In [Corollary 3](http://wstein.org/papers/bib/Serre-Tate-Good_Reduction_of_Abelian_Varieties.pdf) 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. 



Potential good reduction of abelian varieties
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 $L(A\_m)$. 


