Generalizations of Hilbert's Syzygy theorem Hilbert's Syzygy theorem states that a minimal free resolution of a finitely generated graded module over a (standard graded) polynomial ring in $n$ variables $k[x_1, \ldots, x_n]$ does not have more than $n+1$ terms in it.  To what rings other than the polynomial ring has Hilbert's theorem been generalized?  Does it hold for polynomial rings which are not standard graded?  Please give me a reference if the answers to these are known.
 A: As suggested by Jason I put my comments into an answer. Hilbert's theorem means that all modules over $k[x_1,\ldots ,x_n]$ have projective dimension $\leq n$ -- one says that the global dimension (aka homological dimension) of $k[x_1,\ldots ,x_n]$ is $n$. This is a property of the ring $k[x_1,\ldots ,x_n]$, it does not depend on the grading. It applies to many other rings: by a famous theorem of Serre, a local (commutative) ring has finite global dimension if and only if it is regular.
A: Here's a proof that the grading doesn't matter. It's actually relatively straightforward.
Let $R=k[x_1,\cdots,x_n]$ with  $\deg x_i = d_i$. Then the Koszul complex, which is the total complex of the tensor product of the complexes $$0\to R(-d_i)\stackrel{x_i}{\to} R\to 0,$$ forms a graded free resolution of $k$ as an $R$-module of length $n+1$.
Now let $M$ be an arbitrary graded finitely-generated $R$-module, and let $$\cdots \to F_n \to \cdots \to F_0 \to M \to 0$$ be a minimal graded free resolution, where minimal means we pick the smallest number of generators for the kernel at each step and use that to build the next module in the resolution. Now let $e_j$ be a basis element of $F_{i+1}$ and consider the image $(f_1,\cdots,f_k)\in F_i$: each of the $f$ are homogeneous by the fact our resolution is graded, and I claim none of the $f$ can be constant. To see why, if $f_1$ was constant, then the image of $e_j$ would be in the span of the other basis vectors because the images of the $e_j$ span the kernel, contradicting minimality.
What does this give us? If we apply the functor $-\otimes_R k$ to our resolution $F_\bullet$, we get a complex where every map is zero except possibly the map $F_0\otimes_R k\to M\otimes_R k$ and thus the homology is particularly simple. But now we can use the fact that Tor is symmetric to compute the rank of $\operatorname{Tor}^R_\ell(M,k)$ from resolving $M$ or resolving $k$ using the Koszul complex. Since the Tor groups vanish for $\ell > n+1$ from the Koszul complex, we see that swapping to the other resolution, $F_\ell = 0$ for $\ell >n+1$.
