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 [1973] 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.



The Existence of Pure Resolutions, Given a Degree Sequence?
It looks like your problem is having very little knowledge about Koszul complexes. I'm not very good at references, but Eisenbud definitely covers this topic in his CA-AG book.


As for your questions:


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.
3. For a general regular sequence the Koszul complex gives a resolution of the structure sheaf of the zero locus. In your case the zero locus is empty, so the complex is exact.
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.


