Lemma 15.30.11. Let $A$ be a ring. Let $I$ be an ideal generated by an $H_1$-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 an $H_1$-regular sequence in $A/I$. Then $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ is an $H_1$-regular sequence in $A$.

**Proof.**
We have to show that $H_1(A, f_1, \ldots , f_ n, g_1, \ldots , g_ m) = 0$. To do this consider the commutative diagram

Consider an element $(a_1, \ldots , a_{n + m}) \in A^{\oplus n + m}$ which maps to zero in $A$. Because $\overline{g}_1, \ldots , \overline{g}_ m$ form an $H_1$-regular sequence in $A/I$ we see that $(\overline{a}_{n + 1}, \ldots , \overline{a}_{n + m})$ is the image of some element $\overline{\alpha }$ of $\wedge ^2(A/I^{\oplus m})$. We can lift $\overline{\alpha }$ to an element $\alpha \in \wedge ^2(A^{\oplus n + m})$ and substract the image of it in $A^{\oplus n + m}$ from our element $(a_1, \ldots , a_{n + m})$. Thus we may assume that $a_{n + 1}, \ldots , a_{n + m} \in I$. Since $I = (f_1, \ldots , f_ n)$ we can modify our element $(a_1, \ldots , a_{n + m})$ by linear combinations of the elements

in the image of the top left horizontal arrow to reduce to the case that $a_{n + 1}, \ldots , a_{n + m}$ are zero. In this case $(a_1, \ldots , a_ n, 0, \ldots , 0)$ defines an element of $H_1(A, f_1, \ldots , f_ n)$ which we assumed to be zero. $\square$

## 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Your email address will not be published.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: