The Stacks project

110.47 Non flasque quasi-coherent sheaf associated to injective module

For more examples of this type see [Exposé II, Appendix I, SGA6] where Illusie explains some examples due to Verdier.

Consider the affine scheme $X = \mathop{\mathrm{Spec}}(A)$ where

\[ A = k[x, y, z_1, z_2, \ldots ]/(x^ nz_ n) \]

is the ring from Properties, Example 28.25.2. Set $I = (x) \subset A$. Consider the quasi-compact open $U = D(x)$ of $X$. We have seen in loc. cit. that there is a section $s \in \mathcal{O}_ X(U)$ which does not come from an $A$-module map $I^ n \to A$ for any $n \geq 0$.

Let $\alpha : A \to J$ be the embedding of $A$ into an injective $A$-module. Let $Q = J/\alpha (A)$ and denote $\beta : J \to Q$ the quotient map. We claim that the map

\[ \Gamma (X, \widetilde{J}) \longrightarrow \Gamma (U, \widetilde{J}) \]

is not surjective. Namely, we claim that $\alpha (s)$ is not in the image. To see this, we argue by contradiction. So assume that $x \in J$ is an element which restricts to $\alpha (s)$ over $U$. Then $\beta (x) \in Q$ is an element which restricts to $0$ over $U$. Hence we know that $I^ n\beta (x) = 0$ for some $n$, see Properties, Lemma 28.25.1. This implies that we get a morphism $\varphi : I^ n \to A$, $h \mapsto \alpha ^{-1}(hx)$. It is easy to see that this morphism $\varphi $ gives rise to the section $s$ via the map of Properties, Lemma 28.25.1 which is a contradiction.

Lemma 110.47.1. There exists an affine scheme $X = \mathop{\mathrm{Spec}}(A)$ and an injective $A$-module $J$ such that $\widetilde{J}$ is not a flasque sheaf on $X$. Even the restriction $\Gamma (X, \widetilde{J}) \to \Gamma (U, \widetilde{J})$ with $U$ a standard open need not be surjective.

Proof. See above. $\square$

In fact, we can use a similar construction to get an example of an injective module whose associated quasi-coherent sheaf has nonzero cohomology over a quasi-compact open. Namely, we start with the ring

\[ A = k[x, y, w_1, u_1, w_2, u_2, \ldots ]/(x^ nw_ n, y^ nu_ n, u_ n^2, w_ n^2) \]

where $k$ is a field. Choose an injective map $A \to I$ where $I$ is an injective $A$-module. We claim that the element $1/xy$ in $A_{xy} \subset I_{xy}$ is not in the image of $I_ x \oplus I_ y \to I_{xy}$. Arguing by contradiction, suppose that

\[ \frac{1}{xy} = \frac{i}{x^ n} + \frac{j}{y^ n} \]

for some $n \geq 1$ and $i, j \in I$. Clearing denominators we obtain

\[ (xy)^{n + m - 1} = x^ my^{n + m}i + x^{n + m}y^ mj \]

for some $m \geq 0$. Multiplying with $u_{n + m}w_{n + m}$ we see that $u_{n + m}w_{n + m}(xy)^{n + m - 1} = 0$ in $A$ which is the desired contradiction. Let $U = D(x) \cup D(y) \subset X = \mathop{\mathrm{Spec}}(A)$. For any $A$-module $M$ we have an exact sequence

\[ 0 \to H^0(U, \widetilde{M}) \to M_ x \oplus M_ y \to M_{xy} \to H^1(U, \widetilde{M}) \to 0 \]

by Mayer-Vietoris. We conclude that $H^1(U, \widetilde{I})$ is nonzero.

Lemma 110.47.2. There exists an affine scheme $X = \mathop{\mathrm{Spec}}(A)$ whose underlying topological space is Noetherian and an injective $A$-module $I$ such that $\widetilde{I}$ has nonvanishing $H^1$ on some quasi-compact open $U$ of $X$.

Proof. See above. Note that $\mathop{\mathrm{Spec}}(A) = \mathop{\mathrm{Spec}}(k[x, y])$ as topological spaces. $\square$


Comments (0)


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 0273. Beware of the difference between the letter 'O' and the digit '0'.