Degree of an isogeny of abelian varieties related to the rank of its kernel

59 Views Asked by At

Let $\alpha:X\rightarrow Y$ be an isogeny between two abelian varieties over a field $k$, of degree $d$ and separable degree $d_s$. We may assume $k$ to be algebraically closed if necessary.

In Mumford's book "Abelian varieties" (application $3$ in chapter $6$), it is recalled that the cardinality of $\alpha^{-1}\{y\}$ is $d_s$ for almost all points $y\in Y$. Because each of these fibers is just a translate of $\ker(\alpha)$, it follows that the rank of $\ker(\alpha)$, seen as a finite group-scheme over $k$, is $d_s$.

In Moonen's notes on abelian varieties (more precisely here at the top of page $2$), using the finite flatness of $\alpha$ we know that $\alpha_{\star}\mathcal O_X$ is a locally free $\mathcal O_Y$-module of finite rank (ref. Mumford's GIT, chapter $0$ section $5$). This rank can be computed at the generic point of $Y$ and at the closed point $0\in Y$, which gives the equalities $d = \operatorname{rank}_{\mathcal O_Y}(\alpha_{\star}\mathcal O_X)=\operatorname{rank}(\ker \alpha)$.

Both claims seem to enter into contradiction, as we shouldn't have $d=d_s$ in general as far as I understand. Would somebody see which statement is false and why ?