By Stanislaw Balcerzyk, Tadeusz Jozefiak

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**Example text**

3. S = 0, then Rs/PRs ~ (R/P),cs>o where v:R-+ R/P is the natural homomorphism. Deduce that if Q => P is also a prime ideal then R 0 /PR 0 ~ (R/P) 01p. 4. If t is a non-nilpotent element of a ring R, write R, for the ring of fractions of R with respect to the multiplicative set {l,t,t 2 , ... },and denote by S the set {reR:r~Ker(R ~ R,) => Ann(r)}. (i) Prove that Sis the inverse image of the set of non-zero-divisors by the homomorphism R-+ R,. (ii) Prove that the total ring of fractions of R, is isomorphic to Rs.

It follows directly from the definition that 1 e R0 , R0 is a subring of R, co and Rp are R0 -submodules of R. We shall write, briefly, "a graded ring $RP". 4 we used the symbol Rco> or R 0 to denote the field of fractions of a domain, this ambiguity should not cause any misunderstanding. co Given an element r e R = $ Rp, we write r p=O = r0 +r 1 + .. +rm; this means that r0 e R 0 , r 1 e R 1 , ... , rm e Rm. We call rp the homogeneous component of degree p of the element r. The elements of Rp are called homogeneous of degree p and we write deg(a) = p for a e Rp.

The change of coefficients by means of the natural homomorphism R -4 R/I leads to the original R-module M. In particular, for any ideal I c Rand any R-module M, the factor R-module M/IM satisfies the condition I(M/IM) = 0 and therefore may be regarded as an R/I-module. 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.