Let $X$ be a non-singular (complex) variety of dimension $n>1$. Let $D$ and $D'$ be distinct integral divisors in $X$. It seems to me that one should have $$\mathcal{Hom}_{\mathcal O_X}(\mathcal O_D,\mathcal O_D)\simeq \mathcal O_D$$ where the isomorphism is given by the image of $1$ and $$\mathcal{Hom}_{\mathcal O_X}(\mathcal O_D,\mathcal O_{D'})=\mathcal{Hom}_{\mathcal O_{D'}}(\mathcal O_{D\cap D'},\mathcal O_{D'})=0$$ but I would like to have, please, a confirmation.
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$\begingroup$ This is correct, and close to trivial. Use that for any $\mathcal{O}_X$-module $\mathcal{F}$ and closed subscheme $Z$ of $X$, $\mathcal{H}om_{\mathcal{O}_X}(\mathcal{O}_Z,\mathcal{F})$ is the subsheaf of $\mathcal{F}$ annihilated by the ideal of $Z$. $\endgroup$– abxCommented May 23, 2018 at 13:03
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You can use the locally free resolution $$ 0 \to O_X(-D) \stackrel{D}\to O_X \to O_D \to 0 $$ to compute. It gives $$ 0 \to \mathcal{Hom}(O_D,O_{D'}) \to \mathcal{Hom}(O_X,O_{D'}) \stackrel{D}\to \mathcal{Hom}(O_X(-D),O_{D'}). $$ The second arrow can be rewritten as $$ O_{D'} \stackrel{D}\to O_X(D)\vert_{D'}. $$ Thus, if $D$ and $D'$ have no common components, the kernel is zero, while if $D = D'$, the morphism is zero and the kernel is $O_{D'} = O_D$.
