Extension by zero not Quasi-coherent.

Question

Hartshorne's Example 5.2.3 in Chapter 2 states that if $X$ is an integral scheme, and $U$ is an open subscheme with $i:U \rightarrow X$ the inclusion, then if $V$ is any open affine not contained in $U$, $i_{!}(\mathcal{O}_U) \mid_{V}$ will have no sections over $V$. But it will have non-zero stalks, and so cannot come from a module on $V$ and so is not be quasi-coherent.

I get everything but for the fact that the extension by zero will have no sections over $V$. My main problem is that Hartshorne states that $i_{!}(\mathcal{O}_U)$ is the sheaf associated to the presheaf $$P(W) \mapsto \begin{cases}\mathcal{O}_U(W) &:\text{if } W \subseteq U \\ 0 &: \text{otherwise}. \end{cases}$$

So a section over $V$ would be a map $s$ from $V$ to the disjoint union of the stalks of $P$ at points in $V$ subject to the compatibility condition. I am having trouble showing that this map vanishes for all $p$. Clearly $s$ must vanish on $V-U$. For $p$ in $U \cap V$, then there must be some $W \subseteq V$ and $g \in P(W)$ so that $s(q)=g_q$ for all $q \in W$. And now no ideas spring to mind. Ideally I'd like $W$ to intersect with $V - U$ and then get that this section must vanish on a neighborhood of $p$, but I can't seem to get this to work out. The problem is that $V-U$ is closed in $V$, and $W$ is open in $V$, so they need not intersect. I can get that $W$ and $U$ intersect, but that isn't really helpful. Any tips?

Answer 1

Consider the short exact sequence $$0 \longrightarrow i_!(\mathcal{O}_X\rvert_U) \longrightarrow \mathcal{O}_X \longrightarrow j_*(\mathcal{O}_X\rvert_Z) \longrightarrow 0$$ where $Z = X \setminus U$, and $j \colon Z \hookrightarrow X$ is the inclusion map, from Exercise II.1.19(c). Taking sections on $V$, we have the exact sequence $$0 \longrightarrow \Gamma(V,i_!(\mathcal{O}_X\rvert_U)) \longrightarrow \Gamma(V,\mathcal{O}_X) \longrightarrow \Gamma(V \cap Z,\mathcal{O}_X\rvert_Z)$$ But the map $\Gamma(V,\mathcal{O}_X) \to \Gamma(V \cap Z,\mathcal{O}_X\rvert_Z)$ is injective, since the composition $$\Gamma(V,\mathcal{O}_X) \longrightarrow \Gamma(V \cap Z,\mathcal{O}_X\rvert_Z) \longrightarrow \mathcal{O}_{V,z}$$ where $z \in V \cap Z$ is injective (see Prop. 3.29 in Görtz/Wedhorn). Thus, $\Gamma(V,i_!(\mathcal{O}_X\rvert_U)) = 0$ by exactness.

EDIT: Thank you to MooS for the injectivity argument!

Answer 2

Denote that sheaf by $\mathcal F$ and let $V = \operatorname{Spec}(A)$. Let $\eta$ be the generic point of $X$ (which also happens to be the generic point of $U$ and $V$). If $\mathcal F = M^\sim$ for some $A$-module, then $0 \neq K(X) = \mathcal O_{X,\eta} = \mathcal F_\eta = \operatorname{Frac}(A) \otimes M$, hence any localization of $M$ does not vanish, but $\mathcal F$ has vanishing stalks.

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This has shown, that the module is not quasi-coherent.

If you are in particular interested in the proof, that $\mathcal F$ has no sections on $V$, look at this: Let $s=(s_x)_{x \in V}$ be a section on $V$. As you have said, there is some $W \ni x$ and some $g \in P(W)$ with $s_q=g_q$ for any $q \in W$. Note that, if we want $s \neq 0$, we need $P(W) \neq 0$, i.e. $W \subset U$, thus we have $P(W)=\mathcal O_U(W) = \mathcal O_X(W) \subset K(X)$. So $q \in K(X)$, hence we have $g_q = g_x$ for any $q \in W$. This shows: the vanishing set and the non-vanishing set of $s$ are both open, hence one of them is $V$. Since $s$ has vanishing stalks on $V - U$, we deduce $s=0$.

Note that we do not need $V$ to be affine, but of course we need $V$ to be affine to conclude that $\mathcal F$ is not quasi-coherent.

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The key algebraic statement of the argument is the following: If $f \in M$ for a torsion-free module over an integral domain $A$, we have $f=0$ if and only if $\frac{f}{1} = 0 \in S^{-1}M$ with $S=A \setminus \{0\}$. Hence, for torsion-free quasi-coherent sheaves over integral schemes, you can test the vanishing of sections at the generic point. In fact you can test the vanishing on any point, i.e. $f=0$ if and only if $f_x=0$ for some point $x$.

This is a property, that $i_{!} \mathcal O_U$ certainly does not admit. Hence it is not quasi-coherent.

Answer 3

Here is an alternative line of reasoning:

I think in the specific case when $X$ is integral, the pre sheaf $F = j_! (O_U)$ is already a sheaf. (Unless I overlooked something - be skeptical!)

So sheafification doesn't add any global sections, and the claim that there are no sections over $V$ is obvious for the pre sheaf.

Let's check that $F$ is a sheaf:

Let $W_a$ be an open cover of $W$. There are two cases:

1) $W \subseteq U$:

In general the restriction of $F$ to $U$ is the restriction of $O_X$ to $U$. So this case is done.

2) $W \not \subseteq U$:

$F(W) = 0$, so the map $F(W) \to \Pi F(W_i)$ is injective tautologically. We have to check exactness at the middle of $F(W) \to \Pi F(W_i) \to \Pi F(W_{ij})$, here we will use integrality (both reducedness and irreducibility).

There must be some $W_a \not \subseteq U$. Suppose that there is some $W_b \subseteq U$ (if not, the middle term is just 0, so again is tautologically exact). Since $X$ was integral, the restriction of the structure sheaf from $W_a$ to $W_a \cap W_b \not = \emptyset$ induces an injection (note that both of these sheaves are in $U$, so $F$ agrees with the restriction of the structure sheaf there).

But now suppose that $f_i \in \Pi F(W_i)$. $f_a = 0$, and $f_a = f_b$ in $W_b \cap W_a \not = \emptyset$. But this implies that $f_b = 0$ in $W_a \cap W_b$, which implies that $f_b = 0$ in $W_b$, by injectivity of the restriction map.

So in fact $f_i = 0$, hence is in the image of $F(W)$.