# Download Cogalois Theory by Toma Albu PDF

By Toma Albu

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Up-1 in E . Consequently, these elements are exactly all the roots vk E F. We now multiply together the in R of f , and then w = equalities [ku = vi, k = 0 , . . , p - 1, and obtain qup = qa = wP, where 40 Conversely, assume that a # FP, and prove that XP - a is irreducible in F [ X ] . Suppose the contrary, and let f be an irreducible factor of degree n, 1 Q n < p, of XP - a. If u is a fixed root of f in R, then the roots of f in R all have the form [u, where C E R is a p t h root of unity.

Prove that E = F ( i ) , [ E : F] = 2, and Cog ( E / F ) is a countably infinite group. 6. (Dummit [55]). Let p be an odd prime, let E = Q(Cp), and let F = E n R . Showthat [ E :F] = 2 and ICog(E/F)I = 2p. 7. (May [81]). Let F be a field, and let p be a prime different from the characteristic of F . Let u E R with UP E F \ FP, and let E = F ( u ) . If p = 2, further assume that E # F(C4). Prove that Cog(E/F) = ((u)t(E*)F*)/F*. 8. (May [81]). Let F = Q , p = 2, u = C4 = i, and E = F ( u ) . Show that 1 i E Cog(E/F) \ ((u)t(E*)F*)/F*.

However, the finite extension E/F is not G-Kneser, since (3) We do not know whether or not the separability condition in the Kneser Criterion can be weakened. 0 -4pplications of the Kneser Criterion to algebraic number fields will be given in Chapter 9. 3. Exercises to Chapter 2 1. Prove that pn ( F( X I , . . ,X,)) m,nEN*. = pn ( F ) for any field F and any 2. Let p be any positive prime number, and let n be any positive integer. Prove that the extension IF', ( X I , . . ,Xn)/IFp ( X r , . .