# Hartshorne II Ex 5.9(a) or R. Vakil Ex 15.4.D(b): Saturated modules

The question is basically like this:

Prove that if $S_{\cdot}$ is a finitely generated (in degree 1) graded ring over a field $k$ and $M_{\cdot}$ is finitely generated, then the saturation map $M_{n}\rightarrow\Gamma(\text{Proj}S_{\cdot},\widetilde{M(n)_{\cdot}})$ is an isomorphism for large $n$.

The hint in [Hartshorne] says we can follow the hint in the proof of Theorem 5.19. But here is what I have tried to do but I couldn't solve it. Will be glad if someone can give the answer.

Discussion

The only reason why I would need a large $n$ is because of Serre's theorem A, which says that there exists $n_{0}$ such that the quasicoherent sheaf $\widetilde{M(n)_{\cdot}}$ is finitely generated by its global sections.

The module $M_{n}$ seen as a submodule of the finitely generated notherian module $M_{\cdot}$, is also finitely generated, so we let $M_{n}=(m_{1},...,m_{n})$.

By Serre's theorem, $\widetilde{M(n)_{\cdot}}$ is finitely generated by sections, we let $$\{s_{1},...,s_{k}\}\subset \Gamma(\text{Proj}S_{\cdot},\widetilde{M(n)_{\cdot})}$$ be such a generating set.

Now I have problems relating the two generating sets. The saturation map

$$\varphi:M_{n}\rightarrow \Gamma(\text{Proj}S_{\cdot},\widetilde{M(n)_{\cdot}})$$

is given as follows: if $m_{n}\in M_{n}$, and suppose $\{X_{1},...,X_{t}\}$ generates $S_{\cdot}$ in degree $1$, on $D_{+}(X_{i})$ we have

$$\varphi^{i}:m_{n}\mapsto \frac{m_{n}}{1}\in ((M_{\cdot})_{X_{i}})_{n}$$

where the last expression means the submodule of $(M_{\cdot})_{X_{i}}$ in degree $n$.

Then they glue to give a single global section, which I call it $\varphi(m_{n})$. My goal is to show that they generate $\Gamma(\text{Proj}(S_{\cdot},\widetilde{M(n)_{\cdot}})$. This way I can show that the saturation map is surjective, but I can't continue further.

Since you haven't responded to my comment, I'll give you a sketch for how the proof goes.

We first reduce to the case when $S$ is integral and $M=S$. This is similar to how it goes in the proof of Thm. 5.19. The idea is that since $S$ is noetherian, we can apply the dévissage argument from Ch. I, Thm. 7.4 that gives a filtration $$0 = M^0 \subset M^1 \subset \cdots \subset M^\ell = M$$ of graded submodules of $M$, where $M^i/M^{i-1} \cong (S/\mathfrak{p}_i)(n_i)$ for each $i$, where $\mathfrak{p}_i \in \operatorname{Proj}(S)$ and $n_i \in \mathbf{Z}$. We want to prove the claim by induction on the length $\ell$ of this filtration. If $\ell=0$ there is nothing to show; in the inductive case, we have a short exact sequence of graded modules $$0 \longrightarrow M^{\ell-1} \longrightarrow M \longrightarrow (S/\mathfrak{p}_\ell)(n_\ell) \longrightarrow 0$$ which induces the commutative diagram $$\require{AMScd} \begin{CD} 0 @>>> M^{\ell-1} @>>> M @>>> (S/\mathfrak{p}_\ell)(n_\ell) @>>> 0\\ @. @VV{\alpha_d}V @VV{\alpha_d}V @VV{\alpha_d}V\\ 0 @>>> \Gamma(X,\widetilde{M}^{\ell-1}(d)) @>>> \Gamma(X,\widetilde{M}(d)) @>>> \Gamma(X,\widetilde{S/\mathfrak{p}_\ell}(n_\ell+d)) \end{CD}$$ by exactness of $\widetilde{(-)}$, left-exactness of $\Gamma(X,-)$, and the naturally of $\alpha_d$ that you showed in part $(a)$. By inductive hypothesis the leftmost vertical arrow is an isomorphism for $d$ large enough. By the snake lemma, if we show the rightmost vertical arrow is an isomorphism for $n_\ell+d$ large enough, the middle vertical arrow will also be an isomorphism for $d$ large enough. But we can replace $X$ in $\Gamma(X,\widetilde{S/\mathfrak{p}_\ell}(n_\ell+d))$ with $\operatorname{Proj}(S/\mathfrak{p}_\ell)$, so we've reduced to showing the following:

Let $S$ be a graded integral domain, finitely generated by $S_1$ as an $S_0$-algebra, where $S_0$ is a finitely generated integral domain over $k$. Let $X = \operatorname{Proj}(S)$. Then, for $\alpha_d\colon S_d \to \Gamma(X,\mathcal{O}_X(d))$ is an isomorphism for $d \gg 0$.

But this is exactly what I showed in the other question, which I reproduce (with slight edits) here:

Letting $S' = \bigoplus_{n \ge 0} \Gamma(X,\mathcal{O}_X(n))$, and following the steps in Hartshorne's proof of Theorem 5.19, we arrive at the fact that $S'$ is a finitely generated $S$-module, and that $S' \supset S$. Let $\{z_i\}$ be our generators. Then, as Hartshorne proves earlier, $yz_i \in S_{\ge n}$ for all $y \in S_{\ge n}$ for some $n$. In particular, if we take $d_0 = \max \deg z_i + n$, then $S'_d \subset S_d$ for all $d \ge d_0$, and so we are done.

There are a lot of details in the proof that are just verbatim the same as in the proof of Thm. 5.19, so I'd encourage you to try to fill in the details by reading said proof. But feel free to comment if this wasn't clear enough!

• Thanks! I was actually working on the problem and I think the solution should be correct, though I was wondering if it is always true that $S\cong S'$ for all $d$, because I don't quite get the proof at the end. Is $S'_{d}=S_{d}$ simply because every element in the LHS is a linear combination of the form $y_{i} z_{i}$ where $y_{i}\in S_{\geq n}$? Commented Feb 16, 2015 at 12:46
• You are right with your last question, that $S'_d = S_d$ since every element in the LHS is a linear combination of the form $y_iz_i$ where $y_i \in S_{\ge n}$. This is the hardest part of the proof though, in some sense: Hartshorne uses the finiteness of integral closure that only works for finitely generated $k$-algebras that are domains. On the other hand, it is definitely not true that always $S \cong S'$ for all $d$; take for example the example in Exercise $5.14(a)$. Commented Feb 16, 2015 at 21:22