By Paul J. McCarthy

ISBN-10: 0486666514

ISBN-13: 9780486666518

*Math.*reports. Over 2 hundred workouts. Bibliography.

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**Extra info for Algebraic Extensions of Fields**

**Sample text**

Let a E G(L/k). By Proposition 2 there is an element T E G(K/k) such that T(b) = a(b) for all bEL. In particular, T(a) = a(a) = a and so a is in the fixed field of G(L/k). It follows from Theorem 4 that a E k. II We shall now prove the converse of Theorem 20, that is, if K is a Galois extension of k then K/k is normal and separable. Since K/k is Galois it is algebraic. Let a E K and letf(x) = lrr (k,a). Let a = ah a 2, ••• , an be the distinct images of a under the various k-automorphisms of K.

Since F0 = k the result is true when r = 0. _1 • Since Kt! k is normal and separable we have Kt = k(c) and every conjugate of c in Cis also in Kt. th root of unity in C and consider Kt( ') = k(c,,). - I) Irr (k,c), which is in k[x]. Thus Kt(,)/k is normal. Since Kt(') is obtained from Kt by adjoining a radical we may assume from the start that ' E Kt . ss Recall that F,. ) where a,. , b,. , b~2 >, • •• , b~m) be the k-conjugates of b,. in C and let m f(x) =IT (xn" - b~i)). ). Then K, is a splitting field over k of f(x) Irr {k,c) e k[x] and so Ktf k is normal.

Note that the order of Xi is y i· Now suppose that for non-negative integers {31 , identity element of G"". Then for i = 1, ... , r, 1 = (X1P 1 • • • ••• , {J,, x1P 1 • • • Xrp, is the 'f ~, Xrflr)(gi) = 'i and since is a primitive yith root of unity this implies that y i divides fJi· Thus x1 , ••• , Xr form a basis of G". II Let G and H be Abelian groups and A a finite cyclic group. By a pairing of G and H into A we mean a mapping cp from G x H into A such that cp(glg2,h) = cf>(gt,h)cf>(g2,h) and c/>(g,hlh2) = cf>(g,hl)c/>(g,h2) for all g, gh g 2 E G and h, h1, h2 E H.

### Algebraic Extensions of Fields by Paul J. McCarthy

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