This is exercise 2.2.27 / 72 from Cohen Macaulay Rings, Bruns & Herzog
Let R be a Noetherian ring over which every finite module has a finite free resolution. Show R is a factorial domain.
1
-
2$\begingroup$ Could you tell us why you want to know the answer to this question? After all, the FAQ makes it clear that MO is not for homework. Also, could you tell us what you have tried already? There was a discussion recently on Meta about questions from textbooks, which might be useful to read through: tea.mathoverflow.net/discussion/1294/…. $\endgroup$– David WhiteCommented Feb 6, 2012 at 13:59
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This is well-known and references are presumably easy to find. Let me just give a few pointers.
It is enough to show that prime ideals of height one are principal. Let $I$ be such a prime ideal. Thanks to our hypotheses on $R$, the ideal $I$ has a finite free resolution. Hence, if $I$ is a projective module, then it is stably free, and thus principal. In order to show that $I$ is projective, we look at the cokernel $X$ of $\operatorname{Hom}(I,M)\rightarrow\operatorname{Hom}(I,N)$ when $M\rightarrow N$ is a surjection, and to do so it is enough to localize at a maximal ideal $\mathfrak{m}$. The residual field $k$ of the local ring $R_{\mathfrak{m}}$ admits a finite free resolution by $R$-modules by our hypothesis so admits a finite free resolution by $R_{\mathfrak{m}}$-modules. So the ring $R_{\mathfrak{m}}$ is regular so the ideal $IR_{\mathfrak{m}}$ is principal so $IR_{\mathfrak{m}}$ is a projective $R_{\mathfrak{m}}$-module so $X_{\mathfrak{m}}$ vanishes. So $X$ is zero, so $I$ is projective and hence principal. And so $R$ is factorial.
