An Introduction to Galois Theory by Andrew Baker

By Andrew Baker

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Pm , u)/K(p0 , . . , pm ). 6(ii), K(p0 , . . 15, u is algebraic over K. 17. Definition. For an extension L/K, let Lalg = {t ∈ L : t is algebraic over K} ⊆ L. 18. Proposition. For an extension L/K, Lalg is a subfield containing K and Lalg /K is algebraic. Proof. Clearly K ⊆ Lalg . We must show that Lalg L. Let u, v ∈ Lalg . 13, K(u, v)/K is a finite dimensional extension, hence every element of K(u, v) is algebraic over K. In particular, u + v and uv are in K(u, v) and if u ̸= 0, u−1 is also in K(u, v).

Uk ) : K(u1 , . . , uk−1 )) divides [K(u1 , . . , uk ) : K(u1 , . . , uk−1 )], so the desired result follows. 72. Proposition. Let L/K be a finite extension. Then L/K is separable if and only if (L : K) = [L : K]. Proof. Suppose that L/K is separable. If K E L, then for any u ∈ L, u is algebraic over E, and in the polynomial ring E[X] we have minpolyE,u (X) | minpolyK,u (X). As minpolyK,u (X) is separable, so is minpolyE,u (X), and therefore L/E is separable. Clearly E/K is also separable.

Then I(t) = {f (X) ∈ K[X] : f (t) = 0} ⊆ K[X] is an ideal which is principal and has an irreducible monic generator q(X) ∈ K[X]. In fact, q(X) = minpolyK,t (X). Proof. It is easy to see that I(t) ▹ K[X] and therefore I(t) = (q(X)) for some monic generator q(X). To see that q(X) is irreducible, suppose that q(X) = q1 (X)q2 (X) with deg qi (X) < deg q(X). Now as q1 (t)q2 (t) = 0, we must have q1 (t) = 0 or q2 (t) = 0, hence q1 (X) ∈ I(t) or q2 (X) ∈ I(t). These possibilities give q(X) | q1 (X) or q(X) | q2 (X) and so deg q(X) deg q1 (X) or deg q(X) deg q2 (X), contradicting the above assumption that deg qi (X) < deg q(X).

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