0
$\begingroup$

Let $(R,\frak{m})$ be a hypersurface (i.e., $R=Q/(f)$, where $Q$ is a regular local ring, $0\not=f\in Q$). If $M$ is a MCM $R$-module, is it possible for the injective dimension of $M$ over $R$ to be $\infty$? What about the Gorenstein injective dimension in this case?

Since a noetherian local ring is regular iff it has finite global dimension = max projective dimension of all finite modules = max injective dimension of all modules (Matsumura, p.155), over a regular local ring we have that of course $M$ will have finite injective (and hence Gorenstein injective dimension). But what about over $R$?

Thank you!

$\endgroup$
1
$\begingroup$

If $M$ is not free it will have infinite injective dimension. This is because $R$ is Gorenstein (since a hypersurface), so the projective dimension of a finitely generated module is finite iff its injective dimension is. By Auslander-Buchsbaum, finite projective dimension for a MCM module is equivalent to freeness.

$\endgroup$
  • $\begingroup$ Of course, thanks. For completeness, I'll also add that in a Gorenstein ring $R$, the Gorenstein injective dimension of any $R$-module is always finite (cf. 6.2.7 of Christensen's "Gorenstein Dimensions"). $\endgroup$ – math-grad May 22 '14 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.