Locally constant sheaves over an irreducible space is constant.

Question

In Hartshorne's Algebraic Geometry Chapter II Proposition 6.15: If $X$ is an integral scheme, the homomorphism $CaCl X \rightarrow Pic X$ is an isomorphism.

In the proof he wants to prove that $\mathcal L \otimes \mathcal K= \mathcal K$. It is clear that on an open cover $\{U_i\}$ $(\mathcal L \otimes \mathcal K)|_{U_i}\cong \mathcal K$.

From this he concludes that $\mathcal L \otimes \mathcal K \cong \mathcal K$, which follows from a general fact that if "$X$ is irreducible, a sheaf whose restriction to each open set of a covering of $X$ is constant, (*) to is in fact a constant sheaf".

Can someone please give a proof of the above fact that locally constant sheaves over an irreducible space is actually constant .

• I guess Harthsorne wants to say that it is isomorphic to a constant sheaf.

Answer 1

Hint. If $X$ is irreducible, it has a generic point, that is a point $\xi$ contained in every non empty open set. Then, recall that a constant sheaf is a sheaf whose sections (viewed as functions) are locally constant.

Answer 2

It suffices to show that for any such sheaf $\mathcal{F}$ and any global section $s\in\Gamma (X,\mathcal{F} )$, there exists an open cover $\{U_{\alpha}\}$ of $X$ and a set $A$ so that $s\vert_{U_{\alpha}}\in A$ for each $\alpha$. This is equivalent to saying that $\mathcal{F}$ is the sheaf associated to the constant presheaf $\underline{A}$.

Now since we have had an open cover $\{U_{\alpha}\}$ so that $\mathcal{F}\vert_{U_{\alpha}}$ is constant, there exists an open cover $\{V_{\gamma}\}$ of $X$ which is a refinement of $\{U_{\alpha}\}$ so that the restriction of $s$ to $V_{\gamma}$ lies in $A_{\alpha}$ if $V_{\gamma}\subseteq U_{\alpha}$. Since $X$ is irreducible, any two open subsets has non-empty intersection, and since $\mathcal{F}\vert_{U_{\alpha}}$ is constant for all $\alpha$, we have for $x\in U_{\alpha\beta }=U_{\alpha }\cap U_{\beta}$, $\mathcal{F}_x =A_{\alpha} =A_{\beta}$, hence the sets are all the same, and we have constructed such an open cover and find such a set, which implies that $\mathcal{F}$ is constant.

Answer 3

It is quite simple. First for arbitrary non-empty open subsets $V\supseteq{}W$ of $X$ the restriction $r_{V,W}$ on $\mathcal{L}\otimes{}\mathcal{K}$ is an isomorphism what we derive from the fact that the restrictions $r_{V\cap{}U_{i},W\cap{}U_{i}}$ are isomorphism ($W\cap{}U_{i}$, $V\cap{}U_{i}$ are non-empty as $X$ is irreducible) and $\mathcal{L}\otimes{}\mathcal{K}$ is a sheaf. Next for a taken $U=U_{i_{0}}$ we extend the isomorfism $\mathcal{L}|_{U}\otimes{}\mathcal{K}|_{U}\cong{}(\mathcal{L}\otimes{}\mathcal{K})|_{U}\cong\mathcal{K}|_{U}$ to the isomorfism $\mathcal{L}\otimes{}\mathcal{K}\cong\mathcal{K}$ as in the following expression $(\mathcal{L}\otimes{}\mathcal{K})(V)\xrightarrow{r_{_{V,V\cap{}U}}}(\mathcal{L}\otimes{}\mathcal{K})(V\cap{}U)\overset{\sim}{\rightarrow}\mathcal{K}(V\cap{}U)\xrightarrow{\varrho^{-1}_{V,V\cap{}U}}\mathcal{K}(V)$ (the restriction $\varrho_{V,V\cap{}U}$ on $\mathcal{K}$ is an isomorphism because $\mathcal{K}$ is constant and $V\cap{}U$ is non-empty)