Recall the following theorem, due to Hilbert. Let ${k}$ be a field, and ${S=k[x_0,...,x_n]}$ considered as a graded ring, and take ${M}$ any finitely generated graded ${S}$-module. Then there is a free resolution of ${M}$ of length at most ${n+2}$. This theorem is important because applying it in the case ${M=S/I}$ for ${I}$ a homogeneous ideal (i.e. to a projective variety), it tells us that the Hilbert polynomial exists and gives an explicit bound on its degree.

I had always assumed this theorem was in some sense fairly deep or hard, but in Chan’s scheme theory course today we saw that it’s actually a small amount of some fairly basic homological algebra.

Step 1: Do the case $M=k$, where each ${x_i}$ acts as the zero homomorphism. I would rather not do the details of this in a blog post, but it isn’t very hard, and you if you try to write down a resolution that’s as efficient as possible, you should end up with a resolution of length ${n+2}$, which usually seems to be given the intimidating name `the Koszul complex’.

Step 2: Let ${M}$ now be arbitrary, and consider a minimal free resolution ${... F_n \rightarrow F_{n-1} \rightarrow ... \rightarrow F_0 \rightarrow M \rightarrow 0}$ (in the sense that the first map has the smallest possible number of generators, then the second one does, and so on). By minimality, for all ${k\geq 1}$ a basis element of ${F_k}$ must be mapped to an element with no constant term in any coefficient. Indeed, if ${e \in F_{k+1}}$ is mapped to ${(p_1,...,p_l) \in F_k}$, each ${p_i}$ is homogeneous (since the maps are maps of graded modules) and if some ${p_i}$ is constant, since ${k}$ is a field and ${(p_1,...,p_l)}$ is in the kernel of ${F_k \rightarrow F_{k-1}}$, we see that the basis vector ${e_i}$ maps to an element that is expressible as a linear combination of the image of the other ${e_j}$s, contradicting minimality. What this observation buys us is that if we take such a complex and tensor it with the ${S}$-module ${k}$, we get a complex

$\displaystyle ... \rightarrow F_n \otimes_S k \rightarrow ... \rightarrow F_1 \otimes_S k \rightarrow F_0 \otimes_S k \rightarrow M \otimes_S k \rightarrow 0,$
and all of the maps save the first one are the zero map.

Step 3: Use the fact that ${A \otimes B = B \otimes A}$, or more generally that ${Tor_i(A,B) = Tor_i(B,A)}$. This really is the key ingredient. Knowing that the above maps are zero, we see that the rank of ${F_j}$ is just the rank of ${Tor_j^S(M,k)}$. And this is the rank of ${Tor_j^S(k,M)}$, which can be computed by as the homology of the complex obtained by tensoring the Koszul complex with ${M}$. In particular, we see that it vanishes for ${i>n+1}$, so Hilbert’s theorem is proved.