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

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

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

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 .

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01MJ. Beware of the difference between the letter 'O' and the digit '0'.



View Section 27.9 as pdf