# Extension by zero not Quasi-coherent.

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?

• I think this result is false. One would need to know about the behavior of the sections near the boundary. Perhaps that's where you need integral. – Future Mar 4 '16 at 3:54

## 4 Answers

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 GĂ¶rtz/Wedhorn). Thus, $\Gamma(V,i_!(\mathcal{O}_X\rvert_U)) = 0$ by exactness.

EDIT: Thank you to MooS for the injectivity argument!

• Let $p: \{z\} \hookrightarrow V \cap Z$ be the inclusion of some point. Consider the composition $$\mathcal O_X \to j_*j^{-1}\mathcal O_X \to j_*p_*p^{-1}j^{-1}\mathcal O_X.$$ The latter ist just the push-forward of the stalk at $z$, I wrote it this way to make clear where the sheaf map comes from. Taking sections on $V$, we get $$\mathcal O_X(V) \to \mathcal O_{X|Z}(V \cap Z) \to \mathcal O_{X,z}.$$ The map to the stalk is well known to be injective for integral schemes, hence the first map is injective. – MooS Mar 4 '16 at 8:35
• For the sake of completion, the injectivity of the map to the stalk $\mathcal O(V) \to \mathcal O_{X,z}$ is part of proposition $3.29$ in the book of Görtz and Wedhorn. – MooS Mar 4 '16 at 8:52
• Wonderful! Thanks Takumi. I had been looking at that sequence, but couldn't reason why the later map was injective. Thanks MooS for the verification. – BigMathTimes Mar 4 '16 at 9:15
• Thank you @MooS I was looking for an argument like that! – Takumi Murayama Mar 4 '16 at 16:21
• Great explanations, @MooS! – Georges Elencwajg Oct 17 '19 at 9:34

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.

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.

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.

• Thanks for the answer. I was fine with the argument on quasi-coherence. I only somewhat followed your argument on the vanishing of $s$. Once we have $s_q = g_q$ for all $q$ in $\mathcal{O}_U(W)$, I think we can use the injectivity of the map to the stalk. If $g_p$ vanished for any $p$ in $W$, then $g$ would have to be zero. But it's not, so $g_p$ never vanishes on $W$. Then the support of $s$ must include $W$, and so is open. It's also closed, so we get our conclusion. – BigMathTimes Mar 4 '16 at 9:14
• Yes, thats the same argument. I called it: The vanishing set (complement of support) and the non-vanishing set (support) are both open. You called it: The support is open and closed. This is the same thing :) – MooS Mar 4 '16 at 9:21
• Yes but I was unsure what you meant by "So $q \in K(X)$, hence we have $g_q =g_x$ for any $q \in W$." ? – BigMathTimes Mar 4 '16 at 9:22
• $q \in \mathcal O_X(W)$ can be viewed as an element of the fraction field $K(X)$. For any $x \in W$, $q_x$ is also an element of the fraction field, namely the same element. This is the same as saying: The integer $5$ is the same, whether we view it as an element of $\mathbb Z, \mathbb Q$ or some localization between. The argument then proceeds like this: We deduce that $q$ is either zero at any point or zero at no point of $W$. Thus, the vanishing set and the support are both open (since they contain $W$, if they contain $x$). – MooS Mar 4 '16 at 9:28

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 $F(W)$.

Here is a simple argument that if $$U,V\subset X$$ are open subsets of an irreducible scheme $$X$$ such that $$V\not\subset U$$, then any $$s\in \Gamma (V,i_{!}(\mathcal{O}_U))$$ is zero (even if $$V$$ is not affine).
I shall use the nice characterization of the sections of $$i_{!}(\mathcal{O}_U))$$ over an arbitrary open subset $$V\subset X$$ as those sections $$s\in \Gamma(U\cap V,\mathcal O_U)$$ whose support $$S=\operatorname {supp}s\subset U\cap V$$ is closed in $$\boldsymbol V$$.
(Notice that the weaker condition that $$s$$ be closed in $$U\cap V$$ is automatic!)
But given an irreducible scheme $$U$$, the support of a section $$s\in \Gamma(W,\mathcal O_U)$$ over any open subset $$W\subset U$$ is empty or the whole of $$W$$ .
Since for us the whole of $$W=U\cap V$$ cannot be both open and closed in the irreducible space $$V$$, we deduce that the support of $$s$$ is empty and thus that $$s=0$$.

Reference
The nice chararacterisation alluded to above can be found in Iversen's Cohomology of Sheaves, Definition II.6.1, page 106 .