Lemma 37.58.7. A regular immersion of schemes is perfect. A Koszul-regular immersion of schemes is perfect.

**Proof.**
Since a regular immersion is a Koszul-regular immersion, see Divisors, Lemma 31.21.2, it suffices to prove the second statement. This translates into the following algebraic statement: Suppose that $I \subset A$ is an ideal generated by a Koszul-regular sequence $f_1, \ldots , f_ r$ of $A$. Then $A \to A/I$ is a perfect ring map. Since $A \to A/I$ is surjective this is a presentation of $A/I$ by a polynomial algebra over $A$. Hence it suffices to see that $A/I$ is pseudo-coherent as an $A$-module and has finite tor dimension. By definition of a Koszul sequence the Koszul complex $K(A, f_1, \ldots , f_ r)$ is a finite free resolution of $A/I$. Hence $A/I$ is a perfect complex of $A$-modules and we win.
$\square$

## 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Your email address will not be published.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)