# Dimension of an abelian subvariety in the proof of the Poincaré's Reducibility Theorem

I am trying to understand the proof of the Poincaré's Reducibility Theorem, that I'm reading from the book "Abelian varieties" of J.S. Milne (see Proposition 10.1) and from the book "Abelian varieties" of D. Mumford (page 173 of the old edition). The theorem says:

Let $X$ be an abelian variety and let $Y$ be an abelian subvariety of $X$, $0\neq Y\neq X$. Then there exists an abelian subvariety $Y'\subset X$ such that the map \begin{align*} Y\times Y' &\to X\\ (y,y') &\mapsto y+y' \end{align*} is an isogeny.

I have understood the first part of the proof:

Let $i: Y\hookrightarrow X$ be the inclusion map and let $\mathcal{L}$ be an ample line bundle on $X$. Then we have: \begin{equation*} X \xrightarrow{\phi_{\mathcal{L}}} X^\vee\xrightarrow{i^\vee}Y^\vee \end{equation*} where $\phi_{\mathcal{L}}$ is an isogeny in this case. Let $Y'$ be the connected component of $\ker(i^\vee\circ\phi_{\mathcal{L}})=\phi_{\mathcal{L}}^{-1}(\ker i^\vee)$ passing through $0$. Then $Y'$ is an abelian variety and \begin{equation*} \dim Y' =\dim\ker i^\vee \ge \dim X^\vee-\dim Y^\vee=\dim X-\dim Y. \end{equation*}

Now, in the last equation, I dont' understand why the equality $\dim Y' =\dim\ker i^\vee$ holds. I guess that \begin{equation*}\dim Y'=\dim\phi_{\mathcal{L}}^{-1}(\ker i^\vee)= \dim \ker i^\vee \end{equation*} where the last equality holds since $\phi_{\mathcal{L}}$ is an isogeny, but I don't see why $\dim Y'=\dim\phi_{\mathcal{L}}^{-1}(\ker i^\vee)$. Can you help me? Thank you!

• In general (over an algebraically closed field), all the connected components are isomorphic, since they can be mapped one into the other via translations. In particular, they all have the same dimension. – Daniele A May 26 '16 at 13:51