# A question about mapping cone and resolutions

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?

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.,

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

1. 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 $$F_\bullet^*$$.

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