A�ne Algebraic Varieties

Undergraduate Seminars: Toric Varieties

Lena JiFebruary 3, 2016

Contents

1. Algebraic Sets 12. The Zariski Topology 33. Morphisms of A�ne Algebraic Sets 54. Dimension 6References 6

1. Algebraic Sets

Let k be an algebraically closed field, for instance the complex numbers C. Wedefine a�ne n-space over k to be the set of n-tuples of elements in k. That is,

An = {(a1, . . . , an) | ai 2 k}.We denote by k[x1, . . . , xn] the polynomial ring in n variables with coe�cients in

k. Polynomials f 2 k[x1, . . . , xn] can be viewed as maps from An to A by evaluatingf at each point, and so we can consider the set of zeroes of a polynomial. Moregenerally, for a collection of polynomials {fi}i2I , we define their zero set to be

V ({fi}i2I) = {(a1, . . . , an) 2 An | fi(a1, . . . , an) = 0 8i 2 I}.

Definition 1.1. A subset of An of the form V ({fi}i2I) is called an a�ne algebraic

set.

Note that these are referred to as a�ne algebraic varieties in Smith, et al.However, we will follow Fulton and adopt the following definition.

Definition 1.2. An algebraic set V ✓ An is irreducible if, for any expressionV = V1 [ V2 where Vi are algebraic sets in An, V1 = V or V2 = V .

Definition 1.3. An a�ne algebraic variety is an irreducible a�ne algebraic set.

Example 1.4.

(1) V (xy) ✓ C2 is an a�ne algebraic set, but it is not irreducible. (Figure 1)(2) V (y2 � x

2 � x

3) ✓ C2 is an a�ne algebraic variety. (Figure 2)

Here are many more examples of a�ne algebraic sets. Due to artistic limitations,pictures are over R.

Example 1.5. A point (a1, . . . , an) 2 An is an a�ne algebraic variety becauseV (x1 � a1, . . . , xn � an) = {(a1, . . . , an)}.

1

2 /(, ,} f }, , )\

Figure 1. V (xy) Figure 2. V (y2 � x

2 � x

3)

Figure 3. V (x2 + y

2 � z

2) Figure 4. V (y2 � x(x2 � 1))

Example 1.6. A hypersurface in An is the zero set of a single nonconstant poly-nomial, for example the quadratic cone V (x2 + y

2 � z

2) ✓ C2. A hypersurface inA2 is called an a�ne plane curve. The a�ne variety given in Figure 2 is an a�neplane curve, as is the elliptic curve V (y2 � x(x2 � 1)).

Example 1.7. This very nice heart is a hypersurface given by the solutions of theequation (x2 + 9

4y2 + z

2 � 1)3 � x

2z

3 � 980y

2z

3 = 0.

Example 1.8. The Whitney umbrella is defined by the equation x

2 � y

2z = 0.

A�ne Algebraic Varieties 3

Example 1.9. The torus with major radius R and minor radius r is defined bythe equation (x2 + y

2 + z

2 +R

2 � r

2)2 = R

2(x2 + y

2).

Example 1.10. The special linear group SL(n,C) = {A 2 Mn(C) | det(A) = 1}is a hypersurface in Mn(C) ⇠= Cn2

. This follows from the fact that the determinantis a polynomial in n

2 variables; for example when n = 3, then

det

0

@a b c

d e f

g h i

1

A = aei+ bfg + cdh� ceg � bdi� afh.

Example 1.11. The unit sphere S

n�1 ✓ Cn is an a�ne algebraic variety definedby the equation x

21 + . . . + x

2n = 1. However, the unit open ball (in the Euclidean

topology) and defined as the set {(a1, . . . , an) 2 Cn | a21 + . . .+ a

2n < 1} is not; if a

polynomial vanishes on an open subset of Cn in the Euclidean topology, then it isuniformly 0.

2. The Zariski Topology

Recall that a collection T of subsets of a space X defines a topology on X if

(1) X and ; are in T ;(2) the union of any subcollection of elements of T is contained in T ;(3) the intersection of any finite subcollection of elements of T is in T .

We would like to use a�ne algebraic sets to define the closed sets of a topologyon An, and so we must check that

(1F) An and ; are a�ne algebraic sets;

(2F) the arbitrary intersection of a�ne algebraic sets is itself an a�ne algebraicset;

(3F) the finite union of a�ne algebraic sets is itself an a�ne algebraic set.

Let’s verify these! If a 2 A is nonzero, then the polynomial equation a = 0 hasno solutions, and so V (a) = ;. However, the equation 0 = 0 is satisfied by every

point in An, and so V (0) = An. So condition (1F) is satisfied.

For (2F), let {V↵}↵2A be a collection of a�ne algebraic sets, where each V↵ =

V ({fi↵}i↵2I↵). Then the intersectionT

↵2A V↵ is the common zero set of {fi↵}i↵2I↵

over all ↵ 2 A, i.e.\

↵2A

V↵ = V

[

↵2A

{fi↵}i↵2I↵

!.

The twisted cubic curve in Figure 5 illustrates this, as it is given by the intersectionof two surfaces: V (x2 � y) \ V (x3 � z) = V (x2 � y, x

3 � z).

So it remains to show (3F). By induction, it is enough to check the union of

two a�ne algebraic sets.

4 /(, ,} f }, , )\

Figure 5. An intersection Figure 6. A union

Proposition 2.1 ([3] Exercise 1.2.1). The union of two a�ne algebraic sets in An

is an a�ne algebraic set.

Proof. Let V ({fi}i2I) and V ({gj}j2J) be a�ne algebraic sets. We claim that

V ({fi}i2I) [ V ({gj}j2J) = V ({figj}(i,j)2I⇥J).

Certainly ✓ holds, since if p = (a1, . . . , an) 2 V ({fi}i2I)[V ({gj}j2J), then fi(p) =0 for all i or gj(p) = 0 for all j. In either case, then figj(p) = 0 for all i and all j,and so p 2 V ({figj}(i,j)2I⇥J).

Now let q 2 V ({figj}(i,j)2I⇥J) and suppose that q 62 V ({fi}i2I) [ V ({gj}j2J).Then there exist i and j with fi(q) 6= 0 and gj(q) 6= 0. But this implies figj(q) 6= 0,a contradiction, so q must be in V ({fi}i2I)[V ({gj}j2J) and we have shown ◆. ⇤

An example of this is the union of the x-axis and yz-plane in Figure 6: V (x) [V (y, z) = V (xy, xz).

Definition 2.2. The topology on An where the closed sets are of the form V ({fi}i2I)is called the Zariski topology.

The Zariski topology on An is very di↵erent from the Euclidean topology. Opensubsets in this topology are very big; in fact they are dense and quasi-compact.Additionally, two non-empty open sets will always intersect, and so the Zariskitopology is not Hausdor↵ on An for n > 0.

Example 2.3. Let k = C. Then the Zariski topology on A1 is the cofinite topologyon C — closed sets are ;, C, and finite sets — since polynomials in one variablehave finitely many roots.

Example 2.4 ([3] Exercise 1.2.2). The Zariski topology on A2 is not the producttopology on A1 ⇥ A1. Recall that the product topology on X ⇥X is generated byopen sets of the form U1⇥U2, where U1, U2 are open subsets of X. So if A1⇥A1 isequipped with the product topology, where each A1 has the Zariski topology, opensets are of the form A2 � {finitely many horizontal lines [ vertical lines [ points}

The diagonal of A1 ⇥ A1, defined �A1⇥A1 = {(a1, a2) 2 A1 ⇥ A1 | a1 = a2},is not closed in the product topology, where each A1 is endowed with the Zariskitopology, since A1 is not Hausdor↵. However, it is the zero set of the polynomialx�y 2 A[x, y], so �A1⇥A1 = V (x�y) ✓ A2 is closed in the Zariski topology on A2.

A�ne Algebraic Varieties 5

If V ✓ An is an a�ne algebraic set, then we can endow V with the subspacetopology induced by the Zariski topology on An. Then closed subsets of V are ofthe form V \W , where W ✓ An is an a�ne algebraic set.

3. Morphisms of Affine Algebraic Sets

Definition 3.1. Let V ✓ An and W ✓ Am be a�ne algebraic varieties. A mor-

phism of algebraic varieties is a map F : V �! W given by the restriction ofa “polynomial map” An �! Am (meaning that each of the m components is givenby a polynomial in k[x1, . . . , xn]).

So when we compose F with the inclusion i : W ,! Am, the resulting mapwill be of the form i � F = (F1, . . . , Fm) where each Fi is the restriction to V of a(non-unique) polynomial in k[x1, . . . , xn].

Definition 3.2. A morphism F : V �! W is an isomorphism if it has an inversemorphism. In this case we say that V and W are isomorphic.

Example 3.3. Let C be the plane parabola given by the equation y � x

2 = 0.Then C is isomorphic to A1 via the maps

'|C , where ' : A2 �! A1 A1 �! C

(x, y) 7! x t 7! (t, t2).

Example 3.4 ([3] Exercise 1.3.2). The twisted cubic V = V (x2 � y, x

3 � z) inFigure 5 is isomorphic to A1. Since

V = {(t, t2, t3) 2 A3 | t 2 A},

we can define a morphism A1 �! V by t 7! (t, t2, t3). The restriction to V of theprojection A3 ! A1 onto the first factor, defined (x, y, z) 7! x, gives an inversemorphism.

Proposition 3.5 ([3] Exercise 1.3.1). If F : V �! W is a morphism of a�ne

algebraic sets, then F is continuous in the Zariski topology.

Proof. Any closed subset ofW is of the form V ({fi}i2I)\W where fi 2 k[x1, . . . , xm].

F

�1(V ({fi}i2I) \W ) = F

�1(V ({fi}i2I)) \ F

�1(W ) = V ({fi � F}i2I) \ V

is a closed subset of V , where F : An �! Am is any “polynomial map” that restrictsto F , so F is continuous. ⇤

6 /(, ,} f }, , )\

4. Dimension

Definition 4.1. The dimension of an a�ne algebraic set V is the length of thelongest chain of distinct nonempty a�ne closed subvarieties of V

sup{d | Vd ) Vd�1 ) · · · ) V0}.

Hence the dimension of an a�ne algebraic set is equal to the maximum of thedimensions of its irreducible components (maximal irreducible subsets).

Example 4.2. The quadratic cone V (x2 + y

2 � z

2) has dimension 2.

Example 4.3. The a�ne algebraic set V (xy, xz) has dimension 2 = max{1, 2}.

Definition 4.4. An a�ne algebraic set is equidimensional if all of its irreduciblecomponents have the same dimension.

So in the earlier examples, V (xy, xz) is not equdimensional but V (x2+ y

2� z

2)and V (xy) (Figure 1) are.

Definition 4.5. The codimension of an a�ne algebraic set V ✓ An is definedcodimV = n� dimV .

Example 4.6. Hypersurfaces in An are precisely the a�ne algebraic sets of codi-mension 1.

References

[1] William Fulton. Algebraic Curves, 2008.

[2] Herwig Hauser. Algebraic Surfaces Gallery. http://homepage.univie.ac.at/herwig.hauser/

bildergalerie/gallery.html.[3] Karen Smith, Lauri Kahanpaa, Pekka Kekalainen, and William Traves. An Invitation to Al-

gebraic Geometry. Springer, 2004.