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Nth root of unity in Nth division field of abelian variety?
Actually, $A^4$ does not have to be principally polarized. In fact, it does not have to be isomorphic to its dual. (As an example, you may take an abelian surface $A$ over an algebraically closed field $K$ with $\End(A)=\Z$ and such that $\Hom(A,A^t)$ is generated by the polarization $\lambda: A \to A^t$ with $\ker(\lambda)$ being a product of two cyclic groups of prime order $\ell \ne char(K)$.) It is $(A \times A^t)^4$, which is always principally polarized. 
Jun 4 
revised 
Nth root of unity in Nth division field of abelian variety?
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Jun 4 
answered  Nth root of unity in Nth division field of abelian variety? 
May 15 
comment 
Magnus' embedding theorem
Thank you, Igor! 
May 14 
comment 
Magnus' embedding theorem
@YCor I would appreciate a reference where such a homomorphism is explicitly described. 
May 14 
comment 
Magnus' embedding theorem
Thanks! It seems that what you denote by $F_n$ is $F/F_n$ in my notation. 
May 14 
asked  Magnus' embedding theorem 
May 9 
awarded  Nice Answer 
Dec 30 
comment 
Completion of a local ring of a curve
You are welcome. 
Dec 30 
answered  Completion of a local ring of a curve 
Nov 25 
comment 
Fermat's last theorem over larger fields
The reference above contains an abstract in English. However, the paper is available in English as well: mr.crossref.org/iPage?doi=10.1070%2FIM2001v065n03ABEH000337 . 
Nov 25 
answered  Fermat's last theorem over larger fields 
Sep 30 
awarded  Yearling 
Sep 19 
answered  Endomorphism Ring of Simple Abelian Varieties 
Jun 4 
awarded  Enlightened 
Jun 4 
awarded  Nice Answer 
May 15 
awarded  Good Answer 
Jan 14 
awarded  Nice Answer 
Dec 13 
awarded  ag.algebraicgeometry 
Dec 12 
comment 
MumfordTate groups of products of Hodge structures
Yes, you are right: please see my counterexample below. 