By Stanislaw Balcerzyk, Tadeusz Jozefiak

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**Extra resources for Commutative Noetherian and Krull Rings**

**Example text**

Let M be an R-module, M 0 , M 1 submodules of M, and I c Ran ideal. The following quotients can be constructed: the ideal M 0 : M 1 = {r e R; r M 1 c M 0 } and the submodule M 0 :I = {me M; Im c M 0 ). We have Ann(M0 ) = O:M0 , where 0 denotes the zero submodule. 1 Let M 0 , Mi, M 2 be submodules of a fixed R-module M. M2)+Mi. We leave the proof to the reader as a simple exercise. , if a e U Ann(m). For such an element a, the homomorphism M ~to M 0¢me¥ which carries an element m e M to the element am is not a monomorphism.

To simplify the exposition, we assume in the whole chapter that a ring R is always a ring with a unity and M always denotes an R-module. 1 A module M satisfies the ascending chain condition for submodules if any sequence of submodules Mi c: M 2 c: ... , if there exists a positive integer n such that Mn = Mn+ i = ... 2 A module M satisfies the maximum condition if every nonempty family of submodules of M, ordered by the inclusion relation, has a maximal element. 3 Let R be a ring and let M be an R-module.

Xn). Then the R-module I] Graded Rings and Modules 51 JP/JP+ 1 is isomorphic to a free R-module whose basis consists of monomials X~' ... X~n of degree p. 2. 8 Let U = R[[X1, ... , XnJ] be the ring of formal power series in indeterminates X1 , •. , Xn, with coefficients in the ring R, and write J = (X1 , ••• , Xn). The reader will find it easy to verify that Gr,(U) ~ Gr1 (T) (the isomorphism being that of graded rings), where T denotes the polynomial ring from the preceding example. 9 00 Let R = Et> RP be a graded ring.