A, B, and C are contained in a line L. Notice that condition (2)in Desargues’ Theorem (Theorem 5.9.3) holds trivially since X,Y, Z ∈ L. Draw such aconfiguration where (1) does not hold.

Exercise 5.19. Consider the cubic polynomial f = f(x) = x(x2 + 1) ∈ R[x] andthe associated Weierstrass cubic g = g(x, y) = y2 − f(x). Let E := Z(g) ⊆ RP2,∞ := [0 : 1 : 0] ∈ E. Recall the binary operation � from Definition 5.10.5 making (E,�)an abelian group.

a) Find all points P of E for which P � P =∞.

b) Calculate ∞� [1 :√

2 : 1].

c) Calculate [0 : 0 : 1] � [1 :√

2 : 1].

d) Calculate [1 :√

2 : 1] � [1 :√

2 : 1].

e) Calculate [1 :√

2 : 1] � [1 : −√

2 : 1].

Exercise 5.20. Let g = y2z − f(x, z) be a Weierstrass cubic form. Show that a pointP ∈ E = Z(g) has order two in (E,�) (meaning P 6=∞, but P � P =∞) iff P = (α, 0)for a root α of f .

Exercise 5.21. Suppose g = y2x− f(x, z) is a Weierstrass cubic form (Definition 5.10.1)over a field K such that Z(g) ⊆ KP2 has a singular point (over K). Prove that g isprojectively equivalent to either the nodal cubic or the cuspital cubic (Definition 5.11.1).Prove that the nodal cubic and the cuspital cubic are not projectively equivalent.

Exercise 5.22. For the cuspital cubic g = y2z− x3, calculate the tangent line L(s, s) toZ(g) at the point (s−2, s−3) (s ∈ K∗) and use your formula for L(s, s) to check equation(5.19) in the proof of Theorem 5.11.2 in the case s = t.

Exercise 5.23. Prove Theorem 5.11.3.

Chapter 6

Appendix

In this appendix I have collected together some basic facts, mostly of algebraic nature,that are used in the text. My hope is that these notes should be readable without thisappendix, but I have included this anyway for the reader who wants to have a betteridea of the “general context” of some constructions in the main text.

6.1 Groups and group actions

Definition 6.1.1. A group (G,�) is a set G equipped with a binary operation (g, h) 7→g � h satisfying the following properties:

1. � is associative: (g � h) � k = g � (h� k) for all g, h, k ∈ G.

2. � has an identity : There is an element e ∈ G (necessarily unique; denoted eG ifthere is any chance of confusion) such that e� g = g � e = g for all g ∈ G.

3. � has inverses: For every g ∈ G there is an h ∈ G such that g � h = h� g = e.

A group (G,�) is called abelian (or commutative) iff g�h = h�g for all g, h ∈ G. If (G,�)and (G′,�′) are groups, a group homomorphism (or simply, a map of groups) from (G,�)to (G′,�′) is a function f : G→ G′ such that f(eG) = eG′ and f(g � h) = f(g) �′ f(h)for all g, h ∈ G. An isomorphism of groups is a bijective group homomorphism.

It is customary to use “G” to refer both to a group (G,�) and the “underlying set” G.Usually, when discussing an arbitrary group G, one usually denotes the binary operation� simply by juxtaposition, writing “gh” instead of “g � h.” Similarly, it is customary todenote the identity element e of G by 1, or perhaps 1G if there is any chance of confusion.For g ∈ G, the element h ∈ G satisfying gh = hg = 1 is easily seen to be unique, and isusually denoted g−1 and called the inverse of g.

For an abelian group, one usually denotes the binary operation by (g, h) 7→ g + h,the identity element by 0, and the inverse of g by −g.

Definition 6.1.2. For a group G, a non-empty subset H ⊆ G is called a subgroup iff itsatisfies:

82

6.1 Groups and group actions 83

1. hh′ ∈ H for all h, h′ ∈ H

2. h−1 ∈ H for all h ∈ H.

These conditions are usually summarized by saying that H is closed under addition andinverses.

A subgroup of a group is a group in its own right, with the binary operation inheritedfrom G. If f : G → G′ is a map of groups and H is a subgroup of G, then one checkseasily that the image f(H) of H under f is a subgroup of G′. In particular, the imagef(G) of G is a subgroup of H.

Example 6.1.3. For a group G and an element g ∈ G, the function fg : G→ G givenby fg(h) := ghg−1 is an isomorphism of groups called conjugation by g. In particular, ifH is a subgroup of G, then the image

fg(H) = {ghg−1 : h ∈ H}

(also denoted gHg−1) of H under fg is also a subgroup of G.

Definition 6.1.4. Two subgroups H and H ′ of a group G are called conjugate iff thereis some g ∈ G such that H ′ = gHg−1. A subgroup N of G is called normal iff N is theonly subgroup of G conjugate to N—equivalently N = gNg−1 for every g ∈ G.

Definition 6.1.5. If G is a group and S is a set, then an action (more precisely: leftaction) of G on S is a function

G× S → S

(g, s) 7→ gs

satisfying:

1. 1s = s for all s ∈ S. Here 1 ∈ G is the identity element of G.

2. g(hs) = (gh)s for all g, h ∈ G, s ∈ S.

Let S be a set equipped with an action of a group G (a “G-set”). For s ∈ S, the subset

Stab s := {g ∈ G : gs = s}

of G is called the stabilizer of s (in G) and the subset

Gs := {gs : g ∈ G}

of S is called the orbit (or G-orbit if there is a chance of confusion) of s.

Proposition 6.1.6. For any s ∈ S, the stabilizer Stab s is a subgroup of G. If s, t ∈ Ghave the same orbit, then Stab s and Stab t are conjugate subgroups of G.

Proof. Exercise 6.2

84 Chapter 6 Appendix

6.2 Fields and division rings

Definition 6.2.1. A (right) near-field is a set K equipped with two binary operations:“addition,” denoted (x, y) 7→ x+ y and “multiplication,” denoted (x, y) 7→ xy, satisfying:

1. (K,+) is an abelian group, called the additive group of K. Let 0 ∈ K denote itsidentity element.

2. K∗ := K \ {0} is a group under multiplication, called the multiplicative group ofK. The identity element for this group is denoted 1, or perhaps 1K if there is anychance of confusion.

3. Multiplication on the right distributes over addition:

(x+ y)z = xz + yz (6.1)

for all x, y, z ∈ K.

A right near field is called a division ring iff it satisfies:

3. Multiplication on the left distributes over addition:

x(y + z) = xy + xz (6.2)

for all x, y, z ∈ K.

A field is a division ring K whose multiplicative group K∗ is abelian. An isomorphismf : K→ K′ of near-fields, division rings, or fields is a bijective function with f(0) = 0,f(1) = 1, and satisfying f(x+ y) = f(x) + f(y) and f(xy) = f(x)f(y) for all x, y ∈ K.

For example, the rational numbers Q, the real numbers R, and the complex numbersC are fields with the usual notions of addition and multiplication. For each prime numberp, the set

Fp := Z/pZ

of integers mod p is a field, with addition and multiplication given by the usual additionand multiplication of integers, modulo p.

If K is a field, a subset K′ ⊆ K of K is called a subfield of K iff K′ is closed underaddition and multiplication in K and these operations make K′ into a field. It is clearthat any intersection of subfields is again a subfield, hence any subset S of a field Kis contained in a smallest subfield of K (construct it by intersecting all subfields of Kcontaining S). If K′ is a subfield of K, then K, with its additive abelian group structure,becomes a vector space over K′ by defining scalar multiplication as multiplication in K.

Definition 6.2.2. A number field is a subfield of C (any such subfield contains Q) whichis finite dimensional as a Q vector space.

6.2 Fields and division rings 85

Example 6.2.3. The smallest subfield Q(√

2) of C containing√

2 is the set of all realnumbers r that can be written in the form r = a+ b

√2 for some a, b ∈ Q. Since

√2 /∈ Q,

we have a + b√

2 = a′ + b′√

2 for a, a′, b, b′ ∈ Q iff a = a′ and b = b′. Hence 1,√

2 is abasis for Q(

√2) as a Q vector space, so Q(

√2) is a number field.

Definition 6.2.4. The characteristic of a field K is defined to be the smallest positiveinteger p such that p = 1 + · · · 1 (p times) is equal to zero in K; if there is no such positiveinteger, the characteristic of K is defined to be zero.

From the fact that K∗ = K \ {0} is a group, it follows that the characteristic of K isprime (when non-zero). Any field of characteristic zero contains a subfield isomorphic toQ. Any field of characteristic p contains a subfield isomorphic to Fp. If K is a field ofcharacteristic p and q = pn is a prime power, then f(x) := xq defines an automorphismf : K→ K called the Frobenius automorphism of order q.

Here is a summary of some basic notions and results from the theory of fields. For allof the results on finite fields presented here, I recommend [Her, §7.1] as a reference.

Definition 6.2.5. A field K is called algebraically closed iff any polynomial f(x) ∈ K[x]of positive degree can be factored as

f(x) = Cd∏i=1

(x− αi)

for some C,α1, . . . , αd ∈ K.

Theorem 6.2.6. Every field is a subfield of an algebraically closed field.

Proof. This is perhaps the most fundamental result in the theory of fields. This is provedin almost every abstract algebra textbook, though I can’t find it explicitly stated in [Her]as Herstein likes to stick to finite field extensions at all times (it does follow the existenceand “uniqueness” of splitting fields for polynomials—Theorems 5.H and 5.J in [Her]).One standard reference is [DF, Proposition 30, §13.4].

Theorem 6.2.7. The number of element q := #K in any finite field K is a prime power:q = pn. For each prime power q = pn there is a unique (up to isomorphism) finite fieldFq with q elements. The q elements of Fq are the roots of the polynomial xq − x ∈ Fq[x],which has coefficients in Fp ⊆ Fq. If r = pm for some m ≥ n, then Fq is a subfield of Fr,equal to the fixed locus of the Frobenius automorphism of Fr of order q.

Proof. This is a summary of the first couple of results in [Her, §7.1].

Theorem 6.2.8. (Wedderburn) Any finite division ring is a field.

Proof. See, for example, [Her, Theorem 7.C] (where there are two proofs) or [DF,Exercise 13, §13.6]. Wedderburn’s original proof (1905) had a slight gap; the firstcomplete proof is usually attributed to Leonard Dickson. The most common “modernproof” that one encounters is due to Ernst Witt.

Theorem 6.2.9. The multiplicative group K∗ of any finite field K is cyclic.

86 Chapter 6 Appendix

Proof. See, for example, [Her, Theorem 7.B].

Example 6.2.10. The Dickson near-field J9 is a near-field defined as follows. As anadditive group, (J9,+) is “the” field F9 of order 9: (J9,+) := (F9,+). The multiplication∗ for J9 is defined in terms of the multiplication on F9 (which we denote by juxtaposition,as usual) by the rule

a ∗ b :=

{ab, b ∈ Sa3b, b /∈ S,

where S := {a2 : a ∈ F9} is the set of squares in F9.

For more on near-fields and related algebraic structures, see the survey article [Wei]by Charles Weibel.

6.3 The Implicit Function Theorem

Here is a version of the Implicit Function theorem which is not the most general statementpossible, but which is sufficient for our purposes:

Theorem 6.3.1. (Implicit Function Theorem) Let U be an open subset of Rn+1,f : U → R a continuously differentiable function (for example, a polynomial function),P = (a1, . . . , an, b) a point of U . Assume that the partial derivative of f with respectto xn+1 is non-zero at P . Then there are open subsets V ⊆ Rn and W ⊆ R containing(a1, . . . , an) and f(P ), respectively, and a continuously differentiable function g : V →Wsuch that the graph of g,

Γg = {(x1, . . . , xn, g(x1, . . . , xn)) : (x1, . . . , xn) ∈ V },

is equal to the level set

(f |V )−1(f(P )) = {(x1, . . . , xn+1) ∈ V ×W : f(x1, . . . , xn+1) = f(P )}

of f on V . Furthermore, if f is k times continuously differentiable on V ×W , then so isg.

The role played by the “last” coordinate in the Implicit Function Theorem is nothingspecial—this coordinate is singled out just to make the statement of the theorem as cleanas possible. Usually one applies the theorem to a differentiable function f : U → R on anopen subset U ⊆ Rn+1 for which it is known that fi(P ) 6= 0 for some partial derivativefi of f . One concludes that there is an open neighborhood V of P in Rn+1 such thatV ∩ f−1(f(P )) is mapped bijectively to an open subset W ⊆ Rn via the projection

πi : Rn+1 → Rn

πi(x1, . . . , xn+1) = (x1, . . . , xi−1, xi+1, . . . , xn+1)

and, furthermore, the inverse of πi : V ∩ f−1(f(P )) → W is given by a functiong : W → V ∩ f−1(f(P )) ⊆ Rn+1 which is “at least as differentiable as f .”

There is also an analytic version of the implicit function theorem:

6.4 Exercises 87

Theorem 6.3.2. Let U be an open subset of Cn+1, f : U → C an analytic functionon U (a function equal to a convergent series Taylor series on a neighborhood of eachpoint of U—for example, a function given by a polynomial with complex coefficients, or aratio of two such polynomials whose denominator is non-zero on U), P = (a1, . . . , an, b)a point of U . Assume that the partial derivative fzn+1(P ) is non-zero. Then there areopen subsets V ⊆ Cn and W ⊆ C containing (a1, . . . , an) and f(P ), respectively, and ananalytic function g : V →W such that the graph of g,

Γg = {(z1, . . . , zn, g(z1, . . . , zn)) : (z1, . . . , zn) ∈ V },

is equal to the level set

(f |V )−1(f(P )) = {(z1, . . . , zn+1) ∈ V ×W : f(z1, . . . , zn+1) = f(P )}

of f on V .

It is worth emphasizing that in both of these “implicit function theorems,” the inversefunction g will rarely be a polynomial function (or even a rational function) even whenthe “input” function f is a polynomial.

Example 6.3.3. The function f(x, y) := y2 − x3 is a polynomial (hence infinitelydifferentiable) function f : R2 → R such that fx = 3x2 is non-zero at P = (1, 1) ∈ f−1(0).Consider the open neighborhood V := R2

>0 of P in R2 and the open subset W := R>0

of R. Then the projection π(x, y) := x defines a bijection π : V ∩ f−1(0) → W withinverse g(x) := (x,

√x3). Although g : W → R2 is a perfectly nice infinitely differentiable

function, it is not given by a ratio of polynomials on any neighborhood of π(P ) = 1.

6.4 Exercises

Exercise 6.1. Let G be a group. Show that “being conjugate” (Definition 6.1.4) is anequivalence relation on the set of subgroups of G.

Exercise 6.2. Prove Proposition 6.1.6.

Exercise 6.3. Check that the Dickson near field J9 defined in Example 6.2.10 is actuallya near-field (Definition 6.2.1). Show that J9 does not satisfy the left distributive law(6.2), so that it is not a division ring.

Exercise 6.4. Show, by writing down an explicit isomorphism, that the multiplicativegroup J∗9 of the Dickson near field J9 (Example 6.2.10) this group is isomorphic to thequaternionic group

Q := {±1,±i,±j,±k}.

(Note that, up to isomorphism, there are exactly two non-abelian groups of order 8:Q and the dihedral group D4—you could probably show that J∗9 is isomorphic to Qvery easily by appealing to this classification.) Also note that the fact that J∗9 is notabelian, together with Wedderburn’s Theorem (Theorem 6.2.8), gives another proof thatJ9 cannot be a division ring.

Bibliography

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[C3] H. S. M. Coxeter, The Real Projective Plane. Springer-Verlag, 1993.

[DF] D. Dummir and R. Foote, Abstract Algebra (Third Edition). John Wiley andSons, Inc., 2004.

[H] R. Hartshorne, Algebraic Geometry. Springer-Verlag, 1977.

[H2] R. Hartshorne, Foundations of Projective Geometry. Harvard University lecturenotes, 1967.

[Her] I. N. Herstein, Topics in Algebra. Ginn and Company, 1964.

[vY] J. van Yzeren, A simple proof of Pascal’s hexagon theorem. Amer. Math. Monthly100, No. 10 (1993), 930-931.

[R] B. A. Rosenfeld, A History of Non-Euclidean Geometry. Springer-Verlag, 1988.

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