Table Of ContentGalois Theory
Prof. A.J. Scholl
Michaelmas 2005
LATEXed by Sebastian Pancratz
ii
These notes are based on a course of lectures given by Prof. A.J. Scholl in Part II of the
Mathematical Tripos at the University of Cambridge in the academic year 2005(cid:21)2006.
These notes have not been checked by Prof. A.J. Scholl and should not be regarded as
o(cid:30)cial notes for the course. In particular, the responsibility for any errors is mine (cid:22)
please email Sebastian Pancratz (sfp25) with any comments or corrections.
Contents
1 Polynomials 1
2 Symmetric Polynomials 3
3 Fields and Extensions 7
4 Algebraic Elements and Extensions 11
5 Algebraic and Transcendental Numbers in R and C 15
6 Splitting Fields 21
7 Separability 25
8 Algebraic Closure 27
9 Field Automorphisms and Galois Extensions 31
10 The Characterisation of Finite Galois Extensions 35
11 The Galois Correspondence 37
12 Finite Fields 41
13 Cyclotomic and Kummer Extensions 45
14 Trace and Norm 51
15 Solving Equations by Radicals 55
Chapter 1
Polynomials
Let R be any commutative ring with an identity element 1. The ring of polynomials is
( n )
X
R[X] = a Xi: n ∈ N∧a ,...,a ∈ R .
i 1 n
i=0
More generally, R[X ,...,X ] is the ring of polynomials in several variables. It is
1 n
important to distinguish between a polynomial and the function that it might represent.
If f(X) = Pn a Xi ∈ R[X] then it determines a function R → R,b 7→ Pn a bi =
i=0 i i=0 i
f(b). The function b 7→ f(b) does not in general determine f(X) uniquely, e.g. if
R = F = Z , f(X) = Xp, g(X) = X then by Fermat’s Little Theorem f(b) = g(b) for
p pZ
all b ∈ F but f 6= g.
p
Here, we are mostly interested in the case R = K, a (cid:28)eld. Recall that K[X] is a
Euclidean domain, so if f,g ∈ K[X] with g 6= 0 then there exist unique q,r ∈ K[X]
such that
(i) f = qg+r;
(ii) degr < degg.
A particular case is when g = X−a is linear, then we get f(X) = (X−a)q(X)+f(a),
since r is constant as degr < 1 = deg(X−a). This is known as the remainder theorem.
As a consequence, K[X] is a unique factorisation domain (UFD) and a principal
ideal domain (PID), so that greatest common divisors exist, and if f,g ∈ K[X] then
gcd(f,g) = pf +qg for some p,q ∈ K[X].
Proposition 1.1. Let K be a (cid:28)eld, 0 6= f ∈ K[X]. Then f has at most degf roots in
K.
Proof. If f has no roots, there is nothing to prove. Otherwise, let c ∈ K be a root. So
f(X) = (X −c)q(X) by the division algorithm, and degq = degf −1. So if b is any
root of f, then (b−c)q(b) = 0, so c = b or b is a root of q (as K is a (cid:28)eld). So
|{roots of f}| ≤ 1+|{roots of q}|
≤ 1+degq
= degf
by induction on the degree of f.
Remark. Consider X2 − 1 ∈ R[X], R = Z/8Z. This has four roots ±1,±3, so the
assumption R a (cid:28)eld is essential.
Lecture 1
Lecture 1
Chapter 2
Symmetric Polynomials
For n ≥ 1 let S denote the symmetry group of degree n, i.e. the permutations of
n
{1,...,n}.
De(cid:28)nition. Let R be a ring, X ,...,X be indeterminates. A polynomial
1 n
f(X ,...,X ) ∈ R[X ,...,X ] is symmetric if
1 n 1 n
∀σ ∈ S f(X ,...,X ) = f(X ,...,X ).
n σ(1) σ(n) 1 n
If f,g are symmetric, clearly so are f + g and fg. Also constant polynomials are
symmetric. SothesetofsymmetricpolynomialsisasubringofR[X ,...,X ]containing
1 n
R.
Example. Power sums p = Xr +···+Xr for any r ≥ 0.
r 1 n
Another way of expressing the de(cid:28)nition is as follows. The group S acts on
n
R[X ,...,X ] by σ: f 7→ f(X ,...,X ) = fσ, say. This is a group action,
1 n σ(1) σ(n)
(fσ)τ = fστ. The ring of symmetric polynomials is just the set of elements of
R[X ,...,X ] (cid:28)xed by S , called invariants.
1 n n
Elementary Symmetric Polynomials
Consider
n
Y
(T +X ) = (T +X )(T +X )···(T +X )
i 1 2 n
i=1
= Tn+(X +···+X )Tn−1+···+X ···X
1 n 1 n
De(cid:28)ne s = s to be the coe(cid:30)cient of Tn−r. So
r r,n
X
s = X ···X .
r i1 ir
1≤i1<···<ir≤n
Forexample,ifn = 3thens = X +X +X ,s = X X +X X +X X ,s = X X X .
1 1 2 3 2 1 2 1 3 2 3 3 1 2 3
We always have s = 1 and by convention s = 0 if r > n.
0,n r,n
Theorem 2.1 (Newton, Main Theorem of Symmetric Functions). (i) Every sym-
metric polynomial in R[X ,...,X ] is a polynomial in {s }n with coe(cid:30)cients
1 n r,n r=0
in R.
Lecture 2
4 Symmetric Polynomials
(ii) There are no non-trivial relations between the elementary symmetric polynomials.
The meaning of (ii) is the following. Consider the map
ϑ: R[Y ,...,Y ] → R[X ,...,X ],G(Y ,...,Y ) 7→ G(s ,...,s ).
1 n 1 n 1 n 1 n
This clearly is a homomorphism. (ii) means that kerϑ = {0}, so ϑ de(cid:28)nes an isomor-
phism between R[Y ,...,Y ] and the ring of symmetric polynomials in X ,...,X .
1 n 1 n
Proof. (i) If I = (i ,...,i ) with i ≥ 0 for all k = 1,...,n, write X =
1 n k I
Xi1···Xin. Any polynomial of this form is called a monomial. The polynomials
1 n
inR[X ,...,X ]arejustR-linearcombinationsofmonomials. Thedegree ofX is
1 n I
i +···+i . A polynomial is said to be homogenous if all the monomials occurring
1 n
in it (i.e. with non-zero coe(cid:30)cients) have the same degree. If f ∈ R[X ,...,X ],
1 n
we can uniquely write f = f +f +···+f for some d where each f is either 0 or
0 1 d k
is homogeneous of degree k. Permuting {X ,...,X } does not change the degree
1 n
ofthemonomials, soiff issymmetric, soarethehomogeneouspartsf ,f ,...,f .
0 1 d
So to prove (i), we may assume without loss of generality that f is homogeneous
of degree d, say.
De(cid:28)ne an ordering on the set of all monomials {X } as follows. We say X > X
I I J
if either i > j or if for some p > 1, i = j ,...,i = j and i > j . (This
1 1 1 1 p−1 p−1 p p
is called the lexicographical ordering.) This is a total ordering on {X }, i.e. for a
I
pair I,J exactly one of X > X ,X < X ,X = X holds.
I J I J I J
Suppose f is homogeneous of degree d and symmetric. Consider the monomial
X occurring in f which is greatest for lexicographical ordering, and let c ∈ R
I
be its coe(cid:30)cient. We claim that i ≥ i ≥ ··· ≥ i ; if not, say i < i ,
1 2 n p p+1
then if we interchange Xp and Xp+1 then the new monomial XI0 has exponents
I0 = (i1,...,ip−1,ip+1,ip,ip+2,...), so XI0 > XI. So
X = Xi1−i2(X X )i2−i3···(X ···X )in−1−in(X ···X )in.
I 1 1 2 1 n−1 1 n
Let g = si1−i2si2−i3···sin−1−insin, which is symmetric, homogeneous of degree d
1 2 n−1 n
and its leading term (highest monomial) is X (since the highest monomial in s
I r
is just X X ···X ).
1 2 r
Considerh = f−cg. Ifhisnon-zero,thenhishomogeneousofdegreed,symmetric,
and its highest monomial is less than X . As the number of monomials of given
I
degreeis(cid:28)nite,repeatingthisprocessultimatelyterminatesandwehaveexpressed
f as a polynomial in {s }.
r
(ii) Suppose G ∈ R[Y ,...,Y ] such that G(s ,...,s ) = 0. We want to show by
1 n 1,n n,n
induction on n that G = 0. If n = 1, there is nothing to prove.
If Yk divides G for some k > 0 then H = G/Yk is also a relation
n n
(H(s ,...,s ) = 0). So dividing by powers of Y , we may assume Y - G.
1,n n,n n n
Now substitute X = 0 to get
n
(
s r < n
r,n−1
s (X ,...,X ,0) =
r,n 1 n−1
0 r = n
and G(s ,...,s ,0) = 0. By induction, this means G(Y ,...,Y ,0) =
1,n−1 n−1,n−1 1 n−1
0, in other words, Y | G. So there are no non-zero relations.
n
Lecture 2
5
Example. Consider P X2X . The highest monomial occurring is X2X =
i6=j i j 1 2
X (X X ). We have
1 1 2
XX
s s = X X X
1 2 i j k
i j<k
X X
= X2X +3 X X X .
i j i j k
i6=j i<j<k
So P X2X = s s −3s .
i6=j i j 1 2 3
Considering similarly P X5 leads to Newton’s identities.
i i
Theorem 2.2 (Newton’s Formula). Let n ≥ 1. Let p = Xk +···+Xk. Then for all
k 1 n
k ≥ 1,
p −s p +···+(−1)k−1s p +(−1)kks = 0
k 1 k−1 k−1 1 k
Note that k = p and by convention s = 0 if r > n.
0 r
Proof. Consider F(T) = Qn (1−X T) = Pn (−1)rs Tr. We have
i=1 i r=0 r
F0(T) Xn −Xi
=
F(T) 1−X T
i
i=1
n ∞
1 XX
= − (X T)r
i
T
i=1 r=1
∞
1 X
= − p Tr
r
T
r=1
∴ −TF0(T) = s T −2s T2+···+(−1)n−1ns Tn
1 2 n
∞
X
= F(T) p Tr
r
r=1
= (s −s T +s T2+···+(−1)ns Tn)(p T +p T2+···)
0 1 2 n 1 2
Comparing coe(cid:30)cients of Tk gives the identity
k−1
X
(−1)k−1ks = (−1)rs p
k r k−r
r=0
Discriminant
Consider
Y
∆(X ,...,X ) = (X −X )
1 n i j
i<j
∆(X ,...,X ) = sgn(σ)∆(X ,...,X )
σ(1) σ(n) 1 n
for all σ ∈ S , so
n
Y
∆2(X ,...,X ) = (X −X )2
1 n i j
i<j
Lecture 3
6 Symmetric Polynomials
Y
= (−1)n(n−1)/2 (X −X )
i j
i6=j
is a symmetric polynomial.
Example. Let n = 2. Then
∆2(X ,X ) = (X −X )2 = (X +X )2−4X X
1 2 1 2 1 2 1 2
= s2−4s .
1 2
Let
n
Y
f(T) = (T −α )
i
i=1
= Tn−c Tn−1+···+(−1)nc
1 n
where c = s (α ,...,α ). Then the discrimant of f
r r 1 n
Disc(f) = ∆2(α ,...,α )
1 n
Y
= (α −α )2
i j
i<j
is a polynomial in {c } and vanishes if and only if f has a repeated root.
i
Example. If n = 2, f(T) = T2+bT +c then Disc(f) = b2−4c. (Note that Disc(f) =
s2−4s as above. But then s (α ,α ) = α +α = b and s (α ,α ) = α α = c.)
1 2 1 1 2 1 2 2 1 2 1 2
Lecture 3
