Lemma 15.25.3. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]$ be a graded polynomial algebra over $R$, i.e., $\deg (x_ i) > 0$ but not necessarily equal to $1$. Let $M$ be a graded $S$-module. Assume
$R$ is a local ring,
$M$ is a finite $S$-module, and
$M$ is flat over $R$.
Then $M$ is finitely presented as an $S$-module.
Proof. Let $M = \bigoplus M_ d$ be the grading on $M$. Pick homogeneous generators $m_1, \ldots , m_ r \in M$ of $M$. Say $\deg (m_ i) = d_ i \in \mathbf{Z}$. This gives us a presentation
which in each degree $d$ leads to the short exact sequence
By assumption each $M_ d$ is a finite flat $R$-module. By Algebra, Lemma 10.78.5 this implies each $M_ d$ is a finite free $R$-module. Hence we see each $K_ d$ is a finite $R$-module. Also each $K_ d$ is flat over $R$ by Algebra, Lemma 10.39.13. Hence we conclude that each $K_ d$ is finite free by Algebra, Lemma 10.78.5 again.
Let $\mathfrak m$ be the maximal ideal of $R$. By the flatness of $M$ over $R$ the short exact sequences above remain short exact after tensoring with $\kappa = \kappa (\mathfrak m)$. As the ring $S \otimes _ R \kappa $ is Noetherian we see that there exist homogeneous elements $k_1, \ldots , k_ t \in K$ such that the images $\overline{k}_ j$ generate $K \otimes _ R \kappa $ over $S \otimes _ R \kappa $. Say $\deg (k_ j) = e_ j$. Thus for any $d$ the map
becomes surjective after tensoring with $\kappa $. By Nakayama's lemma (Algebra, Lemma 10.20.1) this implies the map is surjective over $R$. Hence $K$ is generated by $k_1, \ldots , k_ t$ over $S$ and we win. $\square$
Your email address will not be published.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 (0)