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 subtract 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. $\square$

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