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assume that K contains the prime subп¬Ѓeld Fp , i.e., Fp K.

5.1. Finite п¬Ѓelds
If K is a п¬Ѓnite п¬Ѓeld, then K is an Fp -vector space. Our п¬Ѓrst goal is to count the elements of
K. Here is a more general result.
Lemma 5.1. Let F be a п¬Ѓnite п¬Ѓeld with q elements and let V be an F -vector space. Then
dimF V < в€ћ if and only if V is п¬Ѓnite in which case |V | = q dimF V .
Proof. If d = dimF V < в€ћ, then for a basis v1 , . . . , vd we can express each element v в€€ V
uniquely in the form v = t1 v1 + В· В· В· + td vd , where t1 , . . . , td в€€ F . Clearly there are exactly q d
such expressions, so |V | = q d .
Conversely, if V is п¬Ѓnite then any basis has п¬Ѓnitely many elements and so dimF V < в€ћ.
Corollary 5.2. Let F be a п¬Ѓnite п¬Ѓeld and E/F an extension. Then E is п¬Ѓnite if and only
if E/F is п¬Ѓnite and then |E| = |F |[E:F ] .
Corollary 5.3. Let K be a п¬Ѓnite п¬Ѓeld. Then K/Fp is п¬Ѓnite and |K| = p[K:Fp ] .
Our next task is to show that for each power pd there is a п¬Ѓnite п¬Ѓeld with pd elements. We
d
start with the algebraic closure Fp of Fp and consider the polynomial О˜pd (X) = X p в€’ X в€€
Fp [X]. Notice that О˜pd (X) = в€’1, hence by Proposition 3.55 every root of О˜pd (X) in Fp is
simple. Therefore by Corollary 1.34 О˜pd (X) must have exactly pd distinct roots in Fp , say
0, u1 , . . . , upd в€’1 . Then in Fp [X] we have
d
X p в€’ X = X(X в€’ u1 ) В· В· В· (X в€’ upd в€’1 ),
and each root is separable over Fp . Let
F0d = {u в€€ Fpd : u = 0}.
Fpd = {u в€€ Fp : О˜pd (u) = 0} вЉ† Fp , p
d в€’1
Notice that u в€€ F0d if and only if up = 1.
p

1, Fpd is a п¬Ѓnite subп¬Ѓeld of Fp with pd elements and
Proposition 5.4. For each d
F0d = FГ— . Furthermore, the extension Fpd /Fp is separable.
pd
p

Proof. If u, v в€€ Fpd then by the IdiotвЂ™s Binomial Theorem 1.11,
d d d d d
(u + v)p в€’ (u + v) = (up + v p ) в€’ (u + v) = (up в€’ u) + (v p в€’ v) = 0,
d d d
(uv)p в€’ uv = up v p в€’ uv = uv в€’ uv = 0.
d d
Furthermore, if u = 0 then up в€’1 = 1 and so u has multiplicative inverse up в€’2 . Hence Fpd Fp .
Notice that Fp Fpd , so Fpd /Fp is a п¬Ѓnite extension. In any п¬Ѓeld the non-zero elements are
always invertible, hence F0d = FГ— .
pd
p

Fp is called the Galois п¬Ѓeld of order pd .
Definition 5.5. The п¬Ѓnite subп¬Ѓeld Fpd
59
60 5. GALOIS EXTENSIONS FOR FIELDS OF POSITIVE CHARACTERISTIC

Fpd is often denoted GF(pd ). Of course, Fp1 = GF(p1 ) = Fp . Notice also that [Fpd : Fp ] = d.
Proposition 5.6. Let d 1.
d d
Fp is the splitting subп¬Ѓeld for each of the polynomials X p в€’ X and X p в€’1 в€’ 1
(i) Fpd
over Fp .
(ii) Fpd Fp is the unique subп¬Ѓeld with pd elements.
(iii) If K is any п¬Ѓeld with pd elements then there is an monomorphism K в€’в†’ Fp with image
Fpd , hence K в€ј Fpd .
=
Proof. (i) As Fpd consists of exactly the roots of О˜pd (X) in Fp , it is the splitting subп¬Ѓeld.
d
The non-zero elements of Fpd are the roots of X p в€’1 в€’ 1, so Fpd is also the splitting subп¬Ѓeld for
this polynomial.
Fp have pd elements. Notice that the non-zero elements of F form a group K Г—
(ii) Let K
under multiplication. This group is abelian and has pd в€’ 1 elements, so by LagrangeвЂ™s Theorem,
d d
each element u в€€ K Г— has order dividing pd в€’ 1, therefore up в€’1 = 1 and so up = u. But this
means every element of K is a root of О˜pd (X) and so K Fpd ; equality follows since these
subп¬Ѓelds both have pd elements.
(iii) Apply the Monomorphism Extension Theorem 3.49 for K = E = Fp and L = K.
It is worth noting the following consequence of this result and the construction of Fpd .
Corollary 5.7. Let K be a п¬Ѓnite п¬Ѓeld of characteristic p. Then K/Fp is separable.
Example 5.8. Consider the polynomial X 4 в€’ X в€€ F2 [X]. By inspection, in the ring F2 [X]
we п¬Ѓnd that
X 4 в€’ X = X 4 + X = X(X 3 + 1) = X(X + 1)(X 2 + X + 1).
Now X 2 + X + 1 has no root in F2 so it must be irreducible in F2 [X]. Its splitting п¬Ѓeld is a
quadratic extension F2 (w)/F2 where w is one of the roots of X 2 + X + 1, the other being w + 1
since the sum of the roots is the coeп¬ѓcient of X. This tells us that every element of F4 = F2 (w)
can be uniquely expressed in the form a + bw with a, b в€€ F2 . To calculate products we use the
fact that w2 = w + 1, so for a, b, c, d в€€ F2 we have
(a + bw)(c + dw) = ac + (ad + bc)w + bdw2 = (ac + bd) + (ad + bc + bd)w.
Example 5.9. Consider the polynomial X 9 в€’ X в€€ F3 [X]. Let us п¬Ѓnd an irreducible poly-
nomial of degree 2 in F3 [X]. Notice that X 2 + 1 has no root in F3 , hence X 2 + 1 в€€ F3 [X] is
irreducible; so if u в€€ F3 is a root of X 2 + 1 then F3 (u)/F3 has degree 2 and F3 (u) = F9 . Every
element of F9 can be uniquely expressed in the form a + bu with a, b в€€ F3 . Multiplication is
carried out using the relation u2 = в€’1 = 2.
By inspection, in the ring F3 [X] we п¬Ѓnd that
X 9 в€’ X = X(X 8 в€’ 1) = (X 3 в€’ X)(X 2 + 1)(X 2 + X в€’ 1)(X 2 в€’ X в€’ 1).
So X 2 + X в€’ 1 and X 2 в€’ X в€’ 1 are also quadratic irreducibles in F3 [X]. We can п¬Ѓnd their roots
in F9 using the quadratic formula since in F3 we have 2в€’1 = (в€’1)в€’1 = в€’1. The discriminant of
X 2 + X в€’ 1 is
1 в€’ 4(в€’1) = 5 = 2 = u2 ,
so its roots are (в€’1)(в€’1 В± u) = 1 В± u. Similarly, the discriminant of X 2 в€’ X в€’ 1 is
1 в€’ 4(в€’1) = 5 = 2 = u2
and its roots are (в€’1)(1 В± u) = в€’1 В± u. Then we have
F9 = F3 (u) = F3 (1 В± u) = F3 (в€’1 В± u).
There are two issues we can now clarify.
Proposition 5.10. Let Fpm and Fpn be two Galois п¬Ѓelds of characteristic p. Then Fpm
Fpn if and only if m | n.
5.1. FINITE FIELDS 61

Proof. If Fpm Fpn , then by Corollary 5.2,
pn = (pm )[Fpn :Fpm ] ,
so m | n.
m
1. Then for u в€€ Fpm we have up = u, so
If m | n, write n = km with k
n mk m m(kв€’1) m(kв€’1) m
up = up = (up )p = up = В· В· В· = up = u.
Hence u в€€ Fpn and therefore Fpm Fpn .
This means that we can think of the Galois п¬Ѓelds Fpn as ordered by divisibility. Here is the
diagram of subп¬Ѓelds for Fp24 showing extensions with no intermediate subextensions.
Fp24
(5.1)
{{
{{
{{
{{
Fp8 Fp12
{{
{{
{{
{{
Fp4 Fp6
gg
{{ gg
{{ gg
{{ gg
{{
Fp2 Fp3
gg
{{
gg
{{
gg
{{
gg
{{
Fp
Theorem 5.11. The algebraic closure of Fp is the union of all the Galois п¬Ѓelds of charac-
teristic p,
Fp = Fpn .
n>1
Furthermore, each element u в€€ Fp is separable over Fp .
Proof. Let u в€€ Fp . Then u is algebraic over Fp and the extension Fp (u)/Fp is п¬Ѓnite. Hence
by Corollary 5.2, Fp (u) Fp is a п¬Ѓnite subп¬Ѓeld. Proposition 5.10 now implies that Fp (u) = Fpn
for some n. The separability statement follows from Corollary 5.7.
We will require a useful fact about Galois п¬Ѓelds.
Proposition 5.12. The group of units FГ— in Fpd is cyclic.
pd

This is a special case of a more general result about arbitrary п¬Ѓelds.
K Г— is cyclic.
Proposition 5.13. Let K be a п¬Ѓeld. Then every п¬Ѓnite subgroup U
Proof. Use Corollary 1.34 and Lemma 1.45.
Definition 5.14. w в€€ FГ— is called a primitive root if it is a primitive (pd в€’ 1)-th root of
pd
unity, i.e., its order in the group FГ— is (pd в€’ 1), hence w = FГ— .
pd pd

Remark 5.15. Unfortunately the word primitive has two confusingly similar uses in the
context of п¬Ѓnite п¬Ѓelds. Indeed, some authors use the term primitive element for what we have
called a primitive root, but that conп¬‚icts with our usage, although as we will in the next result,
every primitive root is indeed a primitive element in our sense!
Proposition 5.16. The extension of Galois п¬Ѓelds Fpnd /Fpd is simple, i.e., Fpnd = Fpd (u)
for some u в€€ Fpnd .
Proof. By Proposition 5.12, Fpnd has a primitive root w say. Then every element of Fpnd
is a polynomial in w, so Fpnd Fpd (w) Fpnd , hence Fpnd = Fpd (w).
62 5. GALOIS EXTENSIONS FOR FIELDS OF POSITIVE CHARACTERISTIC

Remark 5.17. This completes the proof of the Primitive Element Theorem 3.75 which we
had previously only established for inп¬Ѓnite п¬Ѓelds.
Example 5.18. In Example 5.8 we п¬Ѓnd that F4 = F2 (w) has the two primitive roots w and
w + 1.
Example 5.19. In Example 5.9 we have F9 = F3 (u) and FГ— is cyclic of order 8. Since
9
П•(8) = 4, there are four primitive roots and these are the roots of the polynomials X 2 + X в€’ 1
and X 2 в€’ X в€’ 1 which we found to be В±1 В± u.
We record a fact that is very important in Number Theory.
Proposition 5.20. Let p > 0 be an odd prime.
(i) If p в‰Ў 1 (mod 4), the polynomial X 2 + 1 в€€ Fp [X] has two roots in Fp .
(ii) If p в‰Ў 3 (mod 4) the polynomial X 2 +1 в€€ Fp [X] is irreducible, so Fp2 в€ј Fp [X]/(X 2 +1).
=
Proof. (i) We have 4 | (p в€’ 1) = |FГ— |, so if u в€€ FГ— is a generator of this cyclic group, the
p p
Г— Г—
order of u|Fp |/4 is 4, hence this is a root of X 2 + 1 (the other root is в€’u|Fp |/4 ).
(ii) If v в€€ Fp is a root of X 2 + 1 then v has order 4 in FГ— . But then 4 | (p в€’ 1) = |FГ— |, which is
p p
impossible since p в€’ 1 в‰Ў 2 (mod 4).
Here is a generalization of Proposition 5.20.
Proposition 5.21. Fpd contains a primitive n-th root of unity if and only if pd в‰Ў 1 (mod n)
and p n.

5.2. Galois groups of п¬Ѓnite п¬Ѓelds and Frobenius mappings
Consider an extension of Galois п¬Ѓelds Fpnd /Fpd . By Proposition 5.6(i), Corollary 5.7 and
Proposition 3.73, this extension is Galois and
| Gal(Fpnd /Fpd )| = [Fpnd : Fpd ] = n.
We next introduce an important element of the Galois group Gal(Fpnd /Fpd ).
Definition 5.22. The (relative) Frobenius mapping for the extension Fpnd /Fpd is the func-
d
tion Fd : Fpnd в€’в†’ Fpnd given by Fd (t) = tp .
Proposition 5.23. The relative Frobenius mapping Fd : Fpnd в€’в†’ Fpnd is an automor-
phism of Fpnd that п¬Ѓxes the elements of Fpd , so Fd в€€ Gal(Fpnd /Fpd ). The order of Fd is n,
so Gal(Fpnd /Fpd ) = Fd , the cyclic group generated by Fd .
Proof. For u, v в€€ Fpnd , we have the identities
d d d d d d
Fd (u + v) = (u + v)p = up + v p , Fd (uv) = (uv)p = up v p ,
so Fd is a ring homomorphism. Also, for u в€€ Fpd we have
d
Fd (u) = up = u,
so Fd п¬Ѓxes the elements of Fpd . To see that Fd is an automorphism, notice that the composition
power Fn = Fd в—¦ В· В· В· в—¦ Fd (with n factors) satisп¬Ѓes
d
nd
Fn (t) = tp =t
d
for all t в€€ Fpnd , hence Fn = id. Then Fd is invertible with inverse Fв€’1 = Fnв€’1 . This also shows
d d d
that the order of Fd in the group AutFpd (Fpnd ) is at most n. Suppose the order is k with k n;
kd
then every element u в€€ Fpnd satisп¬Ѓes the equation Fk (u) = u which expands to up = u, hence
d
u в€€ Fpkd . But this can only be true if k = n.
Frobenius mappings exist on the algebraic closure Fp . For d 1, consider the function
d
Fd (t) = tp .
Fd : Fp в€’в†’ Fp ;
5.2. GALOIS GROUPS OF FINITE FIELDS AND FROBENIUS MAPPINGS 63

Proposition 5.24. Let d 1.
(i) Fd : Fp в€’в†’ Fp is an automorphism of Fp which п¬Ѓxes the elements of Fpd . In fact for
u в€€ Fp , Fd (u) = u if and only if u в€€ Fpd .
(ii) The restriction of Fd to the Galois subп¬Ѓeld Fpdn agrees with the relative Frobenius
mapping Fd : Fpnd в€’в†’ Fpnd .
(ii) If k 1, then Fk = Fkd , so Fd has inп¬Ѓnite order in the automorphism group AutFpd (Fp ),
d
hence this group is inп¬Ѓnite.
Proof. This is left as an exercise.
The Frobenius mapping F = F1 is often called the absolute Frobenius mapping since it exists
as an element of each of the groups AutFp (Fp ) and AutFp (Fpn ) = Gal(Fpn /Fp ) for every n 1.
In Gal(Fpnd /Fpd ) = Fd , for each k with k | n there is the cyclic subgroup Fk of order
d
k
| Fd | = n/k.
Fk
Fk d
Proposition 5.25. For k | n, the п¬Ѓxed subп¬Ѓeld of in Fpnd is Fpnd = Fpdk .
d

Fpnd

n/k

Fk
d
Fpnd = Fpdk
k

Fpd
dk
Proof. For u в€€ Fpnd we have Fk (u) = up , hence Fk (u) = u if and only if u в€€ Fpdk .
d d

Here is the subgroup diagram corresponding to the lattice of subп¬Ѓelds of Fp24 shown in (5.1).
64 5. GALOIS EXTENSIONS FOR FIELDS OF POSITIVE CHARACTERISTIC
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