Let $k$ be a field and $S=k[x_1,\dots,x_r]$ the polynomial ring in $r$ indeterminates. Let $M$ be a finitely-generated, graded $S$-module, such that there exists a homogeneous $M$-regular element $\xi \in S$ of degree $1$.
Question: Is it true that $\operatorname{Ext}_S^r(M,S)=0$? How can we see that?
Motivation: Last three lines of the proof of Theorem 20.17 in Eisenbud.
You can use Hilbert Syzygy theorem to state that the projective dimension of $M$ is finite, and is at most $r$. Now you can use Auslander Buchsbaum formula (it works also with graded $\mathbb K[x_1,\ldots, x_r]$ modules): $$ pd(M)=depth(S)-depth(M)=r-depth(M).$$ Now the fact that there exist an homogeneous regular element ensure that $depth(M)$ is at least $1$, so the projective dimension of $M$ is at most $r-1$ and you conclude.