Abstract Algebra 2 (MATH 4140/5140)
Finite Fields
Recall:
• If p is a prime number and f ∈ Zp [x] is an irreducible polynomial of degree n, then
K = Zp [x]/(f ) is a finite field of order pn (i.e., the number of elements is pn ).
Questions 1–3.
1. Is the order of a finite field necessarily a prime power?
2. Does there exist a finite field of order pn for every prime p and every n ≥ 1?
3. If yes, how many finite fields of order pn are there, up to isomorphism?
The following theorem summarizes the most important facts about finite fields, including
the answers to Questions 1–3.
Theorem.
(1) Every finite field F has prime characteristic, and if char(F ) = p, then F is an
extension field of Zp .
Reason:
(2) If F is a finite field of characteristic p, then F is a finite dimensional vector space
over Zp , because
,
hence |F | = pn where
.
(3) For every prime p and integer n ≥ 1 there exists a unique finite field of order pn , up
n
to isomorphism, namely a splitting field for xp − x ∈ Zp [x] over Zp .
Proof. Uniqueness: Let F be a finite field of order pn ; hence F is an extension field
n
of Zp . We want to show that F is a splitting field for xp − x ∈ Zp [x] over Zp . The
uniqueness of splitting fields will then imply that F is unique, up to isomorphism.
n
n
• Every element of F is a root of the polynomial xp − x = x(xp −1 − 1), because
.
• It follows that F is a splitting field for x
pn
− x ∈ Zp [x] over Zp :
n
xp − x splits in F [x], because
.
n
F = Zp (roots of xp − x in F ), because
.
pn
Existence: Let L be a splitting field for x − x ∈ Zp [x] over Zp . We will show that
n
|L| = pn . Let R be the set of roots of xp − x in L.
• |R| = pn , because
.
• R is a subfield of L containing Zp . Reason:
.
. So, |L| = pn .
• Hence, L = R. Reason:
1
2
(4) The multiplicative group of a finite field of order pn is a cyclic group of order pn − 1.
Proof. Let F be a finite field of order pn , and let F ∗ denote its multiplicative group.
Clearly, |F ∗ | = |F \ {0}| = pn − 1. To prove that F ∗ is a cyclic group, let e denote
the least common multiple of the orders of the elements of F ∗ , called the exponent of
F ∗ . First we want to show that e = pn − 1.
• e | pn − 1. Reason:
.
• The roots of the polynomial xe − 1 ∈ Zp [x] in F are
.
• Hence, e = pn − 1. Reason:
.
1
n
By the Fundamental Theorem on Finite Abelian Groups the equality e = p − 1
implies that F ∗ is cyclic.
(5) For every prime p and integer n ≥ 1 there exists an irreducible polynomial of degree
n in Zp [x].
Proof. Let F be a finite field of order pn , and let γ be a generator of the cyclic group
F ∗.
• F = Zp (γ), because
.
• γ is algebraic over Zp , because
let f ∈ Zp [x] be the minimal polynomial of γ over Zp .
;
• deg(f ) = n, because
So, f is an irreducible polynomial of degree n in Zp [x].
.
Corollary 1. If K is a finite field, then every algebraic extension of K is a separable
extension of K.
Proof. Let p = char(K), let M be an algebraic extension of K, and let α ∈ M have minimal
polynomial f ∈ K[x] over K.
• the field K(α) is finite, because
n
• f | xp − x in K[x], because
n
• gcd(f, f 0 ) | (xp − x)0 = −1 in K[x], because
; say |K(α)| = pn .
.
.
Hence, f has no multiple roots.
We saw2 that if K is a field of characteristic 0, and L is a splitting field for xn −1 ∈ K[x] over
K, then the n-th roots of unity in L form a cyclic subgroup of order n in the multiplicative
group of L. The analogous result for fields of prime characteristic is the following.
Corollary 2. Let K be a field of prime characteristic p, let n = pk m be a positive integer
where p - m and k ≥ 0. If L is a splitting field for xn − 1 ∈ K[x] over K, then the n-roots
of unity in L form a cyclic subgroup of order m (= pnk ) in the multiplicative group of L.
Proof. HW.
1An
alternative route is to show directly (using basic properties of orders of elements in abelian groups)
that the exponent e of F ∗ is equal to the order of some element in F ∗ .
2See the handout ‘Splitting Fields for Polynomials – Uniqueness’.