## Tag `07Z2`

Chapter 10: Commutative Algebra > Section 10.56: Proj of a graded ring

Lemma 10.56.9. Let $R$ be a ring. Let $S$ be a graded $R$-algebra. Let $f \in S_{+}$ be homogeneous. Assume that $S$ is of finite type over $R$. Then

- the ring $S_{(f)}$ is of finite type over $R$, and
- for any finite graded $S$-module $M$ the module $M_{(f)}$ is a finite $S_{(f)}$-module.

Proof.Choose $f_1, \ldots, f_n \in S$ which generate $S$ as an $R$-algebra. We may assume that each $f_i$ is homogeneous (by decomposing each $f_i$ into its homogeneous components). An element of $S_{(f)}$ is a sum of the form $$ \sum\nolimits_{e\deg(f) = \sum e_i\deg(f_i)} \lambda_{e_1 \ldots e_n} f_1^{e_1} \ldots f_n^{e_n}/f^e $$ with $\lambda_{e_1 \ldots e_n} \in R$. Thus $S_{(f)}$ is generated as an $R$-algebra by the $f_1^{e_1} \ldots f_n^{e_n} /f^e$ with the property that $e\deg(f) = \sum e_i\deg(f_i)$. If $e_i \geq \deg(f)$ then we can write this as $$ f_1^{e_1} \ldots f_n^{e_n}/f^e = f_i^{\deg(f)}/f^{\deg(f_i)} \cdot f_1^{e_1} \ldots f_i^{e_i - \deg(f)} \ldots f_n^{e_n}/f^{e - \deg(f_i)} $$ Thus we only need the elements $f_i^{\deg(f)}/f^{\deg(f_i)}$ as well as the elements $f_1^{e_1} \ldots f_n^{e_n} /f^e$ with $e \deg(f) = \sum e_i \deg(f_i)$ and $e_i < \deg(f)$. This is a finite list and we see that (1) is true.To see (2) suppose that $M$ is generated by homogeneous elements $x_1, \ldots, x_m$. Then arguing as above we find that $M_{(f)}$ is generated as an $S_{(f)}$-module by the finite list of elements of the form $f_1^{e_1} \ldots f_n^{e_n} x_j /f^e$ with $e \deg(f) = \sum e_i \deg(f_i) + \deg(x_j)$ and $e_i < \deg(f)$. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 13064–13073 (see updates for more information).

```
\begin{lemma}
\label{lemma-dehomogenize-finite-type}
Let $R$ be a ring. Let $S$ be a graded $R$-algebra. Let $f \in S_{+}$
be homogeneous. Assume that $S$ is of finite type over $R$. Then
\begin{enumerate}
\item the ring $S_{(f)}$ is of finite type over $R$, and
\item for any finite graded $S$-module $M$ the module $M_{(f)}$
is a finite $S_{(f)}$-module.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose $f_1, \ldots, f_n \in S$ which generate $S$ as an $R$-algebra.
We may assume that each $f_i$ is homogeneous (by decomposing each $f_i$
into its homogeneous components). An element of $S_{(f)}$ is a sum
of the form
$$
\sum\nolimits_{e\deg(f) =
\sum e_i\deg(f_i)} \lambda_{e_1 \ldots e_n} f_1^{e_1} \ldots f_n^{e_n}/f^e
$$
with $\lambda_{e_1 \ldots e_n} \in R$. Thus $S_{(f)}$ is generated
as an $R$-algebra by the $f_1^{e_1} \ldots f_n^{e_n} /f^e$ with the
property that $e\deg(f) = \sum e_i\deg(f_i)$. If $e_i \geq \deg(f)$
then we can write this as
$$
f_1^{e_1} \ldots f_n^{e_n}/f^e =
f_i^{\deg(f)}/f^{\deg(f_i)} \cdot
f_1^{e_1} \ldots f_i^{e_i - \deg(f)} \ldots f_n^{e_n}/f^{e - \deg(f_i)}
$$
Thus we only need the elements $f_i^{\deg(f)}/f^{\deg(f_i)}$ as well
as the elements $f_1^{e_1} \ldots f_n^{e_n} /f^e$ with
$e \deg(f) = \sum e_i \deg(f_i)$ and $e_i < \deg(f)$.
This is a finite list and we see that (1) is true.
\medskip\noindent
To see (2) suppose that $M$ is generated by homogeneous elements
$x_1, \ldots, x_m$. Then arguing
as above we find that $M_{(f)}$ is generated as an $S_{(f)}$-module
by the finite list of elements of the form
$f_1^{e_1} \ldots f_n^{e_n} x_j /f^e$
with $e \deg(f) = \sum e_i \deg(f_i) + \deg(x_j)$ and
$e_i < \deg(f)$.
\end{proof}
```

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