Here is Hilbert's theorem from Eisenbud:
There are a few things I don't understand:
(1) Why does $H_M(s)=0$ for $s$ large in the case of graded vector spaces?
(2) What is the point of working with twists here? Is it used somehow that the maps between twisted objects have degree $0$ (map homogeneous elements to homogeneous elements of the same degree)?
(3) How does the equality (second display) follow from taking components of degree $s$? That is, why is the sum of dimensions of terms 1 and 3 equal the sum of dimensions of terms 2 and 4?
(4) I don't quite understand why $K$ and $M/x_rM$ are finitely generated modules over $k[x_1,\dots,x_{r-1}]$. Well, a f.g. module over a Noetherian ring is Noetherian, and $K$ is a submodule of a f.g. Noetherian module, so it's finitely generated. And the quotient of a Noetherian module is f.g. But why are they modules over $k[x_1,\dots,x_{r-1}]$?
Best Answer
(1) A finite dimensional graded vector space has a finite basis. The degrees of these basis elements is a finite list of integers. Letting $B$ be the maximum degree, then $H_M(s) = 0$ for all $s > B$ since there are no elements of degree larger than $B$.
(2) Yes, that's right: a morphism of graded modules must preserve degrees, and that's what twisting accomplishes.
(3) This is just a statement about exact sequences of vector spaces, namely that they have Euler characteristic $0$. (See this answer.) This is really just a highbrow version of the rank-nullity theorem from linear algebra. Rank-nullity says that for a vector space $V$ with a subspace $W$, $\dim(V/W) = \dim(V) - \dim(W)$, i.e., $\dim(W) - \dim(V) + \dim(V/W) = 0$. We can express this as a short exact sequence of vector spaces $0 \to W \to V \to V/W \to 0$. One can generalize this result and show that given any exact sequence of vector spaces $0 \to V_1 \to V_2 \to \cdots \to V_n \to 0$, we have $\sum_{i=1}^n (-1)^i \dim(V_i) = 0$.
Applying this to the exact sequence $$ 0 \to K(-1) \to M(-1) \to M \to M/x_r M \to 0 $$ we get \begin{align*} 0 &= H_{K(-1)}(s) - H_{M(-1)}(s) + H_M(s) - H_{M/x_r M}(s) = H_{K}(s-1) - H_{M}(s-1) + H_M(s) - H_{M/x_r M}(s) \end{align*} and rearranging yields the result.
(4) We can consider any $k[x_1, \ldots, x_r]$-module as a $k[x_1, \ldots, x_{r-1}]$-module by considering the inclusion $k[x_1, \ldots, x_{r-1}] \hookrightarrow k[x_1, \ldots, x_r]$. The key observation here is that $x_r$ acts as $0$ on both $K$ and $M/x_r M$, so they are still finitely generated over the subring $k[x_1, \ldots, x_{r-1}]$.