2
$\begingroup$

In Corollary 3 on page 498 of the article "Good reduction of abelian varieties" it says that, under some specified conditions, the minimal subextension $L/K$ of $\overline{K}/K$ over which an abelian variety $A/K$ acquires good reduction, is a Galois extension equal to $K(A_{m})$, the smallest subsextension of $K^{s}$ (a fixed separable closure of $K$) over which the the $m$-torsion points of $A$ become rational. This should hold for all $m \geq 3$ which are coprime to the residue field characteristic. Here, $A$ is an abelian variety with potential good reduction at a discrete valuation $v$. The ring of integers $\mathcal{O}_{v}$ is assumed to be strictly henselian.

As I expect the field $K(A_{m})$ to grow as $m$ grows, I do not understand why the above statement can be true. Does anyone know the answer? Thanks in advance.

$\endgroup$
3
$\begingroup$

The given conditions ensure that $L(A_m)$ is an unramified extension of $L$, since $A$ has good reduction over $L$ and $m$ is prime to the residue characteristic. If one were to merely assume that $K$ was a henselian (but not strictly henselian), such at $\mathbb{Q}_p$, then $L(A_m)$ would indeed grow due to the residue field degree increasing. But the strictly henselian assumption implies that the residue field is algebraically closed, so $L$ has no finite unramified extensions, hence is equal to $L(A_m)$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.