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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

  1. the ring $S_{(f)}$ is of finite type over $R$, and
  2. 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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