TUTORIAL 8 1. Galois Groups of Polynomials Definition. Suppose F(X)

MISJA F.A. STEINMETZ

1. Galois groups of Polynomials Definition. Suppose f(X) ∈ K[X] is a separable polynomial (i.e. no repeated roots) over a field K with splitting field Lf over K (i.e. the field generated over K by the roots of f). Then we refer to the group Gal(Lf /K) as the Galois group of f over K.

Remark. A different choice of splitting field will give a group isomorphic to Gal(Lf /K), so the isomorphism class of this group is well-defined and we can refer to any group isomorphic to Gal(Lf /K) as ‘the Galois group of f’.

Proposition. If f is a separable polynomial of degree n, then Gal(Lf /K) is isomorphic to a subgroup of Sn.

Why? Idea is as follows: if α1, . . . , αn are the roots of f and σ ∈ Gal(Lf /K) then

σ(α1) = αi1 πσ(1) = i1

σ(α2) = αi2 πσ(2) = i2 ......

σ(αn) = αin πσ(n) = in and in this way we get a permutation πσ ∈ Sn. It turns out that σ 7→ πσ is an injective group homomorphism, so we get a subgroup of Sn.

Example 1 Find an explicit representation of the elements of the Galois group of (X2−2)(X2−3) ∈ Q[X] as permutations in S4.

Date: 12 March 2018. 1 2 MISJA F.A. STEINMETZ

We can do better. The group Gal(Lf /K) is not just any subgroup of Sn, but it is always a transitive subgroup.

Deﬁnition. A subgroup G of Sn is transitive if for any i, j ∈ {1, 2, . . . , n} there exists an element π ∈ G such that π(i) = j.

Example 2

Is the subgroup A3 := {e, (1 2 3), (1 3 2)} ⊂ S3 transitive?

Proposition. If f ∈ K[X] is a separable irreducible polynomial of degree n, then its Galois group Gal(Lf /K) is isomorphic to a transitive subgroup of Sn.

Classwork 1 Suppose that L/K is a Galois extension of degree n. Prove that L is the splitting ﬁeld of a separable irreducible polynomial of degree n.

Prove that Gal(L/K) is isomorphic to a transitive subgroup of Sn.

2. Galois groups of Cubics

Deﬁnition. Suppose f ∈ K[X] is a monic cubic polynomial with roots α1, α2, α3. Then we deﬁne the discriminant of f as 2 ∆f = ((α1 − α2)(α2 − α3)(α3 − α1)) . Proposition (Sheet 8, Q2). If f is of the form X3 + aX + b (with no quadratic term), then 3 2 ∆f = −4a − 27b .

Moreover, if g(X) = f(X − c) for some c ∈ K, then ∆f = ∆g. Remark. If f = X3 + rX2 + sX + t, then g(X) = f(X − (r/3)) has no quadratic term. GALOIS THEORY - TUTORIAL 8 3

Given an irreducible cubic polynomial f ∈ K[X], we know that its splitting ﬁeld Lf has degree at least 3 over K (and hence Gal(Lf /K) = [Lf : K] ≥ 3). Moreover, we know that ∼ Gal(Lf /K) is isomorphic to a subgroup of S3. Only possibilities Gal(Lf /K) = A3 or S3. When does each option happen? Proposition. Suppose K is a ﬁeld of characteristic not 2 or 3 and f ∈ K[X] is a monic, cubic, irreducible polynomial. Then p (i) if ∆f ∈ K, then Gal(Lf /K) is isomorphic to A3; p (ii) if ∆f ∈/ K, then Gal(Lf /K) is isomorphic to S3.

Example 3 What is the Galois group of f(X) = X3 − 2 ∈ Q[X]?

Classwork 2 What is the Galois group of f(X) = X3 − 4X + 2 ∈ Q[X]? What are all the subﬁelds K ⊂ Lf such that [K : Q] = 2? 4 MISJA F.A. STEINMETZ

Classwork 3 What is the Galois group of f(X) = X3 − 2X − 4 ∈ Q[X]?

Classwork 4 What is the Galois group of f(X) = X3 + 3X2 + 1 ∈ Q[X]? What are all the subﬁelds K ⊂ Lf such that [K : Q] = 2?

Strand 5.12, King’s College London E-mail address: [email protected] URL: https://www.nms.kcl.ac.uk/misja.steinmetz/