The Stacks project

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

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

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

\[ \widetilde M \times \widetilde N \longrightarrow \widetilde{M \otimes _ R N} \]

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

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 on

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

Comment #4621 by Andrea on

So in Remark 26.9.2, does this imply that ?

Comment #4626 by on

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

In lemma 27.9.1. it seams that suddenly becomes .


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