Lemma 49.12.1. Let $A \to P$ be a ring map. Let $f_1, \ldots , f_ n \in P$ be a Koszul regular sequence. Assume $B = P/(f_1, \ldots , f_ n)$ is flat over $A$. Let $g_1, \ldots , g_ n \in P \otimes _ A B$ be a Koszul regular sequence generating the kernel of the multiplication map $P \otimes _ A B \to B$. Write $f_ i \otimes 1 = \sum g_{ij} g_ j$. Then the annihilator of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ is a principal ideal generated by the image of $\det (g_{ij})$.

[Appendix, Mazur-Roberts]

**Proof.**
The Koszul complex $K_\bullet = K(P, f_1, \ldots , f_ n)$ is a resolution of $B$ by finite free $P$-modules. The Koszul complex $M_\bullet = K(P \otimes _ A B, g_1, \ldots , g_ n)$ is a resolution of $B$ by finite free $P \otimes _ A B$-modules. There is a map of complexes

which in degree $1$ is given by the matrix $(g_{ij})$ and in degree $n$ by $\det (g_{ij})$. See More on Algebra, Lemma 15.28.3. As $B$ is a flat $A$-module, we can view $M_\bullet $ as a complex of flat $P$-modules (via $P \to P \otimes _ A B$, $p \mapsto p \otimes 1$). Thus we may use both complexes to compute $\text{Tor}_*^ P(B, B)$ and it follows that the displayed map defines a quasi-isomorphism after tensoring with $B$. It is clear that $H_ n(K_\bullet \otimes _ P B) = B$. On the other hand, $H_ n(M_\bullet \otimes _ P B)$ is the kernel of

Since $g_1, \ldots , g_ n$ generate the kernel of $B \otimes _ A B \to B$ this proves the lemma. $\square$

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