A question about mapping cone and resolutions

Question

I am studying this papper

https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-44/issue-3/Betti-numbers-of-almost-complete-intersections/10.1215/ijm/1256060413.full

By Daniel Dugger. In Section 2 Dugger proves that if $I$ as a perfect ideal such that $\operatorname{grade}(I,R)=d$ and $ \textbf{x}=x_1,\dots ,x_d$ is a regular sequence contained in $I$ such that $(x_1,\dots ,x_d) \neq I$ and $J=(\textbf{x}):I$, then $J$ is perfect and $\operatorname{grade}(J,R)=d$.

During the proof he consider $0 \to F_d \to F_{d-1} \to \cdots \to F_0 \to R/I $ be a minimal free resolutions of $R/I$ and $K_{\bullet}$ the Koszul complex of $x_1,\dots,x_d$. By comparison theorem the projection $R/(\textbf{x}) \to R/I$ lifts to a map os complexes $\pi:K_{\bullet} \to F_{\bullet}$. He says that a free resolution for $R/J$ can be obtained from the following diagram by dualizing and then taking the mapping cone.

Can someone help-me with this argument?

Answer 1

The keyword here is "linkage" or "liaison". This topic is covered in Section 21.10 of Eisenbud's Commutative Algebra, and especially Theorem 21.23 is relevant. Anyway here are the key points:

1. In this situation (from the theorem), $J/({\bf x})$ is the canonical module of $R/I$, and symmetrically, $I/({\bf x})$ is the canonical module of $R/J$. This is not too bad to prove, basically it uses that $R/({\bf x})$ is Gorenstein so ${\rm Hom}_R(-,R)$ is the duality on Cohen-Macaulay modules that you can use to compute canonical modules; also $J = {\rm Hom}_R(R/I, R/({\bf x}))$ essentially by definition of $J$.

2. This tells you that ${\rm Ext}^d_R(I/({\bf x}),R) = R/J$.

3. You have a short exact sequence

$0 \to I/({\bf x}) \to R/({\bf x}) \to R/I \to 0$.

All 3 modules are Cohen-Macaulay of codimension $d$, so applying ${\rm Hom}_R(-,R)$ gives the short exact sequence

$0 \to {\rm Ext}^d_R(R/I, R) \to {\rm Ext}^d_R(R/({\bf x}), R) \to {\rm Ext}^d_R(I/({\bf x}), R) \to 0$,

i.e.,

$0 \to \omega_{R/I} \to \omega_{R/({\bf x})} \to R/J \to 0$.

4. Since $R/({\bf x})$ is Cohen-Macaulay, the dual $K_\bullet^*$ is a resolution of its canonical module $\omega_{R/({\bf x})}$ and similarly for $F_\bullet^*$.

5. Finally, 3) and 4) imply that the mapping cone of $F^* \to K^*$ is a resolution of $R/J$ (general property of mapping cones).