# Non-quasi-coherent injective sheaves

Hartshorne in his Residues and Duality shows that for locally noetherian (pre)schemes $X$, the injective objects in the category of $\mathcal O_X$-modules decompose (uniquely) into a direct sum of indecomposables. These are also characterized—this is the definition just before II, Theorem 7.11:

Definition. Let $X$ be a locally noetherian prescheme, let $x \in X$, and let $x'$ be a specialization of $x$, i.e., $x' \in \overline{\{x\}}$. Let $I$ be an injective hull of $k(x)$ over the local ring $\mathcal O_{x,X}$. We define $J(x, x')$ to be the restriction of the sheaf $i_x(I)$ to the closed subset $\overline{\{x'\}}$ of $\overline{\{x\}}$. If $x = x'$, we write simply $J(x)$ for $J(x, x) = i_X(I)$. Note that $J(x, x')$ is an indecomposable $\mathcal O_X$-module.

My first problem is with the definition in the case $x \neq x'$. I would say that a restriction should be a sheaf on $\overline{\{x'\}}$, not on the whole $X$; but perhaps some pushforward is silently taking place here.

Even with assuming that, I struggle to see what does this procedure produce. Let us try a simple example, $X = \operatorname{Spec} \mathbb Z$. The case $x = x'$ yields injective quasi-coherent sheaves, which are well-understood, namely, for each (non-zero) prime $p$ there is the skyscraper sheaf associated to the Prüfer $p$-group, and there is the constant sheaf having $\mathbb Q$ everywhere (right?).

Now the only possibility for $x \neq x'$ is that $x = 0$ and $x' = (p)$. What is the sheaf $J(x, x')$ in this case? Is it the skyscraper at $x'$ produced from the $\mathbb Z_{(p)}$-module $\mathbb Q$?

If it is so, then it seems to me that this sheaf would be a subsheaf of the constant sheaf with $\mathbb Q$, which cannot be true, since this would be (by injectivity) a direct summand, but the constant sheaf is indecomposable, so I must be doing something wrong here.

• The restriction to a closed subset is naturally a quotient, not a sub, and an indecomposable injective object can certainly have non-trivial quotients. – tracing Oct 3 '17 at 13:40
• Ah, a beginner's mistake, thanks! So, is my description of the structure of $J(0, (p))$ correct? – Olin Oct 3 '17 at 14:06