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. 



R noetherian is factorial
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.


