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108.43 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 108.43.1. 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 108.43.2. 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. $\square$


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