In the bottom of page 32 in Bruns and Herzog, Cohen-Macaulay Rings, the authors write that if $R$ is a $\mathbb{Z}$-graded ring and $M$ a $\mathbb{Z}$-graded $R$-module, then $M$ is the homomorphic image of an $R$-graded module of the form $\oplus_{i \in \mathbb{Z}} R(i)$. Could someone please explain how we see that? In particular, how can we construct such an epimorphism? Note that $M$ need not be finite.
Asked
Active
Viewed 281 times
1 Answers
2
This is not correct (perhaps you have a misquote?), we need $\oplus_{i \in \mathbb{Z}} R[i]^{\oplus B_i}$ for sets $B_i$. These graded modules are called graded-free. They are easily seen to be projective.
In order to write every graded module $M$ as a quotient of a graded free module, just pick a generating system $B_{-i}$ of $M_i$. Every element $b \in B_{-i}$ corresponds to a homomorphism of graded $R$-modules $R[-i] \to M$, the image of $1 \in R = R[-i]_{i}$ is exactly $b \in M_i$. These extend to homomorphisms $R[-i]^{\oplus B_{-i}} \to M$, and then to a homomorphism $\oplus_i R[-i]^{\oplus B_{-i}} \to M$, which is an epimorphism by construction.
Martin Brandenburg
- 163,620