# The Existence of Pure Resolutions, Given a Degree Sequence?

I have been trying to understand the proof of the following theorem for the last month, I read some basics of sheaves theory and their cohomology, but still can't get the idea of this important theorem, proved by Eisenbud and Schreyer, in their paper "Betti Numbers of Graded Modules and Cohomology of Vector Bundles" in page 26.

The paper can be found here : http://arxiv.org/abs/0712.1843

Let $K$ be any field, and let $d = (d_{0} \prec · · · \prec d_{n} )$ be a sequence of integers. There exists a graded $K[x_{1} , . . . , x_{n} ]$-module of finite length with $\beta_{0} = \prod_{i=0}^{i=n} \binom {d_{i}-d_{0}-1} {d_{i}-d_{i-1}-1}$ generators, whose minmal free resolution is pure with degree sequence $d$.

But first we need this lemma:

$\textbf{Lemma}$

Let $m_{0} , . . . , m_{k}$ be non-negative integers, and let the homogeneous coordinates on $\mathbb{P}_{K}^{m_{j}}$ be $x^{(j)}_{0},...,x^{(j)}_{m_{j}}$. The multilinear forms: $x_{l}= \sum_{\mu_{0}+...+\mu_{k}=l}\prod_{j=0}^{j=k} x_{\mu_{j}}^{(j)}$ for $l=0,...,\sum _{j=0}^{k} m_{j}$ have no common zeros in $\mathbb{P}_{K}^{m_{1}} \times \mathbb{P}_{K}^{m_{2}} \times ... \times \mathbb{P}_{K}^{m_{k}}$.

Now we prove the theorem To simplify the notation we may harmlessly assume that $d_{0} = 0$. Let $m_{0} = m = n − 1$, and for $i = 1,...,n$ set $m_{i} = d_{i} − d_{i-1}− 1$ and set $M = \sum_{j=0}^{k}m_{j} = d_{n} − 1$. Choose $M+1$ homogenous forms of multidegree $(1, . . . , 1)$ without a common zero on $\mathbb{P}:= \mathbb{P}^{m}\times \mathbb{P}^{m_{1}} \times \mathbb{P}^{m_{2}} \times ... \times \mathbb{P}^{m_{n}}$ such as the forms described in the previous lemma.

Let $K : 0 \rightarrow K_{M+1} \rightarrow · · · → K_{0} \rightarrow 0$ be the tensor product of the Koszul complex of these forms on $\mathbb{P}$ and the line bundle $\mathcal{O}_{\mathbb{P}} (0, 0, d_{1}, . . . , d_{n−1})$. so $K_{i} = \mathcal{O}_{P}(−i, −i, . . . , d_{n−1} − i)^{\binom{d_{n}}{i}}$ for $i = 0, . . . , d_{n}$

Let $\pi: \mathbb{P}^{m}\times \mathbb{P}^{m_{1}} \times \mathbb{P}^{m_{2}} \times ... \times \mathbb{P}^{m_{n}} \rightarrow \mathbb{P}^{m}$ be the projection onto the first factor. The complex $K$ is exact because the forms have no common zero. Hence $\textbf{R}\pi_{*}(K)=0$.

If we think of $K$ as a resolution of the zero sheaf $\mathcal{F} = 0$, and factor $\pi$ into the successive projections along the factors of the product $\mathbb{P}^{m_{1}} \times · · · \times \mathbb{P}^{m_{n}}$ , then we may use Proposition 5.3. repeatedly to get a resolution of $\pi_{∗}\mathcal{F} = 0$ that has the form: $0\rightarrow \mathcal{O}^{\beta_{n}}(-d_{n})\rightarrow ... \rightarrow \mathcal{O}^{\beta_{1}}(-d_{1}) \rightarrow \mathcal{O}^{\beta_{0}}$.

Taking global sections in all twists, we get a complex: $0\rightarrow S^{\beta_{n}}(-d_{n})\rightarrow ... \rightarrow S^{\beta_{1}}(-d_{1}) \rightarrow S^{\beta_{0}}$. that has homology of finite length. Since the length of this complex is only $n$, the Lemme d’Acyclicite of Peskine and Szpiro  shows that the complex is actually acyclic. Thus it is a pure minimal resolution, with the desired degree sequence, of a graded module of finite length.

My problem is that i can't follow the steps of the proof, so answering these questions will be a great help to me:

1. What does Koszul complex of multilinear forms on $\mathbb{P}$ mean.
2. Why tensoring with the line bundle $\mathcal{O}_{\mathbb{P}} (0, 0, d_{1}, . . . , d_{n−1})$ implies that $K_{i} = \mathcal{O}_{P}(−i, −i, . . . , d_{n−1} − i)^{\binom{d_{n}}{i}}$ for $i = 0, . . . , d_{n}$.

3. The complex $K$ is exact because the forms have no common zero.

4. How can we get that $\pi_{*}\mathcal{F}=0$, using proposition 5.3 in page 25.
• Most of these are probably not research level unfortunately (for instance, 2 is really easy if you understand 1). I would look at introductory texts to algebraic geometry and commutative algebra (for instance Eisenbud's book). Jan 14 '14 at 14:37

1. This is nothing but the Koszul complex associated to the corresponding sections of $\mathcal{O}_{\mathbb{P}}(1,1,\ldots,1)$.
2. Once again, without any twist the $i$-th term would be a direct sum of $\mathcal{O}_{\mathbb{P}}(-i,-i,\ldots,-i)$. Now apply the twist.
4. $\mathcal{F}$ is by definition the zero sheaf. Thus, $\pi_*\mathcal{F}=0$. This has nothing to do with any propositions. The Proposition you mention is used to construct a resolution of $\mathcal{F}$ of the desired form. Now, $\mathcal{F}$ being zero implies that this resolution is actually an exact complex.