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Tag 00JR

Chapter 10: Commutative Algebra > Section 10.56: Proj of a graded ring

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 $\mathop{\rm Spec}(S_{(f)})$ 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 12969–12981 (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}

    Comments (2)

    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 5 comments on Section 10.56: Commutative Algebra.

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