The Stacks project

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

Comments (2)

Comment #968 by JuanPablo on

Here the definition of the isomorphism should be:

( does not make sense in a module)

And maybe indicate the inverse?

There are also:

  • 6 comment(s) on Section 10.57: Proj of a graded ring

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