Let $S=K[X_1,\ldots,X_n]$ and $M$ be a Cohen-Macaulay $S$-module. This equality holds $$ \operatorname{reg}(M)=\dim(M)+\max\{i\in\mathbb{Z}\colon P_{M}(i)\neq H(M,i)\}. $$
It's been proved in Eisenbud, "The Geometry of Syzygies" in Theorem 4.15. I can't understand it because that book has a lot of geometry in its proofs. Can you help me understand it? Or can anybody help me find a more algebraic version in other books? Thank you.
I'd use Bruns and Herzog, Cohen-Macaulay Rings, Theorem 4.4.3(b) noticing that the only non-zero local cohomology module occurs for $d=\dim M$.
Maybe a simpler approach (at least without local cohomology).
One knows that $\max\{i\in\mathbb{Z}\colon P_{M}(i)\neq H(M,i)\}=\deg H_M(t)$ (see Bruns and Herzog, Cohen-Macaulay Rings, Exercise 4.4.10). We can assume that $K$ is infinite and then there is a maximal $M$-sequence $x_1,\dots,x_d$ consisting of homogeneous elements of degree one. From Eisenbud, Commutative Algebra, Proposition 20.20 we have $\operatorname{reg}M=\operatorname{reg}M/\underline{x}M$. On the other side, $H_{M/\underline{x}M}(t)=(1-t)^dH_M(t)$, so $\deg H_{M/\underline{x}M}(t)=d+\deg H_M(t)$. Since $\dim M/\underline{x}M=0$ we have $\operatorname{reg}M/\underline{x}M=\max\{i:(M/\underline{x}M)_i\ne 0\}=\max\{i:H(M/\underline{x}M,i)\ne0\}=\deg H_{M/\underline{x}M}(t)$, and we are done.