Resource C
Finite Fields
C.1
The prime fields
Recall that, for each integer n > 0, Z/(n) is a commutative ring containing
n elements that can be denoted by 0, 1, . . . , n − 1, with addition and multiplication calculated modulo n. For example, in Z/(7) 2 + 6 = 1, 2 · 6 = 5.
Proposition C.1 Z/(n) is a field if and only if n is prime.
Proof I If n is not prime, say n = ab with 1 < a, b < n. Then ab = 0 in
Z/(n). If n is prime and 1 < r < n, consider the map θ : Z/(n) → Z/(n)
under which
i 7→ ir mod n.
This map is injective; for if 0 ≤ i < j ≤ n − 1 then
θ(i) = θ(i) =⇒ ir ≡ jr
since n - r.
mod n =⇒ n | (j − i)r =⇒ j ≡ i mod n,
J
Definition C.1 If p is prime, we denote the field Z/(p) by Fp .
We call the fields Fp the prime fields.
C.2
Finite fields
We are going to show that there exists just one finite field of each prime-power
order pn , which we shall denote by Fpn .
The proof is in two parts. First we show that two finite fields of the same
order pn are necessarily isomorphic. Then we show that there actually exists
a field with pn elements.
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In this course we shall only consider elliptic curves over the prime fields
Fp . However, the number of points on such a curve in the fields Fpn plays a
central rôle in the more advanced theory of elliptic curves beyond this course.
In particular, the L-function of an elliptic curve, which is the main tool in
the advanced theory, is defined in terms of these numbers.
Every mathematician should be aware of the existence of these finite
fields, which turn up in many branches of mathematics. But they will not
be considered examinable material in this course.
C.3
The characteristic of a field
Recall that the characteristic of a ring is the least n > 1 such that
n 1’s
}|
{
z
1 + 1 + · · · + 1 = 0.
if there is such an n, or 0 if there is not.
Thus Q, R, C are all of characteristic 0, while Fp is of characteristic p.
Proposition C.2 The characteristic of a field is either 0 or a prime.
Proof I Let us write
n 1’s
}|
{
z
n · 1 for 1 + 1 + · · · + 1 .
Suppose the order n is composite, say n = rs. By the distributive law,
n · 1 = (r · 1)(s · 1).
There are no divisors of zero in a field; hence
r · 1 = 0 or s · 1 = 0,
contradicting the minimality of n.
J
Proposition C.3 Every field of characteristic p contains Fp as a subfield,
and can be considered as a vector space over Fp .
Proof I Consider the additive subgroup generated by 1:
h1i = {0, 1, 2 · 1, . . . , (p − 1) · 1}.
It is readily verified that this set is closed under addition and multiplication;
and the map
r mod p 7→ r · 1 : Z/(p) → h1i
is an isomorphism. J
This field is called the prime subfield of F .
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Corollary 25 A finite field F of characteristic p contains pn elements for
some n ≥ 1.
Proof I Suppose dimFp F = n. Let e1 , . . . , en be a basis for the space. Then
each element of F is uniquely expressible in the form
a1 e1 + · · · + an en ,
where a1 , . . . , an ∈ Fp .
C.4
J
Primitive roots
Theorem C.1 If F is a finite field then the group F × is cyclic.
Proof I Suppose |F | = pn . Then |F × | = pn − 1. Hence any x ∈ F × has
degree d | pn − 1.
Note that in this case x is a root of the polynomial equation
xd − 1 = 0.
But 1, x, x2 , . . . , xd−1 also satisfy this equation. Since the equation has at
most d roots, it follows that the roots of the equation form the cyclic group
hxi.
We recall the following result, where φ(n) denotes Euler’s totient function
(the number of numbers in {0, 1, . . . , n − 1} coprime to n).
Lemma 22 The cyclic group Cn has φ(n) elements of order n.
So the number of elements of order d | pn − 1 is either 0 or φ(d).
P
Lemma 23
d|n φ(d) = n.
Proof of Lemma B Recall that Cn has just one subgroup of each degree
d | n, and that this subgroup is cyclic. (If Cn = hgi this is the subgroup
Cd generated by g n/d .) As we saw above Cd contains just φ(d) elements of
order d. The result follows on adding the number of elements in Cd of order
d | p− 1. C
But now we see that F × must contain exactly φ(d) elements of each order
d | pn − 1. In particular there will be φ(pn − 1) > 0 of order pn − 1, and so
F × is cyclic. J
Definition C.2 A generator of F×
p is called a primitive root mod p.
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Example: From the argument above, there are φ(p − 1) primitive roots mod
p. Eg F23 has φ(22) = φ(2) · φ(11) = 10 primitive roots. (More generally, it
has 1 element of order 1, φ(2) = 1 element of order 2, φ(11) = 10 elements of
order 11, and φ(22) = 10 elements of order 22.) ) To find a primitive root,
try 2, 3, 5, . . . successively until we find one:
25 = 32 ≡ 9 mod 23,
210 ≡ 92 ≡ −1 mod 23.
Since 2 is not of order 1, 2 or 10, it must be a primitive root mod 23. The
other primitive roots mod 23 are the powers 2i mod 23 for i coprime to 23.
C.5
The uniqueness of Fpn
.
Theorem C.2 Two fields of order pn are necessarily isomorphic.
Proof I Suppose |F | = |F 0 | = pn .
The elements of F × all satisfy the polynomial
U (x) = xp
n −1
− 1 ∈ Fp [x]
(and so give all pn − 1 roots of this polynomial). We note that U (x) is
separable, ie has no repeated roots, since
U 0 (x) = xp
n −2
and so gcd(U (x), U 0 (x)) = 1.
U (x) will factorise into irreducible polynomials in Fp [x], say
U (x) = f1 (x) · · · fr (x).
Let g be a generator of F . Then g must be a root of one of these polynomials (since every element of F × is a root of U (x)). Suppose fi (g) = 0. Let
us write f (x) for fi (x).
Now let us pass to F 0 [x]. The polynomial U (x) splits in exactly the same
way in F 0 [x], since this only depends on arithmetic in Fp [x]. So we can find
an element g 0 ∈ F 0 such that f (g 0 ) = 0.
We define the map Φ : F → F 0 by
g i 7→ g 0i
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for i = 0, 1, . . . , pn − 2 (and 0 7→ 0).
We shall show that this map is a ring-homomorphism, ie preserves addition and multiplication.
n
For multiplication this follows easily from the fact that g p −1 = 1 and
n
g 0p −1 = 1. (We are not claiming at the moment that g 0 generates F 0 , although we shall find later that this is true.)
Addition is more subtle. Suppose
ga + gb = gc.
Then g is a root of the polynomial xa + xb − xc .
Lemma 24 Suppose the field K is a finite extension of the field k, ie K as a
vector space is finite-dimensional over k. Then each element α ∈ K satisfies
a unique irreducible monic polynomial f (x) ∈ k[x]; and if f (x) ∈ k[x] then
f (α) = 0 if and only if m(x) | f (x)
Proof of Lemma B If dimk K = d then the d + 1 elements 1, α, α2 , . . . αd are
linearly dependent over k, ie α satisfies a polynomial f (x) ∈ k[x] of degree
≤ d.
Let m(x) ∈ k[x] be the monic polynomial of minimal degree satisfied by
α. Then m(x) is irreducible. Suppose f (α) = 0. Dividing f (x) by m(x),
f (x) = m(x)q(x) + r(x), where deg r(x) < deg m(x). But now
r(α) = 0 =⇒ r(x) = 0 =⇒ m(x) | f (x).
C
It follows from this Lemma that f (x) is the minimal polynomial of g, and
that it divides xa + xb − xc , say
xa + xb − xc = f (x)g(x),
where g(x) ∈ Fp [x].
But this is an identity in Fp [x] so it will also hold in F 0 . Substituting
x = g0,
g 0a + g 0b = g 0c .
The same argument shows that
g a + g b = 0 =⇒ g 0a + g 0b = 0.
Thus the map Φ : F → F 0 preserves addition as well as multiplication,
and so is a homomorphism. But a homomorphism between fields is necessarily injective. For suppose Φ(a) = 0, where a 6= 0. Let b be the inverse of
a. Then Φ(1) = Φ(ab) = Φ(a)Φ(b) = 0, and so Φ(c) = 0 for all c.
Finally, since Φ is injective, and |F | = |F 0 |, it follows that Φ is bijective,
ie an isomorphism between F and F 0 . J
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C.6
Existence
Theorem C.3 There exists a field of order pn for each prime-power pn .
Proof I We argue by induction on n. Suppose n = mq, where q is a prime.
By the inductive hypothesis there is a field F = Fpm of order pm .
Lemma 25 Suppose f (x) ∈ k[x] is an irreducible polynomial of degree d.
Then the quotient ring K = k[x]/(f (x)) is a field extension of k, of dimension
d as a vector space over k. Moreover K = k(α) where f (α) = 0, ie each
element θ ∈ K is expressible in the form θ = g(α) with g(x) ∈ k[x].
Proof of Lemma B We know that if p is a prime then Z/(p) is a field. The
argument was based on the Euclidean Algorithm for Z. But the Euclidean
Algorithm holds equally for the ring k[x], showing that if two polynomials
f (x), g(x) ∈ k[x] are coprime then there exist polynomials u(x), v(x) ∈ k[x]
such that
u(x)f (x) + v(x)g(x) = 1.
In particular, if f (x) is irreducible and does not divide g(x) the g(x)
has inverse v(x) modulo f (x). Hence K = k[x]/(f (x)) is a field. Moreover
α = x mod f (x) satisfies f (α) = 0. C
Corollary 26 k[x]/(f (x)) is a vector space over k of dimension d.
For it is easy to see that 1, x, x2 , . . . , xd−1 form a basis for this vector
space. J
n
Consider the factorisation of U (x) = xp −1 −1 into irreducible polynomials
over F :
U (x) = f1 (x) · · · fr (x).
The factors cannot all be of degree 1; for that would imply that all pn roots
of U (x) are in F , which is impossible since |F | = pm < pn .
Suppose f (x) is an irreducible factor of U (x) over F of degree > 1. We
claim that the field K = F [x]/(f (x)) must be of order pn .
By the Lemma, K = F (α), and
f (α) = 0 =⇒ U (α) = 0 =⇒ αp
n −1
= 1.
Lemma 26 If α1 , . . . , αr ∈ k, a field of characteristic p then
n
n
n
(α1 + · · · + αr )p = α1p + · · · + αrp .
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Proof of Lemma B This follows from the binomial theorem, since the cross
terms are all multiples of p, and p = 0 in a field of characteristic p.
C
Lemma 27 pm − 1|pn − 1 ⇐⇒ m | n.
Proof of Lemma B Suppose n = md. Then
xd − 1 = (x − 1)(xd−1 + xd−2 + · · · + 1).
The result follows on setting x = pm . C
m
We know that every c ∈ F satisfies cp = c. It follows from the Lemma
n
that cp = c.
n
Lemma 28 Every element a ∈ K satisfies ap = a
Proof of Lemma B We know that a = g(α) where
g(x) = xd + c1 xd−1 + · · · cd ,
with ci ∈ F . Hence
n
n
n
n
n
cp = αdp + cp1 α(d−1)p + · · · + cpd
= αd + c1 αd−1 + · · · + cd
= g(α) = c.
C
We know K is larger than F , so
|K| = pe
for some e > m.
It follows that there is an element of order pe − 1 in K × . On the other
n
hand each element c ∈ K × satisfies cp −1 = 1. Hence
pe − 1 | pn − 1 =⇒ e | n =⇒ e = n,
since n = mq with q prime. Thus
K = Fp n .
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Exercises 3
In
** 1.
** 2.
** 3.
** 4.
*** 5.
** 6.
*** 7.
** 8.
** 9.
*** 10.
** 11.
*** 12.
*** 13.
*** 14.
*** 15.
Finite Fields
exercises 1–5 determine all the primitive roots in the given field.
F3
F5
F7
F1 1
F29
In exercises 6–10 determine all the solutions of the given equation in
F19
x2 = 1
x3 = 1
x4 = 1
x5 = 1
x2 + 3x + 1 = 0
Draw up the addition and multiplication tables for F4 .
Show from first principles that there is no field with 5 elements.
Show that the map φ : x 7→ xp is an automorphism of Fpn . (This is
known as the Frobenius automorphism.)
Show that the only automorphisms of Fpn are 1, φ, φ2 , . . . , φn−1 .
Show that Fpn contains a subfield isomorphic to Fpm if and only if
m | n.
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