Free resolution of $(x^2, y^2, xy+yz)$

Question

Let $S = k[x,y,z]$ ($k$ field) and let $I$ be the ideal $(x^2,y^2,xy+yz)$. I computed a minimal free resolution of $S/I$, and the dimensions of the free modules in the resolution are 1,3,3,1. (Just to make sure, I confirmed the result with Singular.)

But now, I see the following result in Eisenbud's The Geometry of Schemes (Theorem III-61, p. 133):

If $I$ is the homogeneous ideal of a zero-dimensional subscheme $X \subset \mathbb{P}_K^2$, then any minimal free resolution of the homogeneous coordinate ring $S/I$ has the form $$0 \longrightarrow \sum_{j=1}^{n-1} S(-b_{2j}) \longrightarrow \sum_{j=1}^{n} S(-b_{1j}) \longrightarrow S.$$

(I have corrected what appears to be a typo in Eisenbud, namely, a missing summation over the $S(-b_{1j})$ term.)

The resolution quoted in the theorem has only three nonzero terms, but the one I computed has four. What happened? Do I need to assume that $I$ is saturated for the theorem to apply? (The saturation of $I$ is $(x^2,y)$ which has only 3 terms in the minimal free resolution of dimensions 1,2,1, which is consistent with the theorem above.)

Answer 1

I think your interpretation is correct.

Let $I$ be the homogeneous ideal of zero-dimensional schemes $X \subseteq \mathbb{P}^2$. Let $S = k[x,y,z]$ be the homogeneous coordinate ring of $\mathbb{P}^2$. Then the homogeneous coordinate ring of $X$ is $S/I$ which is of (Krull) dimension $1$. The above quoted resolution is a special case of the Hilbert-Burch theorem. We can apply the theorem if the ideal $I$ is a Cohen-Macaulay ideal, i.e., $S/I$ is Cohen-Macaulay, of codimension $2$. When $\dim S/I = 1$, $I$ is Cohen-Macaulay if and only if $I$ does not have embedded components. Therefore, by saturating with respect to the irrelevant maximal ideal $(x,y,z)$ we have a Cohen-Macaulay ideal.

In other words the original ideal $I$ in your example is not a Cohen-Macaulay ideal, but its saturation is. If you use Singular, you may want to compare the primary decompositions of $I$ and its saturation.