Section 27.9 (01MJ): Quasi-coherent sheaves on Proj

27.9 Quasi-coherent sheaves on Proj

Let $S$ be a graded ring. Let $M$ be a graded $S$ -module. We saw in Lemma 27.8.4 how to construct a quasi-coherent sheaf of modules $\widetilde{M}$ on $\text{Proj}(S)$ and a map

27.9.0.1

$\begin{equation} \label{constructions-equation-map-global-sections} M_0 \longrightarrow \Gamma (\text{Proj}(S), \widetilde{M}) \end{equation}$

of the degree $0$ part of $M$ to the global sections of $\widetilde{M}$ . The degree $0$ part of the $n$ th twist $M(n)$ of the graded module $M$ (see Algebra, Section 10.56) is equal to $M_ n$ . Hence we can get maps

27.9.0.2

$\begin{equation} \label{constructions-equation-map-global-sections-degree-n} M_ n \longrightarrow \Gamma (\text{Proj}(S), \widetilde{M(n)}). \end{equation}$

We would like to be able to perform this operation for any quasi-coherent sheaf $\mathcal{F}$ on $\text{Proj}(S)$ . We will do this by tensoring with the $n$ th twist of the structure sheaf, see Definition 27.10.1. In order to relate the two notions we will use the following lemma.

Lemma 27.9.1. Let $S$ be a graded ring. Let $(X, \mathcal{O}_ X) = (\text{Proj}(S), \mathcal{O}_{\text{Proj}(S)})$ be the scheme of Lemma 27.8.7. Let $f \in S_{+}$ be homogeneous. Let $x \in X$ be a point corresponding to the homogeneous prime $\mathfrak p \subset S$ . Let $M$ , $N$ be graded $S$ -modules. There is a canonical map of $\mathcal{O}_{\text{Proj}(S)}$ -modules

$\widetilde M \otimes _{\mathcal{O}_ X} \widetilde N \longrightarrow \widetilde{M \otimes _ S N}$

which induces the canonical map $M_{(f)} \otimes _{S_{(f)}} N_{(f)} \to (M \otimes _ S N)_{(f)}$ on sections over $D_{+}(f)$ and the canonical map $M_{(\mathfrak p)} \otimes _{S_{(\mathfrak p)}} N_{(\mathfrak p)} \to (M \otimes _ S N)_{(\mathfrak p)}$ on stalks at $x$ . Moreover, the following diagram

$\xymatrix{ M_0 \otimes _{S_0} N_0 \ar[r] \ar[d] & (M \otimes _ S N)_0 \ar[d] \\ \Gamma (X, \widetilde M \otimes _{\mathcal{O}_ X} \widetilde N) \ar[r] & \Gamma (X, \widetilde{M \otimes _ S N}) }$

is commutative where the vertical maps are given by (27.9.0.1).

Proof. To construct a morphism as displayed is the same as constructing a $\mathcal{O}_ X$ -bilinear map

$\widetilde M \times \widetilde N \longrightarrow \widetilde{M \otimes _ S N}$

see Modules, Section 17.16. It suffices to define this on sections over the opens $D_{+}(f)$ compatible with restriction mappings. On $D_{+}(f)$ we use the $S_{(f)}$ -bilinear map $M_{(f)} \times N_{(f)} \to (M \otimes _ S N)_{(f)}$ , $(x/f^ n, y/f^ m) \mapsto (x \otimes y)/f^{n + m}$ . Details omitted. $\square$

Remark 27.9.2. In general the map constructed in Lemma 27.9.1 above is not an isomorphism. Here is an example. Let $k$ be a field. Let $S = k[x, y, z]$ with $k$ in degree $0$ and $\deg (x) = 1$ , $\deg (y) = 2$ , $\deg (z) = 3$ . Let $M = S(1)$ and $N = S(2)$ , see Algebra, Section 10.56 for notation. Then $M \otimes _ S N = S(3)$ . Note that

$\begin{eqnarray*} S_ z & = & k[x, y, z, 1/z] \\ S_{(z)} & = & k[x^3/z, xy/z, y^3/z^2] \cong k[u, v, w]/(uw - v^3) \\ M_{(z)} & = & S_{(z)} \cdot x + S_{(z)} \cdot y^2/z \subset S_ z \\ N_{(z)} & = & S_{(z)} \cdot y + S_{(z)} \cdot x^2 \subset S_ z \\ S(3)_{(z)} & = & S_{(z)} \cdot z \subset S_ z \end{eqnarray*}$

Consider the maximal ideal $\mathfrak m = (u, v, w) \subset S_{(z)}$ . It is not hard to see that both $M_{(z)}/\mathfrak mM_{(z)}$ and $N_{(z)}/\mathfrak mN_{(z)}$ have dimension $2$ over $\kappa (\mathfrak m)$ . But $S(3)_{(z)}/\mathfrak mS(3)_{(z)}$ has dimension $1$ . Thus the map $M_{(z)} \otimes N_{(z)} \to S(3)_{(z)}$ is not an isomorphism.

Comments (5)

Comment #1839 by Otto Koerner on February 15, 2016 at 21:52

In Remark 26.9.2. the Relation M (tensor S) N = S(3) is not correct. Therefore we have here a counterexample to the statemment in the first part of 26.10: S(n)(tensor S)S(m)= S(n+m).

Comment #1876 by Johan on April 01, 2016 at 13:52

Sorry, but I'm afraid I do not understand what you are saying. The tensor product of two graded $S$ -modules $M$ , $N$ is the graded $S$ -module whose underlying $S$ -module is the usual tensor product. So if $M = S(n)$ and $N = S(m)$ , then $M \otimes_S N$ is free of rank $1$ as an $S$ -module. To see that it is indeed $S(n + m)$ you have to compute the degree the generator is in; this I leave up to you.

Comment #4621 by Andrea on October 24, 2019 at 07:09

So in Remark 26.9.2, does this imply that $\widetilde{M} \otimes_{\mathcal{O}_X} \widetilde{N} \not \cong \widetilde{S(3)}$ ?

Comment #4626 by Johan on October 24, 2019 at 20:58

Dear Andrea, yes this is correct. So not only is the canonical map not an isomorphism, but there is no isomorphism period.

Comment #5319 by Thiago on June 19, 2020 at 10:52

In lemma 27.9.1. it seams that $S$ suddenly becomes $R$ .

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27.9 Quasi-coherent sheaves on Proj

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