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Tag 052N

Lemma 10.56.10. Let $R$ be a ring. Let $R'$ be a finite type $R$-algebra, and let $M$ be a finite $R'$-module. There exists a graded $R$-algebra $S$, a graded $S$-module $N$ and an element $f \in S$ homogeneous of degree $1$ such that

1. $R' \cong S_{(f)}$ and $M \cong N_{(f)}$ (as modules),
2. $S_0 = R$ and $S$ is generated by finitely many elements of degree $1$ over $R$, and
3. $N$ is a finite $S$-module.

Proof. We may write $R' = R[x_1, \ldots, x_n]/I$ for some ideal $I$. For an element $g \in R[x_1, \ldots, x_n]$ denote $\tilde g \in R[X_0, \ldots, X_n]$ the element homogeneous of minimal degree such that $g = \tilde g(1, x_1, \ldots, x_n)$. Let $\tilde I \subset R[X_0, \ldots, X_n]$ generated by all elements $\tilde g$, $g \in I$. Set $S = R[X_0, \ldots, X_n]/\tilde I$ and denote $f$ the image of $X_0$ in $S$. By construction we have an isomorphism $$S_{(f)} \longrightarrow R', \quad X_i/X_0 \longmapsto x_i.$$ To do the same thing with the module $M$ we choose a presentation $$M = (R')^{\oplus r}/\sum\nolimits_{j \in J} R'k_j$$ with $k_j = (k_{1j}, \ldots, k_{rj})$. Let $d_{ij} = \deg(\tilde k_{ij})$. Set $d_j = \max\{d_{ij}\}$. Set $K_{ij} = X_0^{d_j - d_{ij}}\tilde k_{ij}$ which is homogeneous of degree $d_j$. With this notation we set $$N = \mathop{\mathrm{Coker}}\Big( \bigoplus\nolimits_{j \in J} S(-d_j) \xrightarrow{(K_{ij})} S^{\oplus r} \Big)$$ which works. Some details omitted. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 13150–13162 (see updates for more information).

\begin{lemma}
\label{lemma-homogenize}
Let $R$ be a ring.
Let $R'$ be a finite type $R$-algebra, and let $M$ be a finite $R'$-module.
There exists a graded $R$-algebra $S$, a graded $S$-module $N$ and
an element $f \in S$ homogeneous of degree $1$ such that
\begin{enumerate}
\item $R' \cong S_{(f)}$ and $M \cong N_{(f)}$ (as modules),
\item $S_0 = R$ and $S$ is generated by finitely many elements
of degree $1$ over $R$, and
\item $N$ is a finite $S$-module.
\end{enumerate}
\end{lemma}

\begin{proof}
We may write $R' = R[x_1, \ldots, x_n]/I$ for some ideal $I$.
For an element $g \in R[x_1, \ldots, x_n]$ denote
$\tilde g \in R[X_0, \ldots, X_n]$ the element homogeneous of minimal
degree such that $g = \tilde g(1, x_1, \ldots, x_n)$.
Let $\tilde I \subset R[X_0, \ldots, X_n]$ generated by all
elements $\tilde g$, $g \in I$.
Set $S = R[X_0, \ldots, X_n]/\tilde I$ and denote $f$ the image
of $X_0$ in $S$. By construction we have an isomorphism
$$S_{(f)} \longrightarrow R', \quad X_i/X_0 \longmapsto x_i.$$
To do the same thing with the module $M$ we choose a presentation
$$M = (R')^{\oplus r}/\sum\nolimits_{j \in J} R'k_j$$
with $k_j = (k_{1j}, \ldots, k_{rj})$. Let $d_{ij} = \deg(\tilde k_{ij})$.
Set $d_j = \max\{d_{ij}\}$. Set $K_{ij} = X_0^{d_j - d_{ij}}\tilde k_{ij}$
which is homogeneous of degree $d_j$. With this notation we set
$$N = \Coker\Big( \bigoplus\nolimits_{j \in J} S(-d_j) \xrightarrow{(K_{ij})} S^{\oplus r} \Big)$$
which works. Some details omitted.
\end{proof}

Comment #3015 by Axel St\"abler on December 11, 2017 a 8:49 am UTC

Typo: The notation switches from lower case $x$'s to upper case $X$'s in the middle of the proof.

Comment #3138 by Johan (site) on February 1, 2018 a 6:30 pm UTC

OK, this wasn't a mistake -- it was intentional. There was a small typo in the proof which may have been the reason you were confused which I fixed here. Does it make sense now?

There are also 6 comments on Section 10.56: Commutative Algebra.

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