By George R. Kempf (auth.)

The legislation of composition contain addition and multiplication of numbers or func tions. those are the fundamental operations of algebra. you'll generalize those operations to teams the place there's only one legislations. the speculation of this ebook used to be began in 1800 via Gauss, while he solved the 2000 year-old Greek challenge approximately developing commonplace n-gons through ruler and compass. the speculation was once additional built through Abel and Galois. After years of improvement the idea used to be installed the current shape by means of E. Noether and E. Artin in 1930. at the moment it used to be referred to as glossy algebra and targeting the summary exposition of the speculation. these days there are too many examples to enter their info. i believe the coed should still examine the proofs of the theorems and never spend time searching for ideas to difficult routines. The workouts are designed to elucidate the idea. In algebra there are 4 easy constructions; teams, earrings, fields and modules. We current the speculation of those simple buildings. expectantly it will supply a very good introduc tion to fashionable algebra. i've got assumed as historical past that the reader has discovered linear algebra over the genuine numbers yet this isn't necessary.

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F n- 1 54 6 More field theory Proof. We have done everything but compute the Galois group. The Frobenius F gives an injective homomorphism from G to G. As G is finite, it is surjective. So it is an automorphism. Flcz/pz) = Idcz/pZ) by (*) with n = 1. Thus F E Aut(Gj(ZjpZ)). Let H be the subgroup of Aut(Gj(ZjpZ)) generated by F. Let K Thus by Galois Theory H = {g E Glh(g) = 9 for all hE H} = {g E G\F(g) = {g E GlgP = {O} U {g E G*lgP-l = ZjpZ = Gal(Gj(ZjpZ)) and #H = n. 1 By Galois theory, determine all fields H C G.

F = II~l (X - ei) where ei E E such that all the ei are distinct. 1 the number of the homomorphism = n = deg(F(e)jF). 1 If k E F(e) - F, then there exists a homomorphism cP : F(e) with cplP = IdF such that cp(k) :f:. k. Proof. If not, # of cp for F(e)j F(k) = # --t E of cp for F(e)j F. Notice that degF(F(e)j F(k)) deg(F(k)j F) = deg(F(e)j F) and, hence, deg(F(k)jF) = 1. Thus F(k) = F contradiction. 2 Let E :2 F be a splitting field of some separable polynomial in F[X]. Let e be an element of E - F.

J Thus H . tHt is a normal subgroup of Altn =f. H. Hence it equals Altn by maximality of H. Now any element of H ·tHr 1 has order 2. Therefore any element of Altn has order two. The contradiction proves that H does not exists. Hence Altn is simple as it is not abelian. It remains to prove the claims. 4 SOLVABLE AND SIMPLE GROUPS. 41 For a), let Symn operate on X = set of conjugates of H by conjugation. As Altn fixes H, we have a surjection Symn/ Alt n -> X but H is not normal in Symn. So this is an isomorphism.