A sheaf of modules which is not quasi-coherent I am reading Hartshorne’s book “Algebraic Geometry”. I am trying to understand example 5.2.3 on page 111. There, he supposed that $X$ is an integral scheme, $U$ is an open subscheme of $X$ and $V=Spec(A)$ is an open affine subset of $X$ which is not contained in $U$. Then, he claimed that $j_!(\mathcal{O}_U)|_V$ has no global sections over $V$. My question is that why is this true?
 A: The definition of this sheaf can be found on p.68 of the book. In particular, for an open set $V\not\subseteq U$ by definition $j_!\mathcal{O}_U(V)=0.$ Hence, the global sections over your $V=\mathrm{spec} (A)$ are zero because $V$ is not contained in $U$.
Edit: As red_trumpet mentions below, this answer is probably not quite right as written. Instead, we can use some properties integral schemes to conclude. We can apply Hartshorne's exercise 1.19 here. In particular, let $Z=X\setminus U$, let $i:Z\hookrightarrow X$ and let $j:U\hookrightarrow X$. There is a short exact sequence of sheaves
$$0\to j_!\mathcal{O}_U\to \mathcal{O}_X\to i_*(\mathcal{O}_X|_Z)\to 0$$
Passing to sections over $V=\mathrm{spec} A$, we get a left exact sequence
$$
0\to \Gamma(V,j_!\mathcal{O}_U)\to \Gamma(V,\mathcal{O}_X)\to\Gamma(V\cap Z,\mathcal{O}_{X}|_Z).
$$
However, choosing any $x\in V\cap Z$ there is a factorization
$$\Gamma(V,\mathcal{O}_X)\to \Gamma(V\cap Z,\mathcal{O}_X|_Z)\to \mathcal{O}_{X,x}=\mathcal{O}_{V,x}.$$
The important point is: on an integral scheme, $\Gamma(V,\mathcal{O}_X)\to \mathcal{O}_{V,x}$ is injective$^*$, whence the map $\Gamma(V,\mathcal{O}_X)\to \Gamma(V\cap Z,\mathcal{O}_X|_Z)$ is also injective. Hence, by exactness $\Gamma(V,j_!\mathcal{O}_U)=0$.
$(*)$ This is because on an integral scheme, there is a generic point $\xi$ and for any affine $V=\mathrm{spec} A$, $\mathcal{O}_{V,\xi}=K(A)$ in such a way that $\mathcal{O}_V(V)\to \mathcal{O}_{V,\xi}$ is the inclusion of $A$ into $K(A)$. The stalks of points $x\in \mathrm{spec}A$ fit into $A\to A_x\to K(A)$ and on the scheme side this is
$$\mathcal{O}_V(V)\to \mathcal{O}_{V,x}\to \mathcal{O}_{V,\xi}.$$
A: I will give an explicit construction of $j_!$.
Let $X$ be a topological space, let $U$ be an open subspace of $X$, and let $j : U \hookrightarrow X$ be the inclusion map.
As always, we have an inverse image functor $j^{-1}$ that takes sheaves on $X$ to sheaves on $U$, and $j_!$ is defined to be the left adjoint of $j^{-1}$.
Let $F$ be a sheaf of abelian groups on $U$.
For each $x \in X$, let $G_x = F_x$ if $x \in U$ and $G_x = 0$ if $x \notin U$.
For each open $V \subseteq X$, let $G (V)$ be the set of all $g \in \prod_{x \in V} G_x$ for which there exist an open $V' \subseteq V$ and $f \in F (U \cap V)$ such that:

*

*For all $x \in U \cap V$, $g_x = f_x$.

*For all $x \in V'$, $g_x = 0$.

*$(U \cap V) \cup V' = V$.

It is straightforward to verify that $G$ so defined is a sheaf of abelian groups on $X$ isomorphic to $j_! F$ as usually defined (i.e. the sheaf associated to the presheaf defined by extension by zero etc.).
Now suppose we are in the situation of the original question: so $X$ is an integral scheme, $U$ is an open subscheme, $F = O_U$, and $V$ is an open subscheme of $X$ but not contained in $U$.
By the above, if $g \in \Gamma (V, j_! O_U)$ then $g_x = 0$ for all $x$ in some non-empty open $V' \subseteq V$.
But $X$ is integral, so $U \cap V'$ is dense in $U \cap V$, and the vanishing of a section of $O_U$ is a closed subset of $U \cap V$, so we must have $g_x = 0$ for all $x \in U \cap V$ as well.
Hence $g = 0$, as required.
