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(5.2)
Gal(Fp24 /Fp ) = F ∼ Z/24 Fp24
=
‚‚‚
lll {{
‚‚‚
lll {{
‚‚‚
lll {{
‚‚‚
lll {{
‚‚‚
lll
Fp8 Fp12
F2 ‚‚ F3
lll
‚‚‚ {{
lll
‚‚‚ {{
l
‚‚‚ lll {{
lll
‚‚‚
{{
l
‚‚‚
lll Fp4 Fp6
F4 ‚‚ F6 gg
{{
‚‚‚ gg
‚‚‚ {{ gg
‚‚‚ {{ gg
‚‚‚ {{
‚‚‚
Fp2 Fp3
F8 ‚‚ F12 gg
{{
‚‚‚ gg
{{
‚‚‚ gg
{{
‚‚‚ gg
{{
‚‚‚

Fp
F24 = {id}

5.3. The trace and norm mappings
For an extension of Galois ¬elds Fpnd /Fpd , consider the function TrFpnd /Fpd : Fpnd ’’ Fpnd
de¬ned by
d 2d (n’1)d
TrFpnd /Fpd (u) = u + up + up + · · · + up = u + Fd (u) + F2d (u) + · · · + F(n’1)d (u).
Notice that
d 2d 3d nd
Fd (TrFpnd /Fpd (u)) = up + up + up + · · · + up
d 2d 3d (n’1)d
= up + up + up + · · · + up + u = TrFpnd /Fpd (u).
So by Proposition 5.24(i), TrFpnd /Fpd (u) ∈ Fpd . Therefore we can rede¬ne this mapping to have
codomain Fpd , giving the relative trace TrFpnd /Fpd : Fpnd ’’ Fpd .
Proposition 5.26. The relative trace TrFpnd /Fpd is a surjective Fpd -linear mapping and
whose kernel is an Fpd -vector subspace of dimension n ’ 1.
d
Proof. Clearly TrFpnd /Fpd is additive. For t ∈ Fpd we have tp = t, so linearity follows from
the formula
d 2d (n’1)d d 2d (n’1)d
tu + (tu)p + (tu)p + · · · + (tu)p = tu + tup + tup + · · · + tup .
To see that TrFpnd /Fpd is surjective, notice that TrFpnd /Fpd (u) = 0 if and only if u is a root of
the polynomial
d 2d (n’1)d
X + Xp + Xp + · · · + Xp ∈ Fpd [X]
which has degree p(n’1)d and so has at most p(n’1)d < pnd roots in Fpnd . This means that
ker TrFpnd /Fpd cannot be the whole of Fpnd . TrFpnd /Fpd is surjective since its codomain has
dimension 1.
A multiplicative version of this construction can also be de¬ned. Consider the function
d 2d (n’1)d
NFpnd /Fpd : F— ’’ F— ; NFpnd /Fpd (u) = uup up · · · up = u Fd (u) F2d (u) · · · F(n’1)d (u).
pnd pd
Then we have
d 2d 3d nd
Fd (NFpnd /Fpd (u)) = up up up · · · up
d 2d 3d (n’1)d
= up up up · · · up u
d 2d 3d (n’1)d
= uup up up · · · up = NFpnd /Fpd (u).
65

So by Proposition 5.24(i), NFpnd /Fpd (u) ∈ Fpd . By rede¬ning the codomain we obtain the relative
norm
d 2d (n’1)d
NormFpnd /Fpd : F— ’’ F— ; NormFpnd /Fpd (u) = uup up · · · up .
pnd pd

Proposition 5.27. The relative norm NormFpnd /Fpd is a surjective group homomorphism.
Proof. Multiplicativity is obvious. The kernel of NormFpnd /Fpd consists of the roots in Fpnd
of the polynomial
d (n’1)d
X 1+p +···+p ’ 1 ∈ Fpd [X],
so
pnd ’ 1
d (n’1)d
| ker NormFpnd /Fpd | 1 + p + · · · + p =d .
p ’1
Hence
pnd ’ 1
pd ’ 1.
| im NormFpnd /Fpd | =
| ker NormFpnd /Fpd |
F— , we also have
Since im NormFpnd /Fpd pd

pd ’ 1,
| im NormFpnd /Fpd |
therefore
im NormFpnd /Fpd = F— .
pd



Exercises on Chapter 5

5.1. Show that Proposition 5.13 also applies to an integral domain in place of a ¬eld.
5.2. What happens to Theorem 5.20 if we try to take p = 2.
5.3. Let f (X) ∈ Fpd [X] be an irreducible polynomial with deg f (X) = n. Find the splitting ¬eld
of f (X). Deduce that for any other irreducible polynomial g(X) ∈ Fpd [X] with deg g(X) = n,
the splitting ¬elds of f (X) and g(X) over Fpd agree.
5.4. Find the smallest Galois ¬elds containing all the roots of the following polynomials:
(a) X 8 ’ 1 ∈ F41 [X]; (b) X 8 ’ 1 ∈ F5 [X]; (c) X 8 ’ 1 ∈ F11 [X]; (d) X 8 ’ 1 ∈ F2 [X].
In each case ¬nd a primitive root of this Galois ¬eld.
5.5. Let w ∈ F— be a primitive root. If < d, show that w ∈ F— . Deduce that degFp w = d
/p
pd
and d | •(pd ’ 1).
5.6. Let p > 0 be a prime. Suppose that d 1, and K/Fpd is an extension. For a ∈ K, let
d
ga (X) = X p ’ X ’ a ∈ K[X].
(a) If the polynomial ga (X) is irreducible over K, show that the splitting ¬eld E of ga (X)
over K is separable and Gal(E/K) ∼ Fpd . [Hint: show that if u ∈ E is a root of ga (X)
=
in an extension E/K, then so is u + t for every t ∈ Fp .]
(b) If d = 1, show that ga (X) is irreducible over K if and only if it has no root in K.
(c) If K is a ¬nite ¬eld and d > 1, explain why ga (X) can never be irreducible over K.
CHAPTER 6


A Galois Miscellany

In this chapter we will explore some miscellaneous topics in Galois Theory. Historically,
Galois Theory has always been an important tool in Number Theory and Algebra, stimulating
the development of subjects such as Group Theory, Ring Theory and such diverse areas as
Di¬erential Equations, Complex Analysis and Algebraic Geometry. Many of the ideas we will
meet in this chapter are of great importance in these mathematical areas.

6.1. A proof of the Fundamental Theorem of Algebra
We will prove the Fundamental Theorem of Algebra for the complex numbers C. This proof
is essentially due to Gauss but he did not use the historically more recent Sylow theory. It is
interesting to compare the proof below with others which use the topology of the plane and circle
or Complex Analysis; our proof only uses the connectivity of the real line (via the Intermediate
Value Theorem) together with explicit calculations in C involving square roots.
Theorem 6.1 (The Fundamental Theorem of Algebra). The ¬eld of complex numbers C is
algebraically closed and R = C.
Proof. We know that [C : R] = 2, so C/R is algebraic. Let p(X) ∈ C[X] be irreducible.
Then any root u of p(X) in the algebraic closure C is algebraic over R, so in C[X] we have
p(X) | minpolyR,u (X). The splitting ¬eld of p(X) over C is contained in the splitting ¬eld E of
minpolyR,u (X)(X 2 + 1) over R. Since C E, we have 2 | [E : R] and so 2 | | Gal(E/R)|.
Now consider a 2-Sylow subgroup P Gal(E/R) and recall that | Gal(E/R)|/|P | is odd.
For the ¬xed sub¬eld of P , we have
| Gal(E/R)|
[E P : R] = ,
|P |
which shows that E P /R has odd degree. The Primitive Element Theorem 3.75 allows us to
write E P = R(v) for some v whose minimal polynomial over R must also have odd degree.
But by the Intermediate Value Theorem, every real polynomial of odd degree has a real root,
so irreducibility implies that v has degree 1 over R and therefore E P = R. This shows that
Gal(E/R) = P , hence Gal(E/R) is a 2-group.
As C/R is a Galois extension, we can consider the normal subgroup Gal(E/C) Gal(E/R) for
which | Gal(E/R)| = 2 | Gal(E/C)|. We must show that | Gal(E/C)| = 1, so suppose not. From
the theory of 2-groups, there is a normal subgroup N Gal(E/C) of index 2, so we can consider
the Galois extension E N /C of degree 2. But from known properties of C (see Proposition 3.29),
every quadratic aX 2 + bX + c ∈ C[X] has complex roots (because we can ¬nd square roots
of every complex number). So we cannot have an irreducible quadratic polynomial in C[X].
Therefore | Gal(E/C)| = 1 and E = C.

6.2. Cyclotomic extensions
We begin by discussing the situation for cyclotomic extensions over Q using material dis-
cussed in Section 1.3. Let ζn = e2πi/n , the standard primitive n-th root of 1 in C. In Theo-
rem 1.42, it was claimed that the irreducible polynomial over Q which has ζn as a root was the
n-th cyclotomic polynomial
t
¦n (X) = (X ’ ζn ).
t=1,...,n’1
gcd(t,n)=1

67
68 6. A GALOIS MISCELLANY

Theorem 6.2. Let n 2. Then
• Q(ζn ) = Q[X]/(¦n (X));
• [Q(ζn ) : Q] = •(n);
• Gal(Q(ζn )/Q) ∼ (Z/n)— , where the element tn ∈ (Z/n)— acts on Q(ζn ) by tn · ζn = ζn .
t
=

t
Proof. Since the complex roots of ¦n (X) are the powers ζn with t = 1, . . . , n ’ 1
™™¦

t
and gcd(t, n) = 1, Q(ζn ) is the splitting ¬eld of ¦n (X) over Q and indeed Q(ζn ) = Q(ζn )
t
whenever t has the above properties and so ζn is a primitive n-th root of unity. The main step
in the proof is to show that ¦n (X) ∈ Z[X] is irreducible. To do this we will show that every
t
power ζn as above is actually a Galois conjugate of ζn over Q, hence
¦n (X) = minpolyQ,ζn (X) = minpolyQ,ζn (X)
t


and so is irreducible.
Consider
r
Z(ζn ) = {a0 + a1 ζn + · · · + ar ζn : r 0, aj ∈ Z} ⊆ Q(ζn ).
Then Z(ζn ) is a subring of Q(ζn ) and so is an integral domain. Its group of units contains the
cyclic subgroup ζn of order n.
Let p > 0 be a prime which does not divide n. Let P Z(ζn ) be a maximal ideal which
contains p; then the quotient ring Z(ζn )/P is a ¬eld of characteristic p. In fact, it is a ¬nite
¬eld, say Fpd for some d. Let π : Z(ζn ) ’’ Fpd be the quotient homomorphism.
Z(ζn )— ; this
Inside the group of units of Z(ζn ) is the subgroup of powers of ζn , ζn
is a cyclic subgroup of order n. We claim that when restricted to ζn , π gives an injective
group homomorphism, π : ζn ’’ F— . To see this, suppose that π (ζn ) = 1 for some r =
r
pd
r
1, 2, . . . , n ’ 1; then ζn ’ 1 ∈ P . By elementary Group Theory we can assume that r | n and so
p r. Factoring, we have
r’1
(ζn ’ 1)(ζn + · · · + ζn + 1) ≡ (ζn ’ 1)r (mod P ),
so ζn ’ 1 ∈ P or r ∈ P since maximal ideals are prime. But Z © P = (p) and so r ∈ P , hence
/
ζn ’ 1 ∈ P . Recalling that
n’1
ζn + · · · + ζn + 1 = 0,
we see that n ∈ P and hence p | n, thus contradicting our original assumption on n. So π is
injective.
Writing u = π (u), we can consider the e¬ect of the absolute Frobenius map F : Fpd ’’ Fpd
t t
on ζ n = ζn ,
t t tp
F(ζ n ) = (ζ n )p = ζn .
t tp
This shows that in the Galois extension Fpd /Fp , ζ n is conjugate to ζn ; by iterating this we ¬nd
t k
tp
that ζ n is conjugate to every power of the form ζn .
Now let t = 1, . . . , n ’ 1 and gcd(t, n) = 1. Suppose there is a factorization
¦n (X) = f (X) minpolyQ,ζn (X)
t
for some monic polynomial f (X) ∈ Z[X] and f (ζn ) = 0. Consider the prime power factorization
t = pr1 · · · prm , where the pj are primes with 2 p1 < · · · < pm and rj 1 with. Notice that
m
1
pj n since gcd(t, n) = 1.
Now consider a maximal ideal P1 Z[ζn ] containing p1 . Reducing modulo P1 and working
r1
p
in the resulting extension Fpd1 /Fp1 , we ¬nd that ζ n is conjugate to ζ n1 . By separability and

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