# How to show $\alpha_d : M_d \to \Gamma(X, \widetilde{M(d)})$ an isomorphism for sufficiently large $d$?

Let $S$ be a (positively) graded ring and $X = \operatorname{Proj} S$. Suppose $S$ is generated by $S_1$ as an $S_0$ - algebra and suppose further that $S_1$ is a finitely generated $S_0$ - module. Thus there are $x_1,\ldots,x_r \in S_1$ such that $S = S_0[x_0,\ldots,x_r]$, hence $S$ is Noetherian.

Let $M$ be a finitely generated $S$ - module; necessarily $M$ is Noetherian since $S$ is. I want to show:

The natural map $\alpha_d : M_d \to \Gamma(X,\widetilde{M(d)})$ is an isomorphism for sufficiently large $d$.

Now I have been trying to get my head around this problem for many days but I don't even have a clear approach. The facts I know are patchy; one fact I do know is that because $S =S_0[x_0,\ldots,x_r]$ with the $x_i$'s in degree $1$, we have that $X= \operatorname{Proj} S$ is covered by finitely many $D_+(x_i)$'s.

However even though I know this I can't say why $\alpha_d$ is injective which is very frustrating: If $\alpha_d(m) = 0$, then there are $n_i$'s for which $x_i^{n_i}m = 0$. How can I conclude $m = 0$?

My question is: What are some general strategies of attacking problems like this?

Edit: Here are some extra thoughts now after the discussion with Martin. Since $M$ is Noetherian we have a filtration $$0 = M^0 \subseteq M^1 \subseteq \ldots \subseteq M^r = M$$ such that for every $0\leq i\leq r-1$, we have $M^{i+1}i/M^{i} \cong (S/p_i)(n_i)$ where $p_i's$ some homogeneous prime ideals of $S$. We can then consider the diagram

$$\require{AMScd} \begin{CD} 0 @>>> M^i @>>> M^{i+1} @>>> (S/p_i)(n_i) @>>> 0\\ @. @VVV @VVV @VVV\\ 0 @>>> \Gamma(X,\widetilde{M^{i}}) @>>> \Gamma(X,\widetilde{M^{i+1}}) @>>> \Gamma(X,\widetilde{S/(p_i)}(n_i)). \end{CD}$$

Now we can apply some snake lemma and induction over here as follows. Suppose we know for all $i$ that there is $d_i$ sufficently large for which $(S/p_i)(n_i)_{d_i} \cong \Gamma(X,\widetilde{S/(p_i)}(n_i))_{d_i}.$ Then in the inductive step we know that the left vertical map is an isomorphism for some $d_i$ sufficiently large, and the right most vertical map an isomorphism for some $d_{i+1}$ sufficiently large. Taking $e = \operatorname{max}(d_i,d_{i+1})$ and using the Snake lemma shows that the middle map is an isomorphism for $d \geq e$.

Conclusion: So as Martin says we can reduce to the case that $M = S/I$. But how can we even do this? Hartshorne claims his proof of Theorem 5.19 shows this but I am not seeing it.

• Duplicate of math.stackexchange.com/questions/56234 but the answer there uses cohomology (which is not necessary here and of course not intended as an exercise in Chapter II of Hartshorne's book). Commented Sep 17, 2013 at 8:28
• Isn't it more natural to use the 5-lemma to conclude the isomorphism? Commented Jul 17, 2020 at 11:45

We have reduced the question to showing $(S/\mathfrak{p})(n)_d \cong \Gamma(X,\widetilde{S/\mathfrak{p}}(n+d))$ for large enough $d$. We can assume $n=0$ since $(S/\mathfrak{p})(n)_d = (S/\mathfrak{p})_{n+d}$. We claim $\Gamma(X,\widetilde{S/\mathfrak{p}}(d)) \cong \Gamma(V(\mathfrak{p}),\widetilde{S/\mathfrak{p}}(d))$. This is a bijection by looking at the definition of $(~)^\sim$, since $$(S/\mathfrak{p}(d))_{(\mathfrak{q})} = ((S/\mathfrak{p})_{\mathfrak{q}})_d = 0$$ for all $\mathfrak{q} \notin V(\mathfrak{p})$, i.e., for all $\mathfrak{q} \not \supset \mathfrak{p}$.

So, by using the identification $\operatorname{Proj} S/\mathfrak{p} \cong V(\mathfrak{p})$, it suffices to show the following special case: 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 $k$. Then, $S_d \cong \Gamma(X,\mathcal{O}_X(d))$ for all $d \gg 0$.

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. 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.

Do you really want to show this by yourself? It is (part of) one of the main theorems in Serre's FAC (when $S_0$ is a field). You can find a proof in EGA II, 2.7.11. (i).

• Dear Martin, this was exercise 2.5.9(b) in Hartshorne that I was trying to do. Hartshorne suggested to do it using the methods in the proof of Theorem 2.5.19. This doesn't assume any cohomology, but even then I don't know how to adapt the proof to my situation here.
– user38268
Commented Sep 17, 2013 at 8:01
• Have you tried to adapt the proof? It uses a filtration argument and reduces thr whole thing to $M=S/I$ and then by a base change argument to $M=S$. Commented Sep 17, 2013 at 8:26
• Dear @Martin, I think we can reduce straight to the case that $M= S/I$; please see my edit above.
– user38268
Commented Sep 17, 2013 at 9:01
• It is enough to consider the case $I=0$, by applying the whole thing to $\mathrm{Proj}(S/I)$. The special case of a polynomial ring is already covered in Hartshorne's book. At the moment I don't know a direct argument for the general case. Commented Sep 17, 2013 at 9:59
• I want to delete my question as it has no answer. To do this, I need you to delete your answer. Is that ok for you?
– user38268
Commented Oct 10, 2013 at 4:06