# The Stacks Project

## Tag 00JR

Lemma 10.56.5. Let $S$ be a graded ring. Let $M$ be a graded $S$-module. Let $\mathfrak p$ be an element of $\text{Proj}(S)$. Let $f \in S$ be a homogeneous element of positive degree such that $f \not \in \mathfrak p$, i.e., $\mathfrak p \in D_{+}(f)$. Let $\mathfrak p' \subset S_{(f)}$ be the element of $\mathop{\mathrm{Spec}}(S_{(f)})$ corresponding to $\mathfrak p$ as in Lemma 10.56.3. Then $S_{(\mathfrak p)} = (S_{(f)})_{\mathfrak p'}$ and compatibly $M_{(\mathfrak p)} = (M_{(f)})_{\mathfrak p'}$.

Proof. We define a map $\psi : M_{(\mathfrak p)} \to (M_{(f)})_{\mathfrak p'}$. Let $x/g \in M_{(\mathfrak p)}$. We set $$\psi(x/g) = (x g^{\deg(f) - 1}/f^{\deg(x)})/(g^{\deg(f)}/f^{\deg(g)}).$$ This makes sense since $\deg(x) = \deg(g)$ and since $g^{\deg(f)}/f^{\deg(g)} \not \in \mathfrak p'$. We omit the verification that $\psi$ is well defined, a module map and an isomorphism. Hint: the inverse sends $(x/f^n)/(g/f^m)$ to $(xf^m)/(g f^n)$. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 12972–12984 (see updates for more information).

\begin{lemma}
\label{lemma-proj-prime}
Let $S$ be a graded ring. Let $M$ be a graded $S$-module.
Let $\mathfrak p$ be an element of $\text{Proj}(S)$.
Let $f \in S$ be a homogeneous element of positive degree
such that $f \not \in \mathfrak p$, i.e., $\mathfrak p \in D_{+}(f)$.
Let $\mathfrak p' \subset S_{(f)}$ be the element of
$\Spec(S_{(f)})$ corresponding to $\mathfrak p$ as in
Lemma \ref{lemma-topology-proj}. Then
$S_{(\mathfrak p)} = (S_{(f)})_{\mathfrak p'}$
and compatibly
$M_{(\mathfrak p)} = (M_{(f)})_{\mathfrak p'}$.
\end{lemma}

\begin{proof}
We define a map $\psi : M_{(\mathfrak p)} \to (M_{(f)})_{\mathfrak p'}$.
Let $x/g \in M_{(\mathfrak p)}$. We set
$$\psi(x/g) = (x g^{\deg(f) - 1}/f^{\deg(x)})/(g^{\deg(f)}/f^{\deg(g)}).$$
This makes sense since $\deg(x) = \deg(g)$ and since
$g^{\deg(f)}/f^{\deg(g)} \not \in \mathfrak p'$.
We omit the verification that $\psi$ is well defined, a module map
and an isomorphism. Hint: the inverse sends $(x/f^n)/(g/f^m)$ to
$(xf^m)/(g f^n)$.
\end{proof}

Comment #968 by JuanPablo on August 30, 2014 a 3:01 am UTC

Here the definition of the isomorphism $\psi$ should be: $\psi(x/g) = (xg^{\deg(f) - 1}/f^{\deg(x)})/(g^{\deg(f)}/f^{\deg(g)})$

($x^{\deg(f)}$ does not make sense in a module)

And maybe indicate the inverse? $\phi((x/f^n)/(g/f^n))=x/g$

Comment #1002 by Johan (site) on September 6, 2014 a 4:38 pm UTC

Yes, thanks! Fixed here.

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

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