In a regular local ring every ideal has a FFR

Question

Let $A$ be a regular local ring. Then every ideal has a finite free resolution.

My thoughts: it's easy to prove that every ideal $I$ has a free resolution. In fact $I$ is finite and there is a surjective map $$f\colon A^n\rightarrow I$$ Now $\ker{f}$ is again finite and i can repeat, but how can i prove that this resolution is definitely 0?

Answer 1

Theorem-2.2.7 [Cohen-Macaulay Rings by Bruns and Herzog] Let $(R,m,k)$ be a Noetherian local ring. Then the following are equivalent:

(a) $R$ is regular

(b) proj dim $M$ is finite for every finite $R$-module $M$

(c) proj dim $k$ is finite.

Answer 2

This is because a regular local ring has finite homological dimension (it is a characterisation of regularity). Indeed $\operatorname{dh}(A)$, the global homological dimension of $A$, is equal to the projective dimension $\operatorname{dp}(k)$ of its residue field $k=A/\mathfrak m$, and the latter has a finite free resolution of length $\dim A$ via the Koszul complex since $\mathfrak m$ is generated by a regular sequence of length $\dim A$.