Lemma 15.30.14. Let $A$ be a ring. Let $f_1, \ldots , f_ n, g_1, \ldots , g_ m \in A$. If both $f_1, \ldots , f_ n$ and $f_1, \ldots , f_ n, g_1, \ldots , g_ m$ are Koszul-regular sequences in $A$, then $\overline{g}_1, \ldots , \overline{g}_ m$ in $A/(f_1, \ldots , f_ n)$ form a Koszul-regular sequence.

Proof. Set $I = (f_1, \ldots , f_ n)$. Our assumptions say that $K_\bullet (A, f_1, \ldots , f_ n)$ is a finite free resolution of $A/I$ and $K_\bullet (A, f_1, \ldots , f_ n, g_1, \ldots , g_ m)$ is a finite free resolution of $A/(f_ i, g_ j)$ over $A$. Then

\begin{align*} A/(f_ i, g_ j) & \cong 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) \end{align*}

The first quasi-isomorphism $\cong$ by assumption. The first equality by Lemma 15.28.12. The second quasi-isomorphism by (the dual of) Homology, Lemma 12.25.4 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. Hence we win. $\square$

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