Hello,


I have the following problem:


if I have an elliptic curve $E$ over some field and consider the diagonal morphism


$\Delta: E \rightarrow E\times E$,


does then hold the following:


the adjunction morphism


$\Delta^\* \Delta\_\* \mathcal{O}\_E\rightarrow\mathcal{O}\_E$


is an isomorphism?


Thank you!



Diagonal morphism and adjunction
The diagonal of an elliptic curve (or more generally of a seprarated scheme) is a closed immersion. Thus the following more general result answers your question affirmatively: Let $i\colon Y \to X$ be a closed immersion and let ${\mathscr G}$ be a quasi-coherent ${\mathscr O}\_Y$-module. Then the adjunction morphism $i^\*i\_\*{\mathscr G} \to {\mathscr G}$ is an isomorphism.


The question is local on $X$ and we may assume that $X = {\rm Spec} A$ is affine. Then $Y = {\rm Spec} (A/{\mathfrak a})$ for an ideal ${\mathfrak a}$ of $A$, and ${\mathscr G}$ corresponds to an $(A/{\mathfrak a})$-module $M$, i.e. to an $A$-module $M$ with ${\mathfrak a} M = 0$ and the adjunction morphism corresponds to the isomorphism $A/{\mathfrak a} \otimes\_A M \to M$.


