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Is it true that if an isogeny between two principally polarized abelian varieties respects the polarization, then it is in fact an isomorphism?

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Can assume field is alg. closed, so polarizations always have the form $\phi_L$ for an ample line bundle $L$. An isogeny $f:X \rightarrow Y$ intertwines $\phi_L$ ($L$ on $Y$) with $\phi_ {f^{\ast}L}$ (see proof of Thm. 1 in section 15 of Mumford's book on abelian varieties), so enough to show $\phi_ {f^{\ast}L}$ and $\phi_L$ have degrees which are off from each other by deg($f)^{2g}$ where $g$ is common dimension. But $L = O(D)$ for some divisor $D$, and $\phi_L$ has degree $((D^g)/g!)^2$ (self-intersection; see sec. 16 of Mumford's book). Since $(f^{\ast}D^g) = {\rm{deg}}(f)^g(D^g)$, QED. –  BCnrd Aug 29 '10 at 22:17

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That should be true, yes.

A polarization of $A$ is given by a bilinear form on $H_1(A, Z)$; this is equivalent to a map $H_1(A,Z) \to H_1(A,Z)^\vee$, which is an isomorphism if the polarization is principal.

A map between two abelian varieties is given by a corresponding linear map $H_1(A_1, Z) \to H_1(A_2, Z)$. The map between the varieties is an isomorphism if the map on $H_1$ is.

The map induced on the bilinear form then is the composition $$ H_1(A_1,Z)\to H_1(A_2,Z) \to H_1(A_2,Z)^\vee \to H_1(A_1,Z)^\vee $$ If the form is respected by this map, then this is an isomorphism. Consequently, the left-hand map must be as well, as claimed.

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