By LAM T. Y.

This quantity deals a compendium of routines of various measure of trouble within the concept of modules and jewelry. it's the better half quantity to GTM 189. All workouts are solved in complete element. each one part starts with an creation giving the overall history and the theoretical foundation for the issues that stick with.

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**Extra resources for Exercises in Modules and Rings**

**Sample text**

14. Let R be the (commutative) ring of real-valued continuous functions on [0, 1], with pointwise addition and multiplication for functions. Let P ={fER: f vanisheson [O,c] for some E=c(f) E (0,1)}. ) Solution. For n 2: 2, let an E P be the piecewise linear function [0, 1] ---7 lR which is zero on [0, 1/n] and whose graph on [1/n, 1] is the line segment joining the two points (1/n, 0) and (1, 1). Note that an+l ~ anR since any function in anR must vanish on [1/(n + 1), 1/n] but an+l doesn't. On the other hand, an E an+lR.

Thus, R8 is a projective, and hence flat, module. By the first part of this exercise, we conclude that R = 8 x T also satisfies the strong rank condition. Ex. 21. If a product ring R = 8 x T satisfies the strong rank condition, show that either 8 or T must satisfy the strong rank condition. Solution. Suppose neither 8 nor T satisfies the strong rank condition. Then there exist an embedding 88'+ 1 ----. 88 for some m, and an embedding r;H ----. T,P for some n. We may assume that m = n. (If, say m > n, add r:;-n to both sides of r;+l----.

This shows that R is a PID. 22)(D) that Pic(R) = {1}. If R is also Dedekind, then (1) shows that R is a PID. Comment. For a more powerful statement than (2), see Ex. 11B(2) below. Ex. 11A. g. ideal in a commutative ring R such that annR(I) = 0. If A1 , ... , An are proper ideals of R, show that U~= 1 (Ai I) ~ I. (2) Show that, if an invertible ideal of R is contained in the union of a finite number of ideals in R, then it is contained in one of them. Solution. (1) We induct on n. To start the induction, we must show first A1 I s;: I.