3E28 Galois Theory [Lecture notes]

65 Pages·2007·0.476 MB·English

by  Shaun Stevens

3E28 Galois Theory
Shaun Stevens
Spring Semester 2007
Contents
0 Recollections 3
1 Fields and polynomials 4
1.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Factorizing polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Splitting fields 12
2.1 Extension fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.4 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.5 Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.6 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 Galois theory 28
3.1 Automorphisms of field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.2 The Galois correspondence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.4 Further results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4 Solving polynomials by radicals 41
4.1 Cubic and quartic polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4.2 Kummer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.3 Soluble groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.4 Galois’s solvability criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.5 An insolvable polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
5 Finite fields 55
1
5.1 Galois groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
5.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
6 Further topics 59
6.1 Constructibility revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
2
0 Recollections
This section will appear as and when we need to recall things for the course, so check here again
for updates.
3
1 Fields and polynomials
1.1 Polynomials
Let F be a field. We consider the ring F[X] of polynomials with coefficients in F in more detail.
If f is a non-zero polynomial in F[X] then we can write
f(X) = a Xn+···+a X +a , a ∈ F, a 6= 0.
n 1 0 i n
Then the degree of f is deg(f) = n.
Note that the zero polynomial 0 does not have degree 0 – it is non-zero constant polynomials which
have degree 0. Depending on your preference, the zero polynomial should either have undefined
degree, or have degree −∞.
In F[X] we have a Division Algorithm: Let f,g be polynomials in F[X], with f 6= 0. Then there
exist polynomials q,r ∈ F[X] (q for quotient, r for remainder) with
g(X) = q(X)f(X)+r(X), and either r = 0 or deg(r) < deg(f).
This can be proved by induction on the degree of g.
For example, if f(X) = X2+1 and g(X) = 2X3+3X2+1 in Q[X] then
g(X) = (X2+1)(2X +3)+(−2X −2),
so q(X) = 2X +3 and r(X) = −2X −2.
Definition 1.1. Let f,g,h be polynomials in F[X]
(i) We say g divides f (and write g|f) if there exists q ∈ F[X] such that f = gq.
(ii) A greatest common divisor of f and g is a polynomial h ∈ F[X] which divides both f and
g, and such that, if q is any polynomial dividing both f and g then q divides h. Note that
gcd(f,g) is well-defined only up to units in F[X], that is up to multiplication by a non-zero
scalar.
(iii) We say that f,g are coprime if gcd(f,g) = 1. [Note that this really means gcd(f,g) is a
constant, since it could be any non-zero constant.]
(iv) We say f is irreducible if it is non-constant and, whenever we have a factorization f = gh,
either g or h is a unit (i.e. either g or h is constant); otherwise f is reducible.
Notice that, if f is an irreducible polynomial and g is any polynomial, then there are only two
possibilities for the greatest common divisor of f and g:
(
f, if f|g,
gcd(f,g) =
1, otherwise.
One important consequence of the Division Algorithm is that F[X] has a Euclidean Algorithm:
Let f,g be non-zero polynomials and q = gcd(f,g); then there exist polynomials u,v ∈ F[X] such
that
q(X) = u(X)f(X)+v(X)g(X).
4
This can be used to prove that any irreducible polynomial f ∈ F[X] is actually prime: that is, if f
divides gh, for some polynomials g,h ∈ F[X], then either f divides g or f divides h. From this it
follows that F[X] is a unique factorization domain: any polynomial in F[X] can be factorized as
a product of irreducible polynomials, and this factorization is unique up to the order of the factors
and multiplication of the factors by non-zero scalars.
Example. Consider again f(X) = X2+1 and g(X) = 2X3+3X2+1 in Q[X]. We can show that
these are coprime in two different ways:
(i) First we notice that f is irreducible: if not then it would have to factorize as a product of
two linear (degree 1) factors, f would have a root in Q; but f(α) = α2+1 ≥ 1, for all α ∈ Q,
so f has no roots in Q.
Now we saw above that
g(X) = (2X +3)f(X)+(−2X −2),
so f does not divide g. Hence (see the remarks after Definition 1.1), gcd(f,g) = 1.
(ii) Alternatively, we can use the Euclidean algorithm:
2X3+3X2+1 = (X2+1)(2X +3)+(−2X −2)
X2+1 = (−2X −2)(−1X + 1)+2
2 2
−2X −2 = 2(−X −1)+0
so gcd(f,g) = 2 (the last non-zero remainder), which is the same as gcd(f,g) = 1.
The advantage of this method is that we can use back-substitution in this calculation to find
polynomials u,v such that uf +vg = 1:
2 = (X2+1)−(−2X −2)(−1X + 1)
2 2
= f(X)−(g(X)−(2X +3)f(X))(−1X + 1)
2 2
= (1X − 1)g(X)+[1+(2X +3)(−1X + 1)]f(X).
2 2 2 2
Dividing by 2 and simplifying, we get
1 = (cid:2)1(X −1)(cid:3)g(X)−(cid:2)1(2X2+X −5)(cid:3)f(X).
4 4
1.2 Factorizing polynomials
We will mostly be concerned with factorizing polynomials in Q[X], but also sometimes over finite
fields F . Note also that factorization of polynomials in C[X] is (in theory) straightforward: the
p
Fundamental Theorem of Algebra says that any polynomial in C[X] factorizes completely, into a
product of linear factors – so the only irreducible polynomials in C[X] are those of degree 1.
Proposition 1.2. Suppose α = r/s ∈ Q, with r,s ∈ Z coprime, is a root of
f(X) = a Xn = ···+a X +a , a ∈ Z.
n 1 0 i
Then r|a and s|a .
0 n
5
Note that this is particularly useful for monic polynomials: If a = 1 then any root of f(X) in Q
n
is actually an integer dividing a .
0
Proof. We substitute α = r/s into f(α) = 0 and clear the denominators:
0 = snf(r/s) = a rn+a rn−1s+···+a rsn−1+a sn.
n n−1 1 0
Now r divides every term except possibly the last; so it divides the last term also, that is r|a sn.
0
But r is coprime to s, so this implies r|a .
0
Similarly, we get s|a rn and, since s is coprime to r, s|a . (cid:4)
n n
We can already use this to check irreducibility of quadratic and cubic polynomials: if such a
polynomial factorizes then one of the factors must have degree 1 so the polynomial must have a
root in Q.
Example. (i) f(X) = 3X2−3X−1 is irreducible in Q[X]: the only possible roots are ±1,±1/3
and we can easily check that
f(1) = −1, f(−1) = 5, f(1/3) = −5/3, f(−1/3) = 1/3.
So f has no roots and is irreducible.
(ii) g(X) = X3−2 is irreducible in Q[X]: the only possible roots are ±1,±2, and we check
g(1) = −1, g(−1) = −3, g(2) = 6, g(−2) = −10,
so g has no roots and is irreducible.
(iii) Note that this does not work for quartics (and higher degree). For example, h(X) = X4 +
X3+2X2+X +1 has no roots in Q (since neither 1 nor −1 is a root) but is reducible since
h(X) = (X2+1)(X2+X +1).
That is, h factorizes as a product of two irreducible quadratics.
The following proposition (valid for any field F) says that irreducibility is unaffected by a linear
change of variables.
Proposition1.3. Letf beapolynomialinF[X], letα ∈ F, letβ ∈ F×, andputg(X) = f(βX+α).
Then
f is irreducible ⇐⇒ g is irreducible.
Proof. Suppose f is reducible, so f = rs, with r,s ∈ F[X] non-constant. Then, putting r0(X) =
r(βX +α) and s0(X) = s(βX +α), we have
g = r0s0, with r0,s0 ∈ F[X] non-constant,
that is, g is reducible. The converse follows by symmetry, since f(X) = g(β−1X −β−1α). (cid:4)
Example. f(X) = 8X3−12X2+6X −3 is irreducible in Q[X]: writing g(X) = X3−2, we have
f(X) = (2X −1)3 −2 = g(2X −1); since g is irreducible (from the Example above), so is f, by
Proposition 1.3.
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Reduction modulo p
In this section we look at polynomials with rational (especially integer) coefficients. Indeed, these
are the same question: if f ∈ Q[X] then we can obtain from it a polynomial with integer coeffi-
cients by clearing the denominators (multiplying by some scalar); since scalars are units, the new
polynomial is irreducible if and only if f is irreducible.
Let p be a prime number; then we have a map
φ : Z → F
p
a 7→ a = a (mod p),
the reduction modulo p map. This is a ring homomorphism, since
φ(a+b) = a+b = a+b = φ(a)+φ(b),
φ(ab) = ab = ab = φ(a)φ(b).
This induces a map on polynomials
φ∗ : Z[X] → F [X]
p
f(X) = a Xn+···+a 7→ f(X) = a Xn+···+a .
n 0 n 0
(That is, we reduce all the coefficients modulo p.) Then a dull exercise shows that φ∗ is also a ring
homomorphism.
Example. With p = 5, we have
f(X) = X4+10X2+11X −2 ∈ Z[X],
f(X) = X4+X −2 ∈ F [X].
5
Proposition 1.4 (Gauss’s Lemma). Let f be a polynomial in Z[X]. If f factorizes (non-trivially)
in Q[X] then it factorizes (non-trivially) in Z[X].
We note that this is not obvious: it could be that f factorizes as a product of two polynomials
with rational coefficients but the denominators somehow cancel when we multiply them together.
Indeed, this does happen:
X2−1 = (2X − 2)(3X + 3).
3 3 2 2
The point here is that, whenever we have something in the denominator of one of the terms, it
appears in the numerator of the other so, by multiplying the two factors by scalars, we can find a
factorization with integral coefficients:
X2−1 = (X −1)(X +1).
The idea of the proof of Gauss’s Lemma is the same.
Proof. Suppose f = gh in Q[X], with g,h non-constant. Let m be the least common multiple of
the denominators of the coefficients of g, so that
g (X) = mg(X) ∈ Z[X].
1
7
Similarly, let n be the least common multiple of the denominators of the coefficients of h, so that
h (X) = nh(X) ∈ Z[X].
1
Then, in Z[X], we have
g (X)h (X) = mnf(X). (∗)
1 1
Suppose a prime number p divides mn. Reducing (∗) modulo p, we get
g (X)h (X) = 0 in F [X].
1 1 p
Now F [X] is an integral domain, so either g = 0 or h = 0; without loss of generality, suppose
p 1 1
g = 0. Then p divides all the coefficients of g so g (X) = pg (X), for some g ∈ Z[X]. But then
1 1 1 2 2
mn
g (X)h (X) = f(X),
2 1
p
which is the same as (∗) but with mn replaced by mn. Continuing, we may remove all the prime
p
factors of mn from (∗), and so we end with
g (X)h (X) = f(X),
∗ ∗
with g ,h ∈ Z[X]. Moreover, g and h are constant multiples of g and h respectively, so neither
∗ ∗ ∗ ∗
is a constant polynomial; that is, the factorization is non-trivial. (cid:4)
A similar proof shows the following:
Proposition 1.5. Suppose f ∈ Z[X] is monic, g ∈ Q[X] is monic, and g|f. Then g ∈ Z[X].
Proof. We have f = gh, with h ∈ Q[X] monic. Let m (respectively n) be the least common
multiple of the denominators of the coefficients of g (respectively h), so that g (X) = mg(X) and
1
h (X) = nh(X) have integer coefficients. Then
1
mbf(X) = g (X)h (X).
1 1
Suppose mn 6= 1 so some prime number p divides mn. Then, reducing modulo p as in the proof
of Gauss’s Lemma, we see that p divides all the coefficients of either g or h ; without loss of
1 1
generality, we may assume it is g . But the leading coefficient of g is m so p divides m also. But
1 1
(cid:16) (cid:17)
then m g(X) = 1g (X) ∈ Z[X], contradicting the definition of m.
p p 1
Hence mn = 1 so m = n = 1 and g ∈ Z[X], as required. (cid:4)
Example. We can use this to prove that f(X) = X4+1 is irreducible in Q[X]. Firstly, f has no
linear factors since, by Proposition 1.2, the only possible roots are ±1 but f(±1) = 2 6= 0. Hence,
if f factorizes in Q[X] it must do so as a product of two irreducible quadratics. By scaling, we may
assumethesetwoquadraticsaremonicandthen, byProposition1.5, theyhaveintegralcoefficients.
So we try writing a factorization:
X4+1 = (X2+aX +b)(X2+cX +d),
8
with a,b,c,d ∈ Z. Expanding out and comparing coefficients, we get the following equations
involving the coefficients:
a+c = 0;
b+ac+d = 0;
bc+ad = 0;
bd = 1.
The last of these gives b = d = ±1 (since b,d are both integers), while the first gives c = −a.
Substituting these into the second gives a2 = ±2, which is absurd.
Remark. Proposition 1.5 implies that factorizing a polynomial f of degree d in Q[X] can be done
in finite time as follows:
• Firstly we can multiply f by a scalar in Q so that it is monic, and, setting m to be the least
common multiple of the denominators of the coefficients of f, we can use Proposition 1.3 to
replace f by mdf(d−1X), which is monic with integer coefficients.
• Any monic factor g of f has integer coefficients, and these coefficients are bounded in terms
of the coefficients of f: If
f(X) = Xd+a XD−1+···+a X +a
d−1 1 0
then put A = max{1,|a |,...,|a |}. Then the coefficients of g are bounded (in absolute
0 d−1
value) by (dA)d so there are only finitely many monic polynomials d which one need try to
divide into f.
This method of proving irreducibility is rather cumbersome already for quartic polynomials, let
aloneforhigherdegree. Thenextpropositiongivesusthefirstinstanceofanirreducibilitycriterion
applicable to polynomials of any degree:
Proposition 1.6 (Eisenstein’s criterion). Let f(X) = a Xn +···+a ∈ Z[X] and suppose there
n 0
exists a prime number p such that
(i) p 6 |a ;
n
(ii) p|a , for i = 0,...,n−1;
i
(iii) p2 6 |a .
0
Then f is irreducible in Q[X].
Proof. Suppose f factorizes in Q[X], hence in Z[X] by Gauss’s Lemma:
(
g(X) = b Xr +···+0,
r
f(X) = g(X)h(X),
h(X) = c Xs+···+0,
s
with r,s ≥ 1, b c 6= 0. Reducing modulo p, we get
r s
f(X) = g(X)h(X).
But f(X) = a Xn 6= 0 and the only possible factorization of this is
n
a Xn = b Xrc Xs.
n r s
Hence p|b and p|c ; but then p2|b c = a , contradicting (iii). (cid:4)
0 0 0 0 0
9
Example. (i) f(X) = X5+3X2+9X+6 is irreducible in Q[X] by Eisenstein’s criterion applied
with p = 3.
(ii) Eisenstein’scriteriongivesanother(quicker)wayofprovingthatg(X) = X4+1isirreducible:
we note that
g(X +1) = (X +1)4+1 = X4+4X3+6X2+4X +2,
which is irreducible by Eisenstein’s criterion applied with p = 2. But then g is irreducible
also, by Proposition 1.3.
(iii) Similarly h(X) = X4+X3+X2+X +1 is irreducible over Q:
h(X +1) = (X +1)4+(X +1)3+(X +1)2+(X +1)+1
= X4 + 4X3 + 6X2 + 4X + 1
+ X3 + 3X2 + 3X + 1
+ X2 + 2X + 1
+ X + 1
+ 1
= X4+5X3+10X2+10X +5,
which is irreducible in Q[X] by Eisenstein’s criterion with p = 5.
Note that (X −1)h(X) = X5−1. This example is a special case of:
Lemma 1.7. Let p be a prime number and put
φ (X) = Xp−1+Xp−2+···+1, the pth cyclotomic polynomial
p
Then φ (X) is irreducible over Q.
p
Xp−1
Proof. We have φ (X) = so
p
X −1
(X +1)p−1 Xp+( p )Xp−1+···+(p)X +1−1
p−1 1
φ (X +1) = =
p
(X +1)−1 X
= Xp−1+pXp−2+···+(p)Xi−1+···+p.
i
Now (p) = p! so p divides (p), for 1 ≤ i ≤ p−1. On the other hand p2 does not divide p so
i i!(p−i)! i
Eisenstein’s criterion is satisfied with p and φ (X +1), so φ (X) is irreducible in Q[X]. (cid:4)
p p
Remark. What are the roots (in C) of φ (X)? They satisfy (X −1)φ (X) = Xp −1 so they are
p p
the pth roots of unity (except 1):
ζ = e2πi/p, ζ2 = e4πi/p,..., ζp−1 = e2(p−1)πi/p.
We can also use the reduction modulo p map to prove irreducibility directly. Suppose f is a monic
polynomial in Z[X] and f = gh in Z[X] is a non-trivial factorization. Then, reducing modulo p,
we get f = gh and, since deg(f) = degf, this is a non-trivial factorization. Hence, if f is reducible
in Q[X] then f ∈ F [X] is reducible for all primes p. The contrapositive of this is quite useful:
p
10