Read e-book online Abstract Algebra with Applications: Rings and Fields PDF

By Karlheinz Spindler

A finished presentation of summary algebra and an in-depth remedy of the functions of algebraic recommendations and the connection of algebra to different disciplines, corresponding to quantity idea, combinatorics, geometry, topology, differential equations, and Markov chains.

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Therefore, M = 0. 21 Let f : M → N be a homomorphism. , an epimorphism, an isomorphism). , an epimorphism, an isomorphism) for any prime ideal p of R. , an epimorphism, an isomorphism) for any maximal ideal m of R. 20. 22 Let M be an R-module and let A, B be submodules of M. Then A ⊆ B if and only if Am ⊆ Bm for any maximal ideal m. Thus A = B if and only if Am = Bm for any maximal ideal m. 21. 23 Let S and T be multiplicative subsets of R such that S ⊆ T . (1) TS := { st | t ∈ T, s ∈ S} is a multiplicative subset of R S .

Therefore H = (X ). (2) This follows directly from (1). Let M be an R-module. If there exists a subset X ⊆ M such that M = (X ), then we call X a generating set of M and an element of X a generator. If X is a finite set, then we call M a finitely generated module; if |X | = 1, then we call M a cyclic module. If X = {x1 , . . 1, M = Rx1 + · · · + Rxn . If M is a cyclic module, then there exists x ∈ M such that M = Rx. Let N be a submodule of M. We also have concepts of a finitely generated submodule and a cyclic submodule.

Proof Let m denote the set of nonunit elements of R. Suppose that R is a local ring with the maximal ideal M. Since M is a proper ideal of R, each element of M is not a unit. Thus M ⊆ m. If x ∈ m, then x is not a unit. So (x) is a proper ideal of R. Then x ∈ (x) ⊆ M. Thus m ⊆ M. Therefore m = M is an ideal of R. Conversely, suppose that m forms an ideal of R. Then m is a proper ideal. For any proper ideal I of R, any element of I is a nonunit. So I ⊆ m. Thus m is the unique maximal ideal. Therefore, R is local.

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Abstract Algebra with Applications: Rings and Fields by Karlheinz Spindler


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