2

I am trying to understand the following question (Proposition 5.12. in ABELIAN VARIETIES, Bas Edixhoven, Gerard van der Geer, and Ben Moonen)

  • If $f: X \to Y$ is an isogeny of degree $d$ between abelian varieties then there exists an isogeny $g: Y \to X$ with $g \circ f = [d]_X$ and $f ◦ g = [d]_Y$.

In the proof, I do not understand why we have this factorization when $f$ is annihilated by multiplication by $d$. Further, it seems that this proposition also holds for algebraic groups (Lemma 1.3.4 in this thesis) which says that

  • Let $A$ and $B$ be smooth, separated, reduced algebraic groups defined over a field $K$, $\alpha$ be a $K$-isogeny from $A$ to $B$, and $R$ be in $A(K)$. Let $d$ be the exponent of the kernel of $\alpha$ (and hence $d$ divides the degree of $\alpha$). Then there exists $\hat{\alpha}$ in $Hom_K(B, A)$ such that $\hat{\alpha}\circ\alpha= [d]_A$.

To prove this, the author view algebraic groups as group functors, however I think we can not have the surjectivity of $\alpha_Z:A(Z)\to B(Z)$ for $K-$schemes $Z$ (maybe this is true because of the assumptions of $A$ and $B$) although $\alpha:A\to B$ is surjective (which, as in Algebraic Groups by J. Milne, means that the underlying map $|A|\to|B|$ is surjective). I really do not know how to get through these, any help would be appreciated.

Khainq
  • 374

1 Answers1

2

Since $f$ is isogeny we have isomorphism $$f: X'=X/\ker f\to Y.$$ Let $h: Y\to X'$ be its inverse. Now consider the map $[d]: X\to X$ multiplication by $d$, which factors through $[d]: X'\to X$, since $d$ kills $\ker f$. Now define $$g: Y\to X$$ as $$g: Y\xrightarrow{h} X'\xrightarrow{[d]} X.$$ Now simply observe that $g\circ f=[d]$.

Similarly, $f\circ g=[d].$

Evans Gambit
  • 2,069