By George R. Kempf (auth.)

ISBN-10: 3322802787

ISBN-13: 9783322802781

ISBN-10: 3528065834

ISBN-13: 9783528065836

The legislation of composition comprise addition and multiplication of numbers or func tions. those are the fundamental operations of algebra. you may generalize those operations to teams the place there's only one legislation. the idea of this booklet was once begun in 1800 by way of Gauss, while he solved the 2000 year-old Greek challenge approximately developing average n-gons via ruler and compass. the idea was once extra constructed through Abel and Galois. After years of improvement the speculation was once installed the current shape by way of E. Noether and E. Artin in 1930. at the moment it was once known as sleek algebra and targeting the summary exposition of the idea. these days there are too many examples to enter their info. i feel the scholar should still research the proofs of the theorems and never spend time searching for strategies to difficult workouts. The routines are designed to elucidate the speculation. In algebra there are 4 easy buildings; teams, earrings, fields and modules. We current the speculation of those simple constructions. with a bit of luck this can provide a great introduc tion to fashionable algebra. i've got assumed as heritage that the reader has discovered linear algebra over the genuine numbers yet this isn't necessary.

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Let Ms be an R-module for s in a set S. Then TIsEsMs and L:sEs Ms are Rmodules. Here the operations are (ms) + (ns) = (ms + ns) and r(ms) = (rms). An easy example is L:s ES R. Any R-module isomorphic to such a sum is called free. Sometimes, L:sE[l,nj R is denoted by R,f)n or Rn. 5 If M is an R-module, then 0 . m all r in Rand (-r) . m = -(r· m) = r· (-m). = 0 for all m in M, r· 0 = 0 for Let M be an A[X]-module where A is a commutative ring. Then M is an A-module by multiplication by constants.

5 Fields 48 Proof. 3. Assume that F is infinite. By induction on n, we may assume that n = 1. Let (31 = (3. Let f and 9 be the irreducible polynomial of a and (3 over F. Let G be a splitting field of fg over F containing E. Then f = 7l"i(X - ai) and 9 = 7l"k(X - (3k) where these (3k are distinct elements of G and (31 = (3 and a1 = a. If k =f 1, the equation ai + x(3k = a + x(3 has a unique solution. As F is infinite we may find r in F such that ai + r(3k =f a + r(3 for all k =f 1. Let 8 = a + r(3.

As (X + Y)P is a homogeneous of degree p it suffices to show but these arc obvious from the power rule for differentiation. o F is called the Frobenius homomorphism and the powers Fi are also homomorphisms. 2 FINITE FIELDS Let G be a finite field. Then G ::::> Z/pZ where p is a prime = char(G). Let n = deg(G/(Z/pZ)). Then G ~ (Z/pZ)ffin. So #G = pn. Now the multiplicative group G-{O} has order pn_l. So if 9 E G-{O}, gpn_l = 1, and hence gpn =9 for all 9 in G. Now the polynomial H = Xpn - X splits in G and is separable.

### Algebraic Structures by George R. Kempf (auth.)

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