GALOIS THEORY - TUTORIAL 8 MISJA F.A. STEINMETZ 1. Galois groups of Polynomials Definition. Suppose f(X) 2 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 2 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 2 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) 2 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 2 f1; 2; : : : ; ng there exists an element π 2 G such that π(i) = j: Example 2 Is the subgroup A3 := fe; (1 2 3); (1 3 2)g ⊂ S3 transitive? Proposition. If f 2 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 2 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 2 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 2 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 2 K[X] is a monic, cubic, irreducible polynomial. Then p (i) if ∆f 2 K; then Gal(Lf =K) is isomorphic to A3; p (ii) if ∆f 2= K, then Gal(Lf =K) is isomorphic to S3: Example 3 What is the Galois group of f(X) = X3 − 2 2 Q[X]? Classwork 2 What is the Galois group of f(X) = X3 − 4X + 2 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 2 Q[X]? Classwork 4 What is the Galois group of f(X) = X3 + 3X2 + 1 2 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/.