2.7 The Galois Group of a Polynomial

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Let F be a field and let f(x) 2 F[x] be a separable polynomial. Let E be a

splitting field over F for the polynomial f(x), and set G = Gal(E/F). Since

E is uniquely determined up to F-isomorphism by f(x), then G is uniquely

determined up to isomorphism by f(x). We’ll call G the Galois group of the

polynomial f(x), and denote it Gal(f(x)).

Proposition 2.7.1 Let f(x) 2 F[x] be a separable polynomial, and let G be

the corresponding Galois group. Assume that f(x) factors into irreducibles

as

f(x) =

Y

fi(x)ei 2 F[x].

Let E be a splitting field over E of f(x), and let _i be the set of roots in E

for fi(x). Then G acts transitively on each _i.

Thus if we set _ =

S

_i, then we have a natural injective homomorphism

G −! S_.

An interesting question which naturally occurs is whether G _ A_, where

we have identified G with its image in S_. To answer this, we introduce the

discriminant of the separable polynomial f(x). Thus let f(x) 2 F[x], where

char F 6= 2, and let E be a splitting field over F for f(x). Let {_1, _1, · · · , _k}

be the set of distinct roots of f(x) in E. Set

2.7. THE GALOIS GROUP OF A POLYNOMIAL 63

_ =

Y

1_j<i_k

(_i − _j) = det

2

6666664

1 _1 _2

1 · · _k−1

1

1 _2 _2

2 · · _k−1

2

· · · · · ·

· · · · · ·

· · · · · ·

1 _k _2k

· · _k−1

k

3

7777775

,

and let D = _2. We call D the discriminant of the polynomial f(x). Note

that D 2 invG(E); since f(x) is separable, D 2 F.

Proposition 2.7.2 Let f(x) 2 F[x], with discriminant D defined as above.

Let G be the Galois group of f(x), regarded as a subgroup of Sn, where

{_1, · · · , _n} is the set of roots in a splitting field E over F for f(x). If

A = G \ An, then invA(E) = F(_).

Corollary 2.7.2.1 Let G be the Galois group of f(x) 2 F[x]. If D is the

square of an element in F, then G _ An.

The following is occasionally useful in establishing that the Galois group

of a polynomial is the full symmetric group.

Proposition 2.7.3 Let f(x) 2 Q[x] be irreducible, of prime degree p, and

assume that f(x) has exactly 2 non-real roots. Then Gf = Sp.

There are straightforward formulas for the discriminants of quadratics

and cubics.

Proposition 2.7.4

(a) If f(x) = x2 + bx + c, then Df = b2 − 4c.

(b) If f(x) = x3 + ax2 + bx + c, then

Df = −4a3c + a2b2 + 18abc − 4b3 − 27c2.

For a general “trinomial,” there is a wonderful formula, due to R.G.

Swan (Pacific Journal, vol 12, pp. 1099-1106, MR 26 #2432. (1962); see

also Gary Greenfield and Daniel Drucker, On the discriminant of a trinomial,

Linear Algebra Appl. 62 (1984), 105-112.), given as follows.

64 CHAPTER 2. FIELD AND GALOIS THEORY

Proposition 2.7.5 Let f(x) = xn +axk +b, and let d = g.c.d.(n, k), N =

n

d , K = k

d . Then

Df = (−1)1

2n(n−1)bk−1[nNbN−k − (−1)N(n − k)N−KkKaN]d.

Exercises 2.6

1. Let f(x) 2 F[x] be a separable polynomial. Show that f(x) is irreducible

if and only if Gal(f(x)) acts transitively on the roots of f(x).

2. Let _ be a primitive n-th root of unity. Show that

1 + _i + · · · + _(n−1)i =

_

n if n|i

0 if not.

3. Show that

D(xn−1) = det

2

6666664

n 0 · · · 0

0 0 · · · n

0 0 · · n 0

· · · · · ·

· · n · · ·

0 n · · · 0

3

7777775

= (−1)1

2 (n−1)(n−2)nn.

4. Prove that

D(xn−a) = an−1nn(−1)1

2 (n−1)(n−2).

5. Compute the discriminant of x7 − 154x + 99. (This polynomial has

PSL(2, 7) as Galois group.)

6. Find an irreducible cubic polynomial whose discriminant is a square

in Q. (One example is x3 − 9x + 9.)

7. Compute discriminants of x7 − 7x + 3, x5 − 14x2 − 42.

8. Let f(x) 2 Q[x] be irreducible, and assume that Gal(f(x)) _= Q8, the

quaternion group of order 8. Prove that deg f(x) = 8.

2.7. THE GALOIS GROUP OF A POLYNOMIAL 65

9. Prove that for each n _ 1, the Galois group over the rationals of the

polynomial f(x) = x3 − 32nx + 33n−1 is cyclic of order 3.

10. If f(x) = x6 − 4x3 + 1, prove that Gf

_= D12.

11. Let G be the Galois group of the polynomial x5 − 2 2 Q[x]; thus, if K

is the splitting field, then K = Q( 5

p

2, _), where _ = e2_i/5. Explicitely

construct an element of order 4 in the Galois group, and show what it

does to 5

p

2 and to _.

12. Let G be the Galois group of the polynomial x8 − 2 2 Q[x]. Thus, if

K is the splitting field, then K = Q( 8

p

2, _), where _ = e2_i/8. Show

that G has order 16. Also, compute the kernel of the action of G on

the four roots of x4 + 1.

13. Let f1(x) = x8 − 2 and let f2(x) = x8 − 3. Prove that Gf1 6_= Gf2 .

66 CHAPTER 2. FIELD AND GALOIS THEORY

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