Bourbaki

Definition 2

Tag 06MF, permanent. Cite it as [Bourbaki, Tag 06MF].

Algebra, V, p. 58, § 10. Galois extensions

Let N be a Galois extension of the field K. The group of all K-automorphisms of N will be called the Galois group of N over K and denoted by Gal(N/K).

Let N be a finite Galois extension of K. Then N is a finite separable and quasi-Galois extension of K. Therefore (V, p. 32, Prop. 4 and V, p. 54, Prop. 3), the order of Gal(N/K) is equal to [N : K]. We shall prove later that if N is a Galois extension of K such that Gal(N/K) is finite, then N is of finite degree over K (V, p. 66, Th. 3).

Let Ω be an algebraically closed extension of K and let A be the set of roots in Ω of a separable polynomial f ∈ K[X]. Then the field N = K(A) is a Galois extension of K. It is clear that every K-automorphism of N leaves A stable, and since A generates N over K, the mapping a ↦ σ|A is an isomorphism of Gal(N/K) onto a subgroup Γ of the symmetric group S_A of the set A, which will be called the Galois group of the polynomial f. From Remark 3 (V, p. 55) it follows that if x and y belong to A, the following properties are equivalent:

a) x and y are conjugate over K, b) x and y belong to the same orbit under Γ, c) x and y are roots of the same irreducible factor of f.

In particular f is irreducible if and only if A is non-empty and Γ operates transitively on A.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed edition, 2003, Springer. Not translated.