Download FINITE FIELDS Although the result statements are largely the same

FINITE FIELDS
BRIAN OSSERMAN
Although the result statements are largely the same as in Artin (§15.7), the proofs are quite
different, and follow a different order.
Proposition. Let F be a field.
(i) If F has characteristic 0, then there is a unique field inclusion of Q into F .
(ii) If F has characteristic p > 0, then there is a unique field inclusion of Fp into F .
Proof. Both of these are determined by sending 0 to 0 and 1 to 1, and letting additivity and
multiplicativity determine the rest.
Now, if a field F is finite, it can’t contain Q, so it immediately follows that F has characteristic
p for some prime p, and that furthermore, F is an extension of Fp in a unique way. If [F : Fp ] = r,
then a basis of F has r elements, so we see that F has pr elements. The main classification theorem
for finite fields is the following:
Theorem.
(i) Every finite field F has characteristic p for some prime p, and then it has pr
elements for some positive integer r.
(ii) If a finite field F has q = pr elements, then F is a splitting field for the polynomial xq − x
over Fp . In particular, any two fields with q elements are isomorphic to one another.
(iii) For every prime p and positive integer r, there is a finite field with pr elements.
Proof. We have already proved (i). For (ii), the multiplicative group F × has order q − 1, so for all
nonzero α ∈ F , we have αq−1 = 1, i.e. α is a root of xq−1 − 1. Of course, 0 is a root of xq − x, so
we conclude that every element of F is a root of xq − x. But since F has q elements, we must have
the factorization
Y
xq − x =
(x − α),
α∈F
so F is a splitting field of xq − x. It then follows from our theorem on splitting fields that any two
fields of order q are isomorphic.
(iii) Let L be a field in which xq − x splits. We observe that the derivative of xq − x is qxq−1 − 1 =
−1, since q = pr and we are in characteristic p. Thus, by our proposition on multiple roots of
polynomials, we have that xq − x cannot have multiple roots in any extension field of Fp . In
particular, xq − x does not have multiple roots in L, so L contains exactly q roots of xq − x. If
we let F be the set of roots of xq − x in L, we will show that F is a subfield of L, so that F is
the desired field with q elements. Note that F can also be described as the set of α ∈ L such that
αq = α. It is clear that F contains 0 and 1, and also that it is closed under multiplication, and also
under division.
We next check that F is closed under addition. By Exercise 11.3.8 of Artin, we have the Frobenius
homomorphism F → F sending α to αp . Composing the Frobenius homomorphism with itself r
r
times, we find that the map α 7→ αp is also a homomorphism. This means that (α + β)q = αq + β q ,
so if α, β ∈ F , so is α + β. Finally, we see that in characteristic p, we have −α = (p − 1)α, so
it follows from F being closed under addition that it is also closed under taking additive inverses.
Thus, F is a subfield, as desired.
The theorem justifies the following notation:
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Notation. Given q ∈ Z a prime power, let Fq be the finite field with q elements.
We next consider which finite fields are contained in one another. Note that if Fpr is an extension
of Fps , then by the theorem, there is no ambiguity about how it is realized as an extension: Fps
s
must be the subfield consisting of the roots of xp − x. The basic result is then the following:
Proposition. Fix a prime p, and r ∈ Z>0 .
(i) Given also s ∈ Z>0 , the field Fpr can be realized as an extension of Fps if and only if s|r.
(ii) Given also s ∈ Z>0 , with s|r, and α ∈ Fpr , then α ∈ Fps if and only if the degree of α over
Fp divides s.
r
(iii) The irreducible factors in Fp [x] of the polynomial xp − x consist precisely of the irreducible
polynomials in Fp [x] having degree dividing r.
Proof. (i) if Fpr can be realized as an extension of Fps , then it is a finite-dimensional vector space
over Fps , so if the dimension is equal to d, we have pr = (ps )d = psd , so s divides r. Conversely, If
s
r
r = sd for some d ∈ Z, and if αp = α, then we note that αp is obtained by raising α to the ps
r
s
power d times, so we conclude that αp = α. Phrased differently, any root of xp − x is a root of
r
s
xp − x also, so if we let K/Fpr be a splitting field of xp − x, we see that all the roots already lie in
r
Fpr , since Fpr is a splitting field for xp − x. We conclude that Fps can be realized as the subfield
s
of Fpr consisting of roots of xp − x.
(ii) First note that α ∈ Fps if and only if Fp (α) ⊆ Fps . If the degree of α over Fp is d, then
Fp (α) = Fpd , so the desired statement then follows from (i).
r
(iii) Fpr consists precisely of roots of xp − x, so if f (x) is an irreducible factor, then Fpr contains
some root α of f (x). Then by (ii) with s = r, the degree of α, which is by definition the degree
of f (x), must divide r. Conversely, if f (x) is an irreducible polynomial of degree d|r, and we let
F = Fp (α) where α is a root of f (x), then F ∼
= Fpd , so by (i) we can realize F as a subfield of Fpr ,
r
and thus f (x) has a root in Fpr . But since all the elements of Fpr are roots of xp − x, this means
r
that f (x) and xp − x have a nonconstant common factor. Since f (x) is irreducible, this implies
r
that f (x) divides xp − x.
We next consider the structure of the multiplicative group of a finite field. We need the following.
Lemma. Let G be a finite abelian group, and let m be the maximal order among all elements of
G. Then every element of G has order dividing m.
Note that this lemma is false if G is not abelian – for instance, it fails already for S3 .
Proof. Suppose that we have g, g 0 ∈ G with order d, d0 respectively. We first show that if d and d0
0
are relatively prime, then ord(gg 0 ) = dd0 . Certainly, (gg 0 )dd = 1. On the other hand, if (gg 0 )n = 1,
then (g 0 )n = (g n )−1 , so both (g 0 )n and g n have the same order. But the order of (g 0 )n divides d0
and the order of g n divides d, so by relative primality we conclude that both (g 0 )n and g n have
order 1, i.e. (g 0 )n = g n = 1. But then d and d0 both divide n, and again using relative primality
we conclude that n is a multiple of dd0 . Thus, gg 0 has order dd0 , as claimed.
Now, suppose that d, d0 are not necessarily relatively prime. We claim there exists g 00 ∈ G having
order equal to the least common multiple of d and d0 . To see this, choose e, e0 so that e|d, e0 |d0 , e
and e0 have no common factors, and the least common multiple of d and d0 is equal to ee0 . (Exercise:
0 0
check this is always possible) Then g d/e has order e, and g 0d /e has order e0 , so by the relatively
0
0
prime case above, we conclude that g 00 = g d/e g 0d /e has order equal to ee0 , as desired.
Finally, given any h ∈ G, let n be the order of h. If n doesn’t divide m, then the least common
multiple of m and n is strictly greater than m, and by the above, there would be an element of
this order, contradicting the hypothesis that m is the maximal order of an element in G. Thus, n
divides m, as desired.
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Corollary. Let F be a field, and G a finite subgroup of F × . Then G is cyclic.
In particular, if F is a finite field, then F × is cyclic.
Proof. Let m be the maximal order among all elements of G. This is certainly at most |G|. However,
according to the previous lemma, every element of G has order dividing m, so it follows that if we
have α ∈ G, then αm = 1. Thus, every element of G is a root of the polynomial xm − 1, so we
conclude that |G| 6 m, and hence that |G| = m, and G is cyclic.
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