Lemma 10.56.1. Let $S$ be a graded ring. Let $M$ be a graded $S$-module.
If $S_+M = M$ and $M$ is finite, then $M = 0$.
If $N, N' \subset M$ are graded submodules, $M = N + S_+N'$, and $N'$ is finite, then $M = N$.
If $N \to M$ is a map of graded modules, $N/S_+N \to M/S_+M$ is surjective, and $M$ is finite, then $N \to M$ is surjective.
If $x_1, \ldots , x_ n \in M$ are homogeneous and generate $M/S_+M$ and $M$ is finite, then $x_1, \ldots , x_ n$ generate $M$.
Proof. Proof of (1). Choose generators $y_1, \ldots , y_ r$ of $M$ over $S$. We may assume that $y_ i$ is homogeneous of degree $d_ i$. After renumbering we may assume $d_ r = \min (d_ i)$. Then the condition that $S_+M = M$ implies $y_ r = 0$. Hence $M = 0$ by induction on $r$. Part (2) follows by applying (1) to $M/N$. Part (3) follows by applying (2) to the submodules $\mathop{\mathrm{Im}}(N \to M)$ and $M$. Part (4) follows by applying (3) to the module map $\bigoplus S(-d_ i) \to M$, $(s_1, \ldots , s_ n) \mapsto \sum s_ i x_ i$. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #5356 by Peter Bruin on
Comment #5595 by Johan on
There are also: