Lemma 47.21.3. A regular local ring is Gorenstein. A regular ring is Gorenstein.

**Proof.**
Let $A$ be a regular ring of finite dimension $d$. Then $A$ has finite global dimension $d$, see Algebra, Lemma 10.110.8. Hence $\mathop{\mathrm{Ext}}\nolimits ^{d + 1}_ A(M, A) = 0$ for all $A$-modules $M$, see Algebra, Lemma 10.109.8. Thus $A$ has finite injective dimension as an $A$-module by More on Algebra, Lemma 15.69.2. It follows that $A[0]$ is a dualizing complex, hence $A$ is Gorenstein by the remark following the definition.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)