Lemma 10.57.5. 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 10.57.3. 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 $(xf^ m)/(g f^ n)$. $\square$

Comment #968 by JuanPablo on

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? $\phi((x/f^n)/(g/f^n))=x/g$

There are also:

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