FINITE FIELDS MARCO ADAMO SEVESO

Contents 1. Finite fields 2. Direct limits of fields References

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1. Finite fields Suppose that F is a finite field and consider the canonical homomorphism Z → F. Since F is a field its kernel is a prime ideal of Z and then, since F is finite, we see that it is of the form pZ for some rational prime p. It follows that we have Fp ,→ F and F is a finite dimensional vector space. In particular, fixing a basis we see that q := #F = p[F :Fp ] . Hence a finite field has order pn for some prime p and some n ≥ 1. Theorem 1.1. For every prime p and every n ≥ 1 there exists a unique (up to isomorphism) field of order pn , denoted by Fpn . Furthermore, Fpn is uniquely characterized as the splitting field of the polynomial n

fpn (X) := X p − X ∈ Fp [X] and, if Rfpn is the set of roots of this polynomial in a splitting field of fpn (X), then Rfpn = Fpn . Proof. We have already remarked that a finite field F is such that q := #F = pn , where n = [F : Fp ]. Since #F × = q − 1 we have that xq−1 − 1 = 0 for every x ∈ F × . We deduce that xq − x = 0 for every x ∈ F .

(1)

Hence F ⊂ Rfq (the set of roots of fq which could be taken, for example, in a splitting field) and, since #Rfq ≤ deg (fq ) = q = #F , we find F = Rfq . It follows that Q fq (X) = x∈F (X − x) in F [X] and, since F = Rfq , it is clear that F is a splitting field of fq (X) ∈ Fp [X]. In particular, if a finite field of order pn exists, it is uniquely determined (up to isomorphism) as the splitting field of fpn (X) ∈ Fp [X]. It remains to prove the existence of a field of order q := pn . Let F be a splitting field of fq (X) ∈ Fp [X] and let Rfq ⊂ F be the set of roots of fq (X). We remark that, since f 0 (X) = qX q−1 − 1 = −1 6= 0, 1

then #Rfq = q. We claim that Rfq is indeed a field, from which it also follows that Rfq is the required field (and that Rfq = F )1. Indeed, let x, y ∈ Rfq . • Rfq is closed with respect to sums: q

(x + y) − (x + y) = xq + y q − x − y = 0. • Rfq is closed with respect to multiplication: q

(xy) − (xy) = xq y q − xy = xy − xy • Rfq is closed with respect to taking the opposite element: q

q

q

(−x) − (−x) = (−1) xq + x = (−1) x + x. q

q

If p 6= 2, then (−1) = −1 and the right hand side is −x + x = 0, while for p = 2 we have (−1) = 1 and the right hand side is x + x = 2x = 0. • Rfq is closed with respect to taking the inverse element:
q −1 x−1 − x−1 = (xq ) − x−1 = x−1 − x−1 = 0. • 0, 1 ∈ Rfq . This is clear.  Let now q = pn with p a prime and n ≥ 1 and consider the Frobenius homomorphism ϕ : Fq → Fq . Since Fq /Fp is normal Theorem 1.2. We have that GFq /Fp = hϕi ' Z/nZ. n

Proof. Set G := hϕi ⊂ GFq /Fp and d := |ϕ|. Since ϕn (x) = xp = x for every x ∈ Fq (by (1)), we have ϕn = 1 and d | n. On the other hand we have d

x = ϕd (x) = xp for every x ∈ Fq . It follows that Fq ⊂ Rfpd (the set of roots of fpd which could be taken, for example, in a splitting field) and
q = pn ≤ #Rfpd ≤ deg fpd = pd . We deduce n ≤ d, so that d = n. On the other hand we have that GFq /Fp is a subset of the set SFq /Fp (L) of all homomorphism σ : Fq → L over Fp , where L is any field containing Fq , and, by [Mi, Corollary 2.8], #SFq /Fp (L) ≤ [Fq : Fp ] for every field L. It follows that, with L = Fq (or any other field L ⊃ Fq ), #GFq /Fp ≤ #SFq /Fp (L) ≤ [Fq : Fp ] = n. Since # hϕi = d = n we deduce that GFq /Fp = hϕi ' Z/nZ.



Corollary 1.3. Let E/F be a finite extension of finite fields with #F = pnF =: q and #E := pnE . (1) Then nF | nE , so that we have #E = q n with n := nE /nF = [E : F ]. (2) Setting ϕq := ϕnF : E → E we have that E/F is a Galois extension with Galois group GE/F =

 ϕq ' Z/nZ. 1Let ϕ : F → F be the Frobenius morphism ϕ (x) = xp and set ϕ := ϕn , so that ϕ (x) = xq . Then we have that x ∈ R fq q q
if and only if ϕq (x) = x, i.e. if and only if x ∈ ker ϕq − 1 . Hence
Rfq = ker ϕq − 1

and, since ϕq − 1 is a morphism of Fp -vector spaces, we deduce that Rfq is an Fp -vector space. Hence, as an alternative to the subsequent computations, one has only to remark that this Fp -vector is multiplicatively closed and that 1 belongs to it in order to deduce that Rfq is a field. 2

(3) Writing FE/F for the lattice of fields K such that F ⊂ K ⊂ E there is an ordering preserving bijection {d ∈ N : d | n} → n FE/F o d d 7→ x ∈ E : xq = x , where the domain is ordered by divisibility. Furthermore we have n o d x ∈ E : xq = x = Fqd = Rfqd . Proof. (1) We have indeed: nE = [E : Fp ] = [E : F ] [F : Fp ] = [E : F ] nF . (2) Since E/Fp is Galois (by Theorem 1.1) then E/F is Galois. We have ϕq (x) = xq , so that ϕq (x) = x if and only if xq = x or, equivalently, if and only if x ∈ Rfq = F (by Theorem 1.1). Hence F = E hϕq i and

 GE/F = GE/E hϕq i = ϕq

 by the Galois correspondence. Since ϕq := ϕnF ∈ GE/Fp ' Z/nE Z has order n = nE /nF we have ϕq ' Z/nZ. (3) Since the subgroup of Z/nZ are in bijection with the divisors of n by the rule d 7→ dZ/nZ, we have that



 d the subgroups of GE/F = ϕq ' Z/nZ are those of the form ϕdq . As above we remark that ϕdq (x) = xq d from which it follows that ϕdq (x) = x if and only xq = x or, equivalently, if and only if x ∈ Rfqd (E), the set of roots of fqd in E. We deduce n o d d x ∈ E : xq = x = Rfqd (E) = E hϕq i as claimed. But thanks to the Galois correspondence we know that h i

 d E : E hϕq i = # ϕdq = n/d. h i d d It follows that E hϕq i : F = d, so that E hϕq i = Rfqd (by Theorem 1.1) and, hence, Rfqd (E) = Rfqd = E hϕq i . d

 Corollary 1.4. Let F be a finite field with #F = pn =: q, let F be an algebraic closure of F and set ϕq := ϕn : F → F . Writing FF for the lattice of fields K such that F ⊂ K ⊂ F and [K : F ] < ∞ there is an ordering preserving bijection N → n FF o d

7→

d

x ∈ F : xq = x ,

where the domain is ordered by divisibility. Proof. We have, by Theorem 1.1, Rfqd = Fqd , the splitting field of fqd . Since Rfqd ⊂ F there is a morphism id : Fqd → F




and we have id Fqd = F Rfqd



= Rfqd because id is an isomorphism onto its image. Hence Rfqd ⊂ F is

d

a subfield with q elements and, if K ∈ FF is another field with q d elements, we have K = Rfqd because, again by Theorem 1.1, it is a splitting field of fqd in F 2.  2More generally, the algebraic closure F of any field F contains precisely one splitting F /F for every f ∈ F [X], i.e. F = F 0 f f f if Ff and Ff0 are both splitting fields of f . Then we are simply applying this remark to the family of polynomials fqd , noticing that they give distinct fields for distinct d by Theorem 1.1 and that, by the same result, all the elements of FF are obtained in this way. 3

It is known that, for every field F there exists an algebraic closure. We can now prove this result for a finite field F , giving an explicit way to construct its algebraic closure. Corollary 1.5. If F is a finite field there exists an algebraic closure of it. Proof.

3

d

If d ∈ N, we let Fqd be the splitting field of X q − X ∈ F [X], where #F = q. Define e := F F d∈N≥1 Fq d

e for the canonical injection (note that Fq = F , by Theorem 1.1). Note that, if d1 | d2 and write jd : Fqd → F there is an F -morphism of fields jd1 |d2 : Fqd1 → Fqd2 which identifies Fqd1 with the set of elements x ∈ Fqd2 d1 such that xq = x (by Corollary 1.3). Note that the set N, ordered by divisibility, is an inductive set, i.e. if we have given d1 , d2 ∈ N there is d such that di | d for i = 1, 2. It follows form the subsequent Lemma 1.6 that we may choose these jd1 |d2 in such a way that jd|d = 1Fqd and, if d1 | d2 | d3 , then jd2 |d3 ◦ jd1 |d2 = jd1 |d3 4. Then we define e ∼, F = lim Fqd := F/ →

where x ∈ Fqd1 and y ∈ Fqd2 are equivalent whenever jd1 |d (x) = jd2 |d (y). It is easily checked that F is a jd e → F is an F -morphism of fields which has to be an inclusion. field and Fqd → F It is clear that F/F is algebraic and the fact that F is an algebraic closure of F follows, once we show that every polynomial f (X) ∈ F [X] has a root in F (see [Mi, Proposition 1.42 and 1.44]). Let Ff /F be a splitting field of f . Being of finite degree we have an F -isomorphism Ff ' Fqd for some d (by Corollary 1.3) and the roots of f in Ff give rise to roots of it in F via Ff ' Fqd → F. 

Lemma 1.6. The morphisms jd1 |d2 : Fqd1 → Fqd2 can be choose in such a way that jd|d = 1Fqd and, if d1 | d2 | d3 , then jd1 |d3 = jd2 |d3 ◦ jd1 |d2 . Proof. Choose an infinite sequence 1 = n1 | n2 | ... | ni | ... such that, for every d ∈ N≥1 , d | ni for some i (for example ni = i!). We define jni |ni = 1Fqni and, assuming by induction that we have defined jni |nj for every i, j ≤ l in such a ways that jni |nk = jnj |nk ◦ jni |nj if i ≤ j ≤ k, we take jnl |nl+1 : Fqnl → Fqnl+1 to be any morphism, as granted by Corollary 1.3 with E/F = Fqnl+1 /F and Theorem 1.1 giving jnl |nl+1 : Fqnl ' K ⊂ Fqnl+1 . If i ≤ l, we set jni |nl+1 := jnl |nl+1 ◦ jni |nl . Then we remark that, if i ≤ j ≤ l, so that jni |nl = jnj |nl ◦ jni |nj , we have jni |nl+1 := jnl |nl+1 ◦ jni |nl = jnl |nl+1 ◦ jnj |nl ◦ jni |nj =: jnj |nl+1 ◦ jni |nj . This gives an infinite family 

jni |nj : Fqni → Fqnj

i≤j

which satisfies the required compatibility. Suppose now that d ∈ N≥1 − {ni : i} and let id be the smallest index such that d | nid . Then we take any morphism jd|nid : Fqd → Fqnid . If we have d1 | d2 there is a unique morphism jd1 |d2 making the following 3We have F ⊂ F and, by [Mi, Corollary 1.46], if F is an algebraic closure of F , then F is an algebraic closure of F (because p p

F ⊂ F since F/Fp is algebraic): hence we may assume F = Fp , but we will not need this reduction in the proof. 4If one uses the existence of F , there is no need to employ Lemma 1.6. By Corollary 1.4 we have that n o d F q d = x ∈ F : xq = x S and the subsequent proof shows that F := d∈N Fqd ⊂ F is algebrically closed, so that F = F . ≥1 We also remark that one only needs to define the sequence of compatible morphisms jni |nj as in the proof of Lemma 1.6. Then in the subsequent proof one take F = lim Fqni →

and a posteriori, once the equality F = F has been checked, we recover every Fqd as a subfield of F as above. 4

diagram commutative: jd1 |d2

→

Fqd1 jd1 |nid ↓ 1

jni

d1

ni q d1

F Indeed we have jnid

1

 |nid

2

jd|nid

1

F q d1




⊂ jd|nid

Fqd2 ↓ jd2 |nid

|ni d2

→

F

ni q d2

2

.




2

−1 jd1 |d2 := jd|n i

d2

Fqd2 (by Corollary 1.3) and then we may take ◦ jnid

1

|nid

2

◦ jd|nid

1

because jd|nid is injective. This shows the existence and uniqueness follows form the injectivity of jd|nid 2 2 and the requirement of making this diagram commutative. The compatibilities jd|d = 1Fqd and jd1 |d3 = jd2 |d3 ◦ jd1 |d2 when d1 | d2 | d3 follow from jnid |nid = 1F nid and jnid |nid = jnid |nid ◦ jnid |nid , in light of q 1 3 2 3 1 2 the above diagram and the injectivity of jd|nid and jd3 |nid .  3

2. Direct limits of fields Suppose that (S, ≤) is a partially ordered set with the property that, given s1 , s2 ∈ S, there is some s ∈ S such that s1 , s2 ≤ s. We suppose that we have given a family fs1 ≤s2 : Ks1 → Ks2 of homomorphisms of fields such that: • fs≤s = 1Ks • fs1 ≤s3 = fs2 ≤s3 ◦ fs1 ≤s2 if s1 ≤ s2 ≤ s3 . Then we may form the disjoint union e := F K s∈S Ks e as follows. If xi ∈ Ks with i = 1, 2, we write x1 ∼ x2 if and and we define an equivalence relation on K i only if there is some s ∈ S such that s1 , s2 ≤ s and fs1 ≤s (x1 ) = fs2 ≤s (x2 )5. Next we define e ∼, K := K/ the set of equivalence classes, and e → K, fs : Ks ⊂ K e e → K the quotient map. If x ∈ K e we write where Ks ⊂ K is the inclusion in the disjoint union and K e [x] ∈ K for its image in K via K → K, i.e. the equivalence class that x represents. Then the following diagrams are commutative for every s1 ≤ s2 6: fs1

→

Ks1 fs1 ≤s2 ↓

fs2

→

Ks2

K k K

We define a ring structure on K by the following rules7: e and s is choosen so that s1 , s2 ≤ s; • [x1 ] + [x2 ] := [fs1 ≤s (x1 ) + fs2 ≤s (x2 )], if xi ∈ Ksi ⊂ K e • [x1 ] · [x2 ] := [fs1 ≤s (x1 ) · fs2 ≤s (x2 )], if xi ∈ Ksi ⊂ K and s is choosen so that s1 , s2 ≤ s. Then K becaomse a field with zero element [0Ks ] = fs (0Ks ) and unit [1Ks ] = fs (1Ks ) (for any s ∈ S) and fs : Ks → K is a homomorphism of fields and, in particular, it is injective8. Furthermore, we have9 S K = s∈S fs (Ks ) 5Exercise: this gives an equivalence relation. 6Exercise: show this is true. 7Exercise: the following ninary operations are well defined and give K a ring structure. 8Exercise: show this is true. 9Exercise: show this is true. 5

∼

and, since fs : Ks → fs (Ks ) and fs1 = fs2 ◦ fs1 ≤s2 , we may abusively write S K = ” s∈S Ks ”, where fs1 ≤s2 : Ks1 ,→ Ks2 . References [La] S. Lang, ”Algebra”, Springer. Chapter V, §5 Finite fields. [Mi] J. S. Milne, ”fields and Galois theory”, www.jmilne.org/math/CourseNotes/FT.pdf. Chapter 4, ”Finite fields”.

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