# Galois Theory, U Glasgow course by John B. Fraleigh

By John B. Fraleigh

Thought of a vintage by way of many, a primary direction in summary Algebra is an in-depth, creation to summary algebra. occupied with teams, jewelry and fields, this article offers scholars an organization beginning for extra really good paintings through emphasizing an knowing of the character of algebraic buildings. The 6th variation of this article maintains the culture of educating in a classical demeanour whereas integrating box conception and a revised bankruptcy 0. New workouts have been written, and former workouts have been revised and converted.

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Subfields of Galois extensions and relative Galois groups Let E/K a Galois extension and suppose that K L E. Then E/L is also a Galois extension whose Galois group Gal(E/L) is sometimes called the relative Galois group of the pair of extensions E/K and L/K. The following is immediate. 14. , Gal(E/L) Gal(E/K), and its order is | Gal(E/L)| = [E : L]. 15. Let K L E. Then L = E Gal(E/L) . Proof. Clearly L E Gal(E/L) . Suppose that u ∈ E − L. Then there is an automorphism θ ∈ Gal(E/L) such that θ(u) = u, hence u ∈ / E Gal(E/L) .

Let t ∈ L. Since the K-vector space L is finite dimensional the powers 1, t, . . , tn , . . must be linearly dependent over K, hence for suitable coefficients cj ∈ K not all zero and some m 1 we have c0 + c1 t + · · · + cm tm = 0. But this means that t is algebraic over K. 16. Let M/L and L/K be algebraic extensions. Then the extension M/K is algebraic. Proof. Let u ∈ M . Then u is algebraic over L, so there is a polynomial p(X) = p0 + p1 X + · · · + pm X m ∈ L[X] of positive degree with p(u) = 0.

Ur ) ∈ F : f (X1 , . . , Xr ), g(X1 , . . , Xr ) ∈ K[X1 , . . , Xr ], g(u1 , . . , ur ) = 0 . g(u1 , . . , ur ) Reordering the ui does not change K(u1 , . . , un ). 8. Let K(u)/K and K(u, v)/K(u) be simple extensions. Then K(u, v) = K(u)(v) = K(v)(u). More generally, K(u1 , . . , un ) = K(u1 , . . , un−1 )(un ) and this is independent of the order of the sequence u1 , . . , un . 9. For a simple extension K(u)/K exactly one of the following conditions holds. (i) The evaluation at u homomorphism εu : K[X] −→ K(u) is a monomorphism and on passing to the fraction field gives an isomorphism (εu )∗ : K(X) −→ K(u).