S is a noetherian graded ring generated by $x_i$ and M is a graded S-mod. $M_{x_i}$ is finite generated $S_{x_i}$ module. Is $M$ a f.g. S-mod?

Question

Suppose $S$ is a noetherian graded ring finitely generated by $x_i$, whose degree are 1, and $M$ is a graded $S$-module. Suppose $M_{x_i}$ is a finitely generated $S_{x_i}$ module. Is $M$ a finitely generated $S$-module?

In fact, this problem is from the exercise 5.9(c) of Chapter II of Hartshorne's Algebraic Geometry, when I try to find that why $\Gamma_*(\mathcal{F})$ is quasi-finitely generated, where $\mathcal{F}$ is a coherent $\mathcal{O}_X$-module.

I think you mean Chapter II, but it's possible you had a different question number in mind. — aeae, Jun 05 '25 at 14:47
Are you sure you want to specify that $M$ is a noetherian module? Because then $M$ is finitely generated over $S$ with no conditions, see here on MSE. — KReiser, Jun 05 '25 at 19:22
@KReiser $M$ is not assumed to be noetherian. I want to show it's noetherian. — cute dunkey, Jun 06 '25 at 00:31
Unfortunately, you cannot do this - see my answer below. But it's okay, since you only need quasi-finitely-generated. — KReiser, Jun 06 '25 at 05:06

score 1 · Accepted Answer · answered Jun 05 '25 at 18:08

No, $M$ need not be finitely generated. Consider $S=k[x_0,x_1]$, let $L = S/(x_0,x_1)$, and let $M=\bigoplus^\infty L$. Then for all $i$, $L_{x_i} = 0$, so $M_{x_i}=0$, which is certainly finitely generated. But $M$ is clearly not finitely generated.

S is a noetherian graded ring generated by $x_i$ and M is a graded S-mod. $M_{x_i}$ is finite generated $S_{x_i}$ module. Is $M$ a f.g. S-mod?

1 Answers1