With "Abelian variety" I mean a integral scheme $X$, proper over an algebraically closed field (complete variety) with a group law $m: X\times X \rightarrow X$ such that $m$ and the inverse map are both morphisms of varieties.

I'm trying to understand the proof of Proposition 10.1 here.

This is the situation:

We have an abelian variety $X$, an abelian subvariety $Y\subset X$ with inclusion map $i: Y\hookrightarrow X$, an ample line bundle $\mathcal{L}$ on $X$. Hence we have: \begin{equation*} X \xrightarrow{\phi_{\mathcal{L}}} X^\vee\xrightarrow{i^\vee}Y^\vee \end{equation*} and we let $Y'$ be the connected component of $\ker(i^\vee\circ\phi_{\mathcal{L}})=\phi_{\mathcal{L}}^{-1}(\ker i^\vee)$ passing through $0$. The claim now is that $Y'$ is an abelian variety.

I'm trying to prove this and those are my thoughts:

$Y'\subset \ker(i^\vee\circ\phi_{\mathcal{L}})\subset\ker(\phi_{\mathcal{L}})=K(\mathcal{L})$ connected component implies, by a general fact from topology, that $Y'\subset K(\mathcal{L})$ is closed. Also, $K(\mathcal{L})\subset X$ is closed so we have:\begin{equation*} Y'\xrightarrow{closed\\ immersion}K(\mathcal{L})\xrightarrow{closed\\ immersion}X\xrightarrow{proper}Spec k. \end{equation*} Where the last arrow is due to the fact that $X$ is complete. Now, closed immersions are proper and the compostitions of proper morphisms is still proper so $Y'$ is proper over $k$. Furthermore, the connected component of $0$ in a topological group is a closed normal subgroub. Hence $Y'\subset K(\mathcal{L})$ has the group structure.

Right? But how can I prove that $Y'$ is integral?

Thank you!

  • 1
    $\begingroup$ It is a general fact that for any closed subgroup, such as an abelian subvariety, the connected component containing the origin $Y^0$ is geometrically irreducible. Moreover, the reduced underlying scheme $Y_{red} ^0$ is geometrically reduced and therefore is an abelian subvariety. So one possible strategy is to show that $Y' = Y_{red}^0$. $\endgroup$ – Future Jun 4 '16 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.