Hilbert's syzygy theorem

Remarks

One proof uses the Koszul complex of $x_1, \ldots, x_n$, which is a free resolution of $K = S/(x_1, \ldots, x_n)$ of length $n$. It shows that $\operatorname{Tor}_i^S(M, K) = 0$ for $i > n$, and the length of a minimal graded free resolution of $M$ is the largest $i$ with $\operatorname{Tor}_i^S(M, K) \ne 0$. The bound $n$ is sharp: for $n = 2$, the resolution $0 \to S \xrightarrow{(-y,\, x)^T} S^2 \xrightarrow{(x \;\; y)} S \to K \to 0$ of $K = S/(x, y)$ has length $2$, and no shorter resolution exists. The same method shows that the global dimension of $K[x_1, \ldots, x_n]$ is $n$.

Going deeper

For a student

The theorem says that the polynomial ring $K[x_1, \ldots, x_n]$ has global dimension $n$. It is closely related to Serre's theorem on regular local rings, since the localizations of $K[x_1, \ldots, x_n]$ at maximal ideals are regular. The minimal free resolution gives the graded Betti numbers of $M$, and Gröbner bases make it computable.