By John B. Fraleigh
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Extra resources for Galois Theory, U Glasgow course
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).