Math 514 homework #5 Due date: 19950913 D. J. Bernstein 19950906 Let R be a ring, let M be a module, and let · : R × M → M be a function. We say that M is a module over R under · if the following identities hold: (1) 1m = m; (2) (r + s)m = rm + sm; (3) (rs)m = r(sm); (4) r(m + n) = rm + rn. Other terminology: M is an R-module; · is an action of R on M. Exercise 13. Let M be an R-module. Prove the following identities: 0m = 0; (−r)m = −(rm); r0 = 0; r(−m) = −(rm). Let M and N be R-modules. Recall that a function f : M → N is a group homomorphism if it takes 0 to 0, + to +, and − to −; i.e., f(0) = 0, f(m + m′) = f(m) + f(m′), and f(−m) = −f(m). A function f : M → N is an R-module homomorphism if (1) it is a group homomorphism and (2) f takes the action on M to the action on N; i.e., f(rm) = r(f(m)) for every r and m. In the following two exercises, R is a ring and a, b are elements of R. The ideals aR and bR are thus submodules of R. Prove or disprove: aR and bR are isomorphic R-modules. Exercise 14. R = Z, a = 2, b = 3. Exercise 15. R = Z × Z, a = (7, 0), b = (0, 7).