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Abstract

We give a short and self-contained proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT (for finite degree extensions) from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below). The argument is essentially taken from Chapter II of Emil Artin’s Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin’s proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin’s argument) was written with Keith Conrad’s help. Theorem Let E/F be an extension of fields, let a1, . . . , an be distinct generators of E/F such that p := (X − a1) · · · (X − an) is in F [X]. Then • the group G of automorphisms of E/F is finite, • there is a bijective correspondence between the sub-extensions S/F of E/F and the subgroups H of G, and we have S ↔ H ⇐⇒ H = Aut(E/S) ⇐⇒ S = E =⇒ [E : S] = |H|, where E is the fixed subfield ofH, where [E : S] is the degree (that is the dimension) of E over S, and where |H| is the order of H. Proof We claim: (a) If S/F is a sub-extension of E/F , then [E : S] = |Aut(E/S)|. (b) If H is a subgroup of G, then |H| = [E : E ]. Proof that (a) and (b) imply the theorem. Let S/F be a sub-extension of E/F and put H := Aut(E/S). Then we have trivially S ⊂ E , and (a) and (b) imply [E : S] = [E : E ]. 2 Conversely let H be a subgroup of G and set H := Aut(E/E). Then we have trivially H ⊂ H, and (a) and (b) imply |H| = |H|. Proof of (a). Let 1 ≤ i ≤ n. Put K := S[a1, . . . , ai−1] and d := [K[ai] : K]. It suffices to check that any F -embedding φ of K in E has exactly d extensions to an F -embedding Φ of K[ai] in E. Let q ∈ K[X] be the minimal polynomial of ai over K. It is enough to verify that φ(q) (the image under φ of q) has d distinct roots in E. But this is clear since q divides p, and thus φ(q) divides φ(p) = p. Proof of (b). In view of (a) it is enough to check |H| ≥ [E : E ]. Let k be an integer larger than |H|, and pick a b = (b1, . . . , bk) ∈ E. We must show that the bi are linearly dependent over E , or equivalently that b⊥ ∩ (E) is nonzero, where •⊥ denotes the vectors orthogonal to • in E with respect to the dot product on E. Any element of b⊥∩(EH)k is necessarily orthogonal to hb for any h ∈ H, so b⊥ ∩ (E) = (Hb)⊥ ∩ (E), where Hb is the H-orbit of b. We will show that (Hb)⊥ ∩ (E) is nonzero. Since the span of Hb in E has E-dimension at most |H| < k, (Hb)⊥ is nonzero. Choose a nonzero vector x in (Hb)⊥ such that xi = 0 for the largest number of i as possible among all nonzero vectors in (Hb)⊥. Some coordinate xj is nonzero in E, so by scaling we can assume xj = 1 for some j. Since the subspace (Hb)⊥ in E is stable under the action of H, for any h in H we have hx ∈ (Hb)⊥, so hx − x ∈ (Hb)⊥. Since xj = 1, the j-th coordinate of hx − x is 0, so hx − x = 0 by the choice of x. Since this holds for all h in H, x is in (E). [A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II: http://projecteuclid.org/euclid.ndml/1175197045.