
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. $\square$

Comment #3015 by Axel St\"abler on

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

Comment #3138 by on

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?

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