NOTES ON FINITE FIELDS

AARON LANDESMAN

CONTENTS

1. Introduction to finite fields 22. Definition and constructions of fields 32.1. The definition of a field 32.2. Constructing field extensions by adjoining elements 43. A quick intro to field theory 73.1. Maps of fields 73.2. Characteristic of a field 83.3. Showing the characteristic of any finite field is a prime 84. Algebraic closures 105. Characterization of finite fields 126. Properties of finite fields 146.1. The multiplicative group of a finite field 146.2. Frobenius 156.3. Containments of finite fields 16Appendix A. Existence of algebraic closures 18Appendix B. Basics of rings 20B.1. Quotients 21References 21

1

2 AARON LANDESMAN

1. INTRODUCTION TO FINITE FIELDS

In this course, we’ll discuss the theory of finite fields. Along theway, we’ll learn a bit about field theory more generally. So, the nat-ural place to start is: what is a field? Many fields appear in nature,such as the real numbers, the complex numbers the rational num-bers, and even finite fields! Before giving a formal definition, let’ssee some examples.

Example 1.1. The rational numbers Q ={ a

b : a, b ∈ Z, b 6= 0}

area field. The key properties are that we can multiply rational num-bers, add rational numbers (via addition of fractions) and furtherthat nonzero rational numbers have inverses. That is, a

b ·ba = 1 when-

ever a 6= 0.

Now, let’s see some examples of finite fields.

Example 1.2. Consider the field F2, the finite field with two ele-ments. Call these elements 0, 1. The addition law is given by 0+ a =a + 0 = a and 1 + 1 = 0. The multiplication law is given by 1 · a = aand 0 · a = 0. 1 is invertible and its inverse is given by 1 since1 · 1 = 1. This can succinctly be described by Z/2Z.

Example 1.3. Next, let’s consider the finite field with 3 elements. Asabove, we can consider Z/3Z. Elements can be added and multi-plied by reducing addition and multiplication in Z modulo 3. Thekey property to check is that nonzero elements have inverses (mean-ing that for any nonzero a there is some b with ab = 1). Indeed,1 · 1 = 1 and 2 · 2 = 1.

Warning 1.4. So far, we have seen that Z/2Z and Z/3Z are fields.However, Z/4Z is not a field! The way to see this is that there is noelement a ∈ Z/4Z with 2a = 1. Indeed, either 2a = 2 or 2a = 0. So,Z/nZ is not in general a field.

Question 1.5. Do you think there exists a finite field of order 4? Doyou think there exists a finite field of order 5? Do you think thereexists a finite field of order 6? For which n ∈ Z does there exist afinite field with n elements?

NOTES ON FINITE FIELDS 3

2. DEFINITION AND CONSTRUCTIONS OF FIELDS

Before understanding finite fields, we first need to understandwhat a field is in general. To this end, we first define fields. Afterdefining fields, if we have one field K, we give a way to constructmany fields from K by adjoining elements.

2.1. The definition of a field. A field is a special type of ring. So, wefirst define a ring:

Definition 2.1. A commutative ring with unit is a set R togetherwith two operations (+, ·) satisfying the following properties:

(1) Associativity: a + (b + c) = (a + b) + c, a · (b · c) = (a · b) · c(2) Commutativity: a + b = b + a, a · b = b · a(3) Additive identity: there exists 0 ∈ R so that a + 0 = a(4) Multiplicative identity: there exists 1 6= 0 ∈ R so that 1 · a = a(5) Additive inverses: For every a ∈ R, there is a additive in-

verse, denoted −a satisfying a + (−a) = 0(6) Distributivity of multiplication over addition: a · (b + c) =

(a · b) + (a · c) .

Remark 2.2. Any mention of “ring” in what follows implicitly means“commutative ring with unit.” There will be no noncommutativerings or rings without units.

Definition 2.3. A field is a ring K such that every nonzero elementhas a multiplicative inverse. That is, for each a ∈ K with a 6= 0, thereis some a−1 ∈ K so that a · a−1 = 1.

Definition 2.4. A finite field is simply a field whose underlying setis finite.

Example 2.5. Given any prime number p, the set Z/pZ forms a fieldunder addition and multiplication. This field is denoted Fp. Nearlyall the axioms are immediate, except possibly for the existence ofmultiplicative inverses.

Exercise 2.6. Verify that every nonzero element has a multiplicativeinverse in two ways:

(1) Use the Euclidean algorithm to show that for any a < p thereexists some b with ab ≡ 1 mod p and conclude that b is aninverse for a. Hint: Use that gcd(a, p) = 1.

(2) Show that ap−1 = 1, so ap−2 is an inverse for a. This is alsoknown as “Fermat’s Little theorem,” not to be confused with“Fermat’s Last theorem,” which is much more difficult. Hint:

4 AARON LANDESMAN

Show that the powers of any element form a subgroup of(Z/pZ)× := Z/pZ − {0} under multiplication. Use La-grange’s theorem (i.e., the order of a subgroup divides theorder of the ambient group) to deduce that this subgroup gen-erated by a has order dividing #(Z/pZ)× = p− 1. Concludethat am = 1 for some m dividing p− 1 and hence ap−1 = 1.

2.2. Constructing field extensions by adjoining elements. We nowexplain how to construct extensions of fields by adjoining elements.Here is a prototypical example:

Example 2.7. Consider the field Q(√

2)

. How should we interpretthis? The elements of this field are of the form

Q(√

2)={

a + b√

2 : a, b ∈ Q}

.

Multiplication works by(a + b

√2) (

c + d√

2)= (ac + 2bd) + (ad + bc)

√2.

Here is another perspective on this field: What is√

2? It is simply aroot of the polynomial x2 − 2. Therefore, we could instead considerthe field

Q[x]/(x2 − 2),

where this means the ring where we adjoin a root of the polynomialx2 − 2. Concretely, Q[x] means polynomials with coefficients in Q,and the notation Q[x]/(x2 − 2) means that in any polynomial f (x),we can replace x2 by 2. So for example, if we had the polynomial x3 +2x2 + 3 this would be considered equivalent to (x2) · x + 2 · (x2) +3 = 2x + 4 + 3 = 2x + 7. In this way, we can replace any polynomialwith a polynomial of degree 1 of the form a + bx. Identifying x with√

2 gives the isomorphism of this ring with the above field Q(√

2)

.

Exercise 2.8. Describe the elements of the fields K as in Example 2.7for K one of the following fields

(1) K = Q(√

3)

,

(2) K = Q(71/5),

(3) K = Q (ζ3), for ζ3 a primitive cube root of unity.In each of the above cases, write K = Q[x]/ f (x) for an appropriatepolynomial f . In each of the above cases, what is the dimension of Kover Q, when K is viewed as a Q vector space?

NOTES ON FINITE FIELDS 5

Definition 2.9. Let K be a field. Define the polynomial ring

K[x] :=

{n

∑i=1

aixi : ai ∈ K

}.

For f ∈ K[x], define

K[x]/( f ) := K[x]/ ∼

where ∼ is the equivalence relation defined by g ∼ h if f | g− h.

Exercise 2.10. Show that K[x]/(x) ' K, where the map is given bysending a polynomial to its constant coefficient.

Lemma 2.11. Let K be a field and let f ∈ K[x] be a monic irreduciblepolynomial. Then K[x]/( f ) is a field.

Proof. Note that K[x]/( f ) is a ring as it inherits multiplication andaddition and all the resulting properties of a ring from K[x]. (Checkthis!) Therefore, it suffices to check that if f is monic and irreducible,then every element has an inverse. In other words, given any g ∈K[x]/( f ), we need to show there is some h with gh = 1. We canconsider g ∈ K[x] as a polynomial of degree less than f . Since fis irreducible, and deg g < deg f , it follows that the two polynomi-als share no common factors. Then, by the Euclidean algorithm forpolynomials (if you have only seen the euclidean algorithm over theintegers, check that the natural analog to the Euclidean algorithm forthe integers works equally well in polynomial rings over arbitraryfields, where the remainder is then a polynomial of degree less thanthe polynomial you are dividing by) we obtain some h, k ∈ K[x] withgh + f k = 1 as elements of K[x]. It follows that gh ∼ 1 in K[x]/( f )because gh− 1 = f k in K[x]. �

Exercise 2.12. Let K be a field and f ∈ K[x] a monic irreducible poly-nomial. Suppose L = K[x]/( f ). Show that dimK L = deg f , wheredeg f denotes the degree of the polynomial f and dimK L denotesthe dimension of L as a K vector space.

Example 2.13. Consider the field F2[x]/(x2 + x + 1). We claim thisis a finite field of order 4. Indeed, this holds because the polynomialx2 + x + 1 is irreducible. To check this, we only need to check it hasno linear factors. It has a linear factor if and only if x2 + x + 1 has aroot in F2. But, when we evaluate it at 0 we get 1 mod 2 and whenwe evaluate it at 1, we get 1 mod 2. So it has no roots, and the claimfollows from Lemma 2.11.

6 AARON LANDESMAN

Exercise 2.14. For any p > 2, show that there are exactly p+12 ele-

ments x ∈ Fp with x = y2 for some y ∈ Fp. We call such x squares.Conclude that there is some x ∈ Fp which is not a square wheneverp > 2. Hint: Show that if x = y2 then we also have x = (−y)2 andfurther that there y and −y are the only two elements of Fp squaringto x.

Example 2.15. Let p > 2 be a prime and let ε ∈ Fp be an elementwhich is not a square (which exists by Exercise 2.14). Then,

Fp[x]/(x2 − ε)

is a finite field of order p2. It is order p2 because it is a two dimen-sional vector space over Fp spanned by the basis 1 and x. It is a fieldbecause x2 − ε is irreducible in Fp[x]. Indeed, to see this, note thatif it were not irreducible, it would factor as a product of two linearfactors, which means it would have a root. But, if it had a root, therewould be some y ∈ Fp so that y2 = ε. However, we chose ε not to bea square, and so no root exists.

NOTES ON FINITE FIELDS 7

3. A QUICK INTRO TO FIELD THEORY

In order to classify finite fields, we’ll need some inputs from fieldtheory. In particular, we’ll need to understand maps of fields and thecharacteristic of a field, which we discuss in this section.

3.1. Maps of fields.

Definition 3.1. Given two fields K and L, a map φ : K → L is a mapof sets sending 1 7→ 1, 0 7→ 0 such that φ(a +K b) = φ(a) +L φ(b) andφ(a ·K b) = φ(a) ·L φ(b).

Remark 3.2. Sometimes, a map of fields is referred to as a homomor-phism or extension. Whenever we have a map of fields, it is requiredto be compatible with the addition and multiplication operations, asdefined above. If we do not wish to require such compatibility, wewill call the map “a map of sets”

Remark 3.3. We shall typically drop the subscript +K, ·K on additionand multiplication when it is clear from context.

Exercise 3.4. Verify from the definition of map that

φ(a−1) = φ(a)−1

and

φ(−a) = −φ(a).

We next prove that maps of fields are injective. If you have notworked much with the notion of injectivity before, you may want totry the following exercise first.

Exercise 3.5. Show that a map of rings is injective (using the defini-tion that f : R → S is injective if f (a) = f (b) implies a = b) if andonly if the only element mapping to 0 is 0. Hint: Consider f (a− b).

Lemma 3.6. Any map of fields is injective.

Proof. By Exercise 3.5 it suffices to show that any c 6= 0 does notsatisfy φ(c) = 0. Suppose there is some such c. But note that 1 =φ(1) = φ(cc−1) = φ(c)φ(c−1) = 0φ(c−1) = 0, a contradiction.Therefore, every nonzero element does not map to 0 and the mapis injective. �

Remark 3.7. Because of Lemma 3.6, a map of fields is also typicallycalled an extension of fields or a field extension.

8 AARON LANDESMAN

Remark 3.8. The property that maps of fields are injective is veryspecial to fields. Indeed, it is not true for groups. For example, themap Z→ {1} is not injective!

Remark 3.9. Using Lemma 3.6, whenever we have a map of fieldsφ : K → L, we can consider L as a vector space over K. The mapK× L→ L corresponding to scalar multiplication is given by

K× L→ L

(a, b) 7→ φ(a) · b

3.2. Characteristic of a field.

Definition 3.10. Let K be a field. If there is some n so that

(3.1) n := 1 + 1 + · · ·+ 1︸ ︷︷ ︸n

is equal to 0 in K, the the minimal such n is defined to be the char-acteristic of K, denoted char(K). If no such n ∈ Z≥0 exists, then wesay K has characteristic 0.

Example 3.11. The rational numbers Q has characteristic 0, but thefield Fp has characteristic p.

Exercise 3.12 (Important exercise). Let p be a prime number and sup-pose K is a field of characteristic p. Show that for any x, y ∈ K, wehave

(x + y)p = xp + yp.

Hint: Expand the left hand side using binomial coefficients, andshow that p divides nearly all of the binomial coefficients.

3.3. Showing the characteristic of any finite field is a prime.

Lemma 3.13. The characteristic of any field is either 0 or prime.

Proof. Note that the characteristic cannot be 1 because 1 6= 0. So, wehave to show that the characteristic is never composite.

Let n be a composite number with n = f g for f , g > 1 two factorsof n.

Exercise 3.14. Suppose a, b ∈ K with ab = 0. Then show either a = 0or b = 0.

By the above exercise, if n = f g = 0, then either f = 0 or g = 0.Say f = 0. But then, we obtain that f < n, and so K does not havecharacteristic n. �

NOTES ON FINITE FIELDS 9

Definition 3.15. For K a field, we say a subset K′ ⊂ K is a subfieldif it is a field and the inclusion K′ ⊂ K is a map of fields (meaning1 7→ 1, 0 7→ 0 and the multiplication and addition are compatible).

Exercise 3.16. Verify similarly that any field of characteristic 0 con-tains Q as a subfield. Hint: Define a map of fields

φ : Q→ Kab7→ ab−1.

Use that b ∈ K is nonzero by the assumption that K has characteristic0 to show this is well defined.

Lemma 3.17. The characteristic of any finite field is prime (and, in partic-ular, never 0).

Proof. By Lemma 3.13, we only need to show the characteristic ofa finite field is nonzero. So, it suffices to show every characteristic0 field is infinite. But, by Exercise 3.16, every characteristic 0 fieldcontains Q as a subfield, and is therefore infinite. �

Lemma 3.18. Any field K of characteristic p > 0 (for p a prime) containsFp as a subfield.

Proof. Inside K, consider the subset {0, 1, 2, . . . , p− 1}. These form pdistinct elements because char K = p. By definition, of n = 1 + 1 + · · ·+ 1︸ ︷︷ ︸

n

,

the elements 0, 1, . . . , p− 1 satisfy the same addition and multiplica-tion rules as Fp ' (Z/pZ). Therefore, when we restrict the multi-plication and addition from K to {0, 1, 2, . . . , p− 1}, we realize Fp asthis subfield. �

Lemma 3.19. Any finite field K has order pn for p a prime and n ∈ Z.

Proof. By Lemma 3.18, K contains Fp as a subfield, so we have aninclusion Fp → K. Observe that this makes K into a vector spaceover Fp. Because K is finite, it must even be a finite dimensionalvector space over Fp. Say it has dimension n. Then we see that as avector space, K ' Fn

p, so |K| = |Fp|n = pn, as desired. �

So, we have established that every finite field has order a primepower. Our next goal is to show that there is a unique finite fieldof order pn. However, for this, we will need to introduce algebraicclosures, which we do now.

10 AARON LANDESMAN

4. ALGEBRAIC CLOSURES

To hit the ground running on finite fields, we’ll need to knowabout “algebraic closures.”

Definition 4.1. An extension of fields φ : K → L is finite if φ makes Linto a finite dimensional vector space over K. An extension of fieldsφ : K → L is algebraic if for every a ∈ L, there is a finite extensionK → La with La ⊂ L a subfield containing a.

In order to get the theory of finite fields off the ground, we willneed the existence of an algebraic closure. It is not too difficult toshow this exists, but to jump to the interesting stuff, we will defer itfor later:

Definition 4.2. A field K is algebraically closed if any finite fieldextension K → L is an isomorphism.

Exercise 4.3. Show that the real numbers are not algebraically closed.Show that the rational numbers are not algebraically closed.

Lemma 4.4. Let K be a field. The following are equivalent.(1) K is algebraically closed.(2) Every monic irreducible polynomial over K has a root.(3) Every monic irreducible polynomial over K factors as a product of

linear polyomials.

Proof. For (1) =⇒ (2), we suppose K is algebraically closed andshow every monic irreducible polynomial over K has a root. Let f beany monic irreducible polynomial over K. Then, K[x]/( f ) is a fieldextension of K. Because K is algebraically closed, the natural mapK → K[x]/( f ) is an isomorphism. Therefore, dimK K[x]/( f ) = 1and so f has degree 1 by Exercise 2.12 (which says deg f = dimK K[x]/( f )),and hence has a root.

Next, if (2) holds, one can prove (3) by induction on the degree ofthe polynomial.

Finally, for (3) =⇒ (1), suppose K is not algebraically closed.We want to show there is some irreducible polynomial over K whichdoes not factor completely. Let L be a finite extension of K with theinclusion K → L not an isomorphism. Since K → L is an injection itis not a surjection, so we may take some y ∈ L \ K. We claim there issome monic irreducible polynomial f ∈ K[x] with f (y) = 0. Indeed,this is the content of the following exercise.

Exercise 4.5. Let K → L be an algebraic extension. Show that anyelement x ∈ L satisfies some monic irreducible polynomial f (x) =

NOTES ON FINITE FIELDS 11

xn + kn−1xn−1 + · · ·+ k0, for ki ∈ K. Hint: By definition of an alge-braic extension, show that the powers of x satisfy some linear depen-dence relation, and obtain the monic irreducible polynomial fromthis relation.

Note that since y /∈ K, the polynomial f with f (y) = 0 has degreemore than 1. Since f is irreducible and has degree more than 1, fdoes not have a root in K, as we wanted to show. �

Exercise 4.6. Show that the complex numbers are algebraically closed(you may assume that every polynomial over the complex numbershas a root).

Definition 4.7. A field extension K → K is an algebraic closure if(1) K → K is algebraic and(2) K is algebraically closed.

Exercise 4.8. Let K → L be an algebraic extension and let L denotean algebraic closure of L. Show that L is also an algebraic closure ofK.

Theorem 4.9 (Existence of algebraic closures). Let K be a field.(1) K has an algebraic closure.(2) Any two algebraic closures of K are isomorphic as field extensions

(meaning that for two algebraic closures K, K′, with K as a subfieldvia the maps φ : K → K, φ′ : K → K′, there is an isomorphismf : K → K′ so that f ◦ φ = φ′).

12 AARON LANDESMAN

5. CHARACTERIZATION OF FINITE FIELDS

Using the existence of an algebraic closure, we are now ready toshow there is a unique finite field of order pn, for every prime p andevery n ≥ 1.

First, we need a preparatory definition and lemma.

Definition 5.1. If K is a field and f := ∑ni=1 aixi ∈ K[x] is a polyno-

mial, we define the derivative of f , denoted f ′, to be ∑ni=1 iaixi−1.

Lemma 5.2. Let K be an algebraically closed field and let f ∈ K[x] be apolynomial. Then, if gcd( f , f ′) = 1, f has no repeated roots. That is, thereis no a ∈ K with (x− a)2 | f .

Proof. Suppose f has a repated root. Call that root r ∈ K. Then since(x− r)2 | f , it follows from the product rule that x− r | f ′. Therefore,x− r | gcd( f , f ′), and so gcd( f , f ′) 6= 1, as desired. �

We can now state and prove our main result.

Theorem 5.3. Let p be a prime and n ≥ 1.(1) There exists a finite field of order pn, notated Fpn . Further, Fpn is

realized as the set of elements of Fp satisfying xpn= x.

(2) Any two finite fields of order pn are isomorphic.

Proof. First let us show there exists a finite field of order pn. Let Fpdenote an algebraic closure of Fp. Define

Fpn :={

x ∈ Fp : xpn= x

}.

We claim Fpn is a field. To check this, the essential points to verifyare that Fpn is closed under multiplication, addition, and inversion.

(1) Addition: We need to show that if xpn= x, ypn

= y then(x + y)pn

= x + y. Indeed, this follows from Exercise 3.12.(2) Multiplication: We need to show that if xpn

= x, ypn= y then

(xy)pn= xy. Indeed, this is clear by commutativity of Fp.

(3) Inversion: Given x 6= 0 with xpn= x, we want to show x−1

(which exists as an element of Fp) satisfies (x−1)pn= x−1. But

indeed,

(x−1)pn= (xpn

)−1 = x−1,

as desired.

Exercise 5.4. Verify the remaining properties such as distributivityand commutativity to show that Fpn is indeed a field. Hint: You maybe able to inherit many of these properties from Fp.

NOTES ON FINITE FIELDS 13

To complete the proof of existence, we have to check that |Fpn | = pn.Indeed, by construction, the elements of Fpn are the set of roots to thepolynomial f (t) = tpn − t. So, it suffices to show this has pn distinctroots in Fpn . Note that f (t) has at most pn roots because it has degreepn. Since Fpn is algebraically closed, f (t) factors as a product of pn

distinct linear factors, by Lemma 4.4. Further, the roots of f (t) are alldistinct by Lemma 5.2 because f ′(t) = pn · tpn−1 − 1 = −1.

So, we have shown existence of finite field over order pn. It re-mains to show uniqueness up to isomorphism.

Let K be some finite field of size pn. We want to construct an iso-morphism K ' Fpn . Because K is finite, hence algebraic over Fp,it follows from Exercise 4.8 that an algebraic closure of K is also analgebraic closure of Fp, and we denote this algebraic closure by Fp.Choose an extension φ : K → Fp. We will show im φ ⊂ Fpn ⊂ Fp.This will complete the proof as it will imply that φ defines a mapbetween two fields of size pn. It is then injective by Lemma 3.6 andhence it will be surjective because it is an injective map between twosets of the same finite size.

So, we will now show im φ ⊂ Fpn . For this, it suffices to showthat any x ∈ K satisfies φ(x)pn

= φ(x). For this, it suffices to showxpn

= x. This clearly holds for x = 0, so after dividing by x, itsuffices to show xpn−1

= 1. But now, note that K× := K − {0} is amultiplicative group of size |K×| = |K| − 1 = pn − 1. Lagrange’stheorem tells us that the order of any element of a group divides theorder of the group. This implies that xm = 1 for some m | pn − 1,which implies that xpn−1 = 1. �

14 AARON LANDESMAN

6. PROPERTIES OF FINITE FIELDS

We next develop several interesting properties of finite fields.

6.1. The multiplicative group of a finite field. As we saw near theend of the proof of Theorem 5.3, because multiplicative inverses ex-ist, for any field K, the nonzero elements K× form a group undermultiplication. The identity element is 1. It turns out that finite fieldshave a particularly nice multiplicative structure.

Proposition 6.1. There is an isomorphism F×pn ' (Z/(pn − 1)Z). Thatis, F×pn is cyclic.

Proof. To show F×pn is cyclic of order pn − 1, since we know it hasorder pn − 1 as a group, it suffices to show there is some element oforder pn − 1.

Exercise 6.2 (Tricky exercise). Verify using that all finite abelian groupsare products of cyclic groups (the fundamental theorem for finiteabelian groups) that if there is no element of order pn − 1 then thereis some m < pn − 1 with xm = 1 for all x ∈ F×pn . Hint: Show that ifG ' ∏i Z/pni

i Z, (where the isomorphism holds by the fundamentaltheorem of finite abelian groups) has some pi = pj for i 6= j then thenevery element of G has order strictly less than |G| = ∏i pni

i . For thisit may help to consider the subgroup Z/pni

i ×Z/pnjj . Then, show

using the Chinese Remainder theorem that if pi 6= pj for any i 6= jthen G is cyclic.

However, we cannot have xm = 1 for all x ∈ F×pn with m < pn − 1because xm − 1 only has m < pn − 1 roots in Fp. Hence, there issome element of F×pn of order exactly pn − 1, and so it is isomorphicto Z/(pn − 1)Z×. �

Exercise 6.3. Using Proposition 6.1 we can now prove results aboutroots of unity modulo primes.

(1) Let p be an odd prime. Using Proposition 6.1, show that −1is a square mod p if and only if p ≡ 1 mod 4.

(2) Let p be an odd prime. Show that there is some x 6≡ 1 mod pso that x3 ≡ 1 mod p if and only if p ≡ 1 mod 3.

(3) Let p be an odd prime. Determine a necessary and sufficientcondition on p modulo n such that there will be n distinctroots of unity modulo p, i.e., there are n distinct residuesx1, . . . , xn mod p with xn

i ≡ 1 mod p.

NOTES ON FINITE FIELDS 15

(4) Given n and p, determine the number of nth roots of unitymodp. That is, determine the number of residues x so thatxn ≡ 1 mod p.

Exercise 6.4. Prove Wilson’s theorem: show that (p− 1)! ≡ −1 modp.

6.2. Frobenius. In what follows, we will let q denote a power of p,say q = pn.

Definition 6.5. The map

Frobp : Fq → Fq

x 7→ xp

is the Frobenius map.

Exercise 6.6. Verify that Frobp is a map Fq → Fq over Fp. That is,show that the natural inclusion i : Fp → Fq respects Frobp in thesense that Frobp ◦i = i. Hint: Show that for any x ∈ Fp, xp = x. SeeExercise 2.6(2) for further help.

Our next goal is to show that the automorphisms of Fq over Fp

(i.e., maps Fq → Fq as vectors spaces over Fp) are precisely id, Frobp, . . . , Frobn−1p ,

where q = pn. First, we show that these are all distinct:

Lemma 6.7. Let q = pn. The maps id, Frobp, . . . , Frobn−1p are distinct as

maps Fq → Fq.

Proof. We wish to show Frobap 6= Frobb

p for 0 ≤ a < b ≤ n − 1.After composing with Frob−a

p , it suffices to show that Frobb−ap is not

the identity for any 0 < b − a ≤ n − 1. That is, we have to showFrobp, . . . , Frobn−1

p are all distinct from id.To show that Frobc

p is not the identity, we have to show there issome x ∈ Fq with xpc 6= x. However, since xpc − x is a polynomial ofdegree pc, there are at most pc such elements. Since |Fq| = pn > pc,there is some element x ∈ Fq with xpc 6= x. �

Hence, we have produced that there are at least p distinct auto-morphisms of Fq given by powers of Frobenius. We next want toshow that these are all the automorphisms of Fq. For this, we willnow give an explicit construction of Fq as a field, by adjoining anelement to Fp, via the method in subsection 2.2.

Lemma 6.8. We can express Fpn in the form Fp[x]/( f ) for f ∈ Fp[x] ofdegree n.

16 AARON LANDESMAN

Proof. Pick y ∈ Fpn to be a generator of F×pn (which is possible byProposition 6.1). Note that y generates Fpn over Fp because all nonzeroelements of Fpn are powers of y. Further, by Exercise 4.5, (recall thissays that any element in an algebraic extension satisfies some monicirreducible polynomial) y satisfies some irreducible monic polyno-mial f over Fp. We obtain that Fp[x]/( f ) is a field by Lemma 2.11.We obtain a map

φ : Fp[x]/( f )→ Fq

x 7→ y.

Exercise 6.9. Verify this is a well defined map.

This map is necessarily injective by Lemma 3.6 but it is also sur-jective because y generates Fq. Therefore it is an isomorphism. Itfollows that f must have degree n since Fp[x] is a dimension deg fvector space over Fp, but it is also isomorphic to Fpn , which is a di-mension n vector space over Fp. �

Corollary 6.10. The automorphisms of Fq over Fp are precisely id, Frobp, . . . , Frobn−1p .

Proof. We have seen in Lemma 6.7 that these are all distinct, so itsuffices to show there are at most n automorphisms of Fq over Fp.However, by Lemma 6.8, we have Fq = Fp[x]/( f ). Note that anymap Fp[x]/( f ) → Fp[x]/( f ) must send x to some root of f , andfurther the map is determined by where it sends x. Since deg f = n,there are at most deg f = n roots of f and hence at most n such maps,as we wanted to show. �

6.3. Containments of finite fields. Let us now determine when Fpm ⊂Fpn for m, n > 0. First, we establish this containment when m | n.

Lemma 6.11. If m | n then Fpm ⊂ Fpn .

Proof. Recall from Theorem 5.3 that Fpm was precisely the set of ele-ments in Fp with xpm

= x. We also have xp2m= xpm

= x. If m | n, saym = dn then iterating this d times we obtain xpn

= xpdm= xp(d−1)m

=· · · = xpm

= x, so x ∈ Fpn . �

In fact, the above case is the only case that n | m, as we will nowsee.

Proposition 6.12. For p a prime and n, m > 0, we have Fpm ⊂ Fpn if andonly if m | n.

NOTES ON FINITE FIELDS 17

Proof. If m | n, then the inclusion holds by Lemma 6.11. Conversely,if Fpm ⊂ Fpn then Fpn is a vector space over Fpm . Say Fpn has dimen-sion d over Fpm . It follows that |Fpn | = |Fpm |d, so pn = (pm)d = pmd

and so m | n. �

18 AARON LANDESMAN

APPENDIX A. EXISTENCE OF ALGEBRAIC CLOSURES

We now guide the reader through a proof of the existence of alge-braic closures in series of exercises.

We first prove the existence of an algebraic closure Theorem 4.9(1),and then show it is unique up to (non-unique) isomorphism. Thekey to proving the existence of an algebraic closure will be Zorn’slemma, which we now recall:

Lemma A.1. Suppose I is a partially ordered set. Suppose any totallyordered subset I′ ⊂ I has a maximum element, i.e., there is some i ∈ I withi ≥ j for all j ∈ I′. Then I contains a maximal element, i.e., there is somei ∈ I so that for any j ∈ I, j 6> i.

Remark A.2. Zorn’s lemma is not a lemma in the conventional sensebecause it is equivalent to the axiom of choice. Therefore, we willnot prove it, but rather take it as an axiom.

We next aim to prove existence of algebraic closures. Logically,if you’d like, you can skip directly to Exercise A.5. However, it mayhelp your understanding of that exercise if you do the prior exercisesfirst.

Exercise A.3. We now prove some basic properties about cardinali-ties of field extensions.

(1) Show that if L is an algebraic extension of a finite field K, then|L| ≤ |Z|. Here |S| denotes the set-theoretic cardinality of aset S.

(2) Show that if L is an algebraic extension of an infinite field K,then |L| = |K|. Hint: Show that K has the same cardinalityas K[x] and defined a map of sets L → K[x] by sending anelement to its minimal polynomial. Show that there are onlyfinitely many elements with a given minimal polynomial anddeduce |K| = |L|.

(3) Conclude that for any infinite field K, if T is a set with |T| >|K| then for any algebaic extension L of K, we have |T| > |L|.

(4) Conclude that for any field K if T is an infinite set with |T| >|K|, then |T| ≥ |L| for any algebraic extension L of K. (By theabove, the only interesting case is the case that K is finite.)

Exercise A.4. Assume K is an infinite field. Using Exercise A.3, solvea slightly simplified version of Exercise A.5 with the modificationthat S is any set so that |S| > |K| (so that there is no intermediate setT in the picture). Therefore, the addition of T is only needed to dealwith finite fields.

NOTES ON FINITE FIELDS 19

Exercise A.5 (Difficult exercise). Use Zorn’s lemma to show an al-gebraic closure of a field K exists as follows: Let T be an infinite setwith |T| > |K| and let S be a set with |S| > |T|.

(1) Consider the partially ordered set

R := {(L, φ) : L is an algebraic extension of K and φ : L ↪→ S is a subset }Check that one can define a partial ordering on R by declaring(L1, φ1) ≤ (L2, φ2) if i : L1 → L2 is an algebraic extension, andφ2 ◦ i = φ1.

(2) Use Zorn’s lemma, Lemma A.1, to show that R has a maximalelement, call it (M, φ).

(3) Show that M is algebraically closed by showing that if i :M → N is any algebraic extension then there is a map ψ :N → S with ψ ◦ i(x) = φ(x). Hint: Use that |N| ≤ |M| ≤|T| < S and |S−M| = |S| > |N −M|.

Exercise A.6. Suppose we have an algebraic extension K ⊂ L andK ⊂ K with K algebraically closed. Show that there is a map ofextensions L→ K in the following steps:

(1) Consider the partially ordered set I of pairs (M, φ) with K ⊂M ⊂ L and φ : M→ K a map of fields. Check that the relation

(M1, φ1) ≤ (M2, φ2)

if M1 ⊂ M2 and φ2|M1 = φ1 defines a partial ordering on suchpairs (M, φ).

(2) Show that any totally ordered subset I′ ⊂ I corresponding toa collection {(Mi, φi)}i∈I′ has a maximum element given bytaking (∪i Mi,∪iφi), with ∪iφi interpreted suitably.

(3) Using Zorn’s lemma obtain a maximal element (M, φ) of I.(4) Verify that the maximum element (M, φ) has M = L and con-

clude there is a map L → K Hint: Suppose L 6= M. Thenthere is some x ∈ L−M. Show that x satisfies some minimalpolynomial over L. Deduce there is a map M(x)→ K restrict-ing to the given map φ : M → K, and hence (M, φ) was notmaximal.

Exercise A.7. Prove Theorem 4.9(2) using Exercise A.6 as follows:(1) Show that for any two algebraic closures K1, K2 of the same

field K there is an injective map between φ : K1 → K2.(2) Show that the injective map φ is an algebraic extension.(3) Conclude that the map produced K1 → K2 is an isomorphism

from the definition of algebraic closure.

20 AARON LANDESMAN

APPENDIX B. BASICS OF RINGS

In this appendix, we review some basic definitions relating to rings.Recall our definition of a (commutative) ring (with unit), Defini-tion 2.1. We repeat this now for your convenience.

Definition B.1. A commutative ring with unit is a set R togetherwith two operations (+, ·) satisfying the following properties:

(1) Associativity: a + (b + c) = (a + b) + c, a · (b · c) = (a · b) · c(2) Commutativity: a + b = b + a, a · b = b · a(3) Additive identity: there exists 0 ∈ R so that a + 0 = a(4) Multiplicative identity: there exists 1 6= 0 ∈ R so that 1 · a = a(5) Additive inverses: For every a ∈ R, there is a additive in-

verse, denoted −a satisfying a + (−a) = 0(6) Distributivity of multiplication over addition: a · (b + c) =

(a · b) + (a · c) .

For us, all rings will be commutative rings with unit, and so wewill simply refer to them as rings from now on. We now recall someelementary properties of rings. Many of these follow directly fromthe analogous properties for groups.

Exercise B.2. Verify, directly from the definition that every ring hasa unique 0 and 1. Show that for any a ∈ R, a has a unique additiveinverse, and so the name −a is justified.

Definition B.3. A map of rings f : R → S is a map of sets such thatf (1R) = 1S, f (0R) = 0S, f (a +R b) = f (a) +S f (b) and f (a ·R b) =f (a) ·S f (b). where the subscripts denote the identity, multiplication,and addition in the corresponding ring.

Definition B.4. A ring map f : R→ S is injective if f (a) = f (b) =⇒a = b. It is surjective if for every s ∈ S there is some r ∈ R withf (r) = s. It is bijective (also known as an isomorphism) if it is bothinjective and surjective. If f : R→ S is bijective, we write R ' S.

Exercise B.5. Show that a ring map f : R→ S is injective if and onlyif f−1(0S) = 0R.

Exercise B.6. Show that a ring map f : R→ S is bijective if and onlyif there is a ring map f−1 : S→ R so that f−1 ◦ f = idR, f ◦ f−1 = idS.Hint: Show that a map is bijective if and only if there is a uniqueelement of R mapping to any given element of S. Use this to definean inverse map.

NOTES ON FINITE FIELDS 21

B.1. Quotients. The following will not be needed in this course. In-deed, we will construct particular quotient rings in this course, butthese quotients will all be of the form K[x]/( f ) for K a field, a situ-ation which is much more concrete than the general case developedbelow. We encourage you to skip the following, but we include it forcompleteness.

Definition B.7. An ideal I of R is a subset I ⊂ R so that(1) 0R ∈ I.(2) For any r ∈ I we also have −s ∈ I.(3) If a, b ∈ I then a + b ∈ I.(4) If r ∈ R and a ∈ I then a ·R b ∈ I.

Definition B.8. Let I ⊂ R be a subring. Construct the quotient R/Ias the set of all elements a ∈ R modulo the equivalence relation a ∼ bif there is some c ∈ I with a = b + c. The equivalence class of a iscalled the coset of a and the coset is notated a + I.

Exercise B.9. Verify that the relation∼ as defined in Definition B.8 isindeed an equivalence relation.

Exercise B.10. Show that if I ⊂ R is an ideal then R/I is again aring. (Under our definition, this includes verifying that the quotientis commutative and has a unit).

REFERENCES