Lemma 15.29.13. Let $A$ be a ring. Let $I$ be an ideal generated by a Koszul-regular sequence $f_1, \ldots , f_ n$ in $A$. Let $g_1, \ldots , g_ m \in A$ be elements whose images $\overline{g}_1, \ldots , \overline{g}_ m$ form a Koszul-regular sequence in $A/I$. Then $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is a Koszul-regular sequence in $A$.

**Proof.**
Our assumptions say that $K_\bullet (A, f_1, \ldots , f_ n)$ is a finite free resolution of $A/I$ and $K_\bullet (A/I, \overline{g}_1, \ldots , \overline{g}_ m)$ is a finite free resolution of $A/(f_ i, g_ j)$ over $A/I$. Then

The first equality by Lemma 15.28.12. The first quasi-isomorphism $\cong $ by (the dual of) Homology, Lemma 12.22.7 as the $q$th row of the double complex $K_\bullet (A, f_1, \ldots , f_ n) \otimes _ A K_\bullet (A, g_1, \ldots , g_ m)$ is a resolution of $A/I \otimes _ A K_ q(A, g_1, \ldots , g_ m)$. The second equality is clear. The last quasi-isomorphism by assumption. Hence 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Your email address will not be published.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)

There are also: