
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

\begin{align*} K_\bullet (A, f_1, \ldots , f_ n, g_1, \ldots , g_ m) & = \text{Tot}(K_\bullet (A, f_1, \ldots , f_ n) \otimes _ A K_\bullet (A, g_1, \ldots , g_ m)) \\ & \cong A/I \otimes _ A K_\bullet (A, g_1, \ldots , g_ m) \\ & = K_\bullet (A/I, \overline{g}_1, \ldots , \overline{g}_ m) \\ & \cong A/(f_ i, g_ j) \end{align*}

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. $\square$

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