User yuri zarhin - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:49:33Z http://mathoverflow.net/feeds/user/9658 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124502/does-the-manin-drinfeld-theorem-hold-over-number-fields/124507#124507 Answer by Yuri Zarhin for Does the Manin-Drinfeld theorem hold over number fields? Yuri Zarhin 2013-03-14T12:37:36Z 2013-03-14T12:37:36Z <p>Here are links to the translations of several Kurchanov's papers.</p> <p><a href="http://mr.crossref.org/iPage/?doi=10.1070%2FIM1978v012n03ABEH002002" rel="nofollow">http://mr.crossref.org/iPage/?doi=10.1070%2FIM1978v012n03ABEH002002</a></p> <p><a href="http://mr.crossref.org/iPage/?doi=10.1070%2FIM1980v014n01ABEH001076" rel="nofollow">http://mr.crossref.org/iPage/?doi=10.1070%2FIM1980v014n01ABEH001076</a></p> <p><a href="http://mr.crossref.org/iPage/?doi=10.1070%2FSM1980v036n04ABEH001848" rel="nofollow">http://mr.crossref.org/iPage/?doi=10.1070%2FSM1980v036n04ABEH001848</a> </p> <p>See also <a href="http://www.mathnet.ru/php/person.phtml?&amp;personid=21921&amp;option_lang=eng" rel="nofollow">http://www.mathnet.ru/php/person.phtml?&amp;personid=21921&amp;option_lang=eng</a> .</p> http://mathoverflow.net/questions/124339/tate-conjecture-for-abelian-varieties-over-a-finitely-generated-extension-of-an-a/124351#124351 Answer by Yuri Zarhin for Tate conjecture for abelian varieties over a finitely generated extension of an algebraically closed field Yuri Zarhin 2013-03-12T21:37:11Z 2013-03-12T22:06:30Z <p>The answer is no: an easy (not interesting) counterexample is provided by "constant" abelian varieties, i.e., when $A$ and $B$ are defined over an algebraically closed field $k$ of characteristic zero while $K$ is finitely generated over $k$. </p> <p>So, let's assume that the $K/k$-traces of $A$ and $B$ are zero, i.e., both $A$ and $B$ do not contain constant abelian subvarieties of positive dimension. Actually, we need more: assume that there are no {\sl isotrivial} abelian subvarieties of positive dimension, i.e., there are no abelian subvarieties (except zero) that become constant after a finite algebraic extension of $K$.</p> <p>Let us assume also that $k$ is the field $C$ of complex numbers. Then $A$ and $B$ become generic fibers of abelian schemes $\mathcal{A}$ and $\mathcal{B}$ over a smooth quasiprojective complex algebraic variety $S$ and the analogue of Tate's conjecture becomes equivalent to a similar question about homomorphisms between the first integral homology groups $H_1(\mathcal{A}_s,Z)$ and $H_1(\mathcal{B}_s,Z)$ of the fibers that commute with the actions(s) of the fundamental group $\pi_1(S,s)$ of the base $S$. Here $s$ is a complex point of $s$ while the corresponding fibers $\mathcal{A}_s$ and $\mathcal{B}_s$ are complex abelian varieties and the question is whether all $\pi_1(S,s)$-equivariant homomorphisms between $H_1(\mathcal{A}_s,Z)$ and $H_1(\mathcal{B}_s,Z)$ come from homomorphisms of abelian varieties $\mathcal{A}_s \to \mathcal{B}_s$?</p> <p>If either $\dim(A)\le 3$ or $\dim(B)\le 3$ then the answer is yes: see Section 4.4 of Delignes's Th\'eorie de Hodge.II 40/PMIHES_1971_<em>40</em>_5_0/PMIHES_1971_<em>40</em>_5_0.pdf">http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1971_<em>40</em>/PMIHES_1971_<em>40</em>_5_0/PMIHES_1971_<em>40</em>_5_0.pdf However, there is a counterexample with $4$-dimensional $A=B$ (Faltings, Inv. Math. 73(1983), 337-347). See also arXiv:math/0504523 [math.AG] . </p> http://mathoverflow.net/questions/122953/smooth-projective-varieties-of-picard-number-one/122966#122966 Answer by Yuri Zarhin for Smooth projective varieties of Picard number one Yuri Zarhin 2013-02-26T10:31:12Z 2013-02-26T10:31:12Z <p>Every abelian variety $X$ of given dimension $g>1$ over the field $\mathbb{C}$ of complex numbers is not a complete intersection, since it is not simply connected. On the other hand, a generic" $g$-dimensional $X$ has endomorphism ring $\mathbb{Z}$ and therefore Picard number 1 (A. Weil).</p> <p>Explicit examples of such $X$ are provided by jacobians $J(C_f)$ of hyperelliptic curves $C_f:y^2=f(x)$ where $f(x)$ a polynomial of degree $n=2g+1$ or $2g+2$ without multiple roots that enjoys the following properties. There is a subfield $K$ of $C$ such that $f(x)$ is an irreducible polynomial in $K[x]$ and its Galois group over $K$ is either the full symmetric group $S_n$ or the alternating group $A_n$. (See arXiv math/9909052 .) </p> <p>For example, if $K$ is the field $\mathbb{Q}$ of rational numbers then $f(x)=x^n-x-1$ is irreducible over $\mathbb{Q}$ (Selmer) and its Galois group is $S_n$ (Serre). This implies that if $n\ge 5$ then the jacobian $J$ of $y^2=x^n-x-1$ has Picard number 1 (and is not a complete intersection).</p> http://mathoverflow.net/questions/122858/an-abelian-hodge-tate-representation-lands-in-a-torus/122892#122892 Answer by Yuri Zarhin for An abelian Hodge-Tate representation lands in a torus Yuri Zarhin 2013-02-25T15:00:34Z 2013-02-25T15:28:39Z <p>What Bogomolov proved is the algebraicity of the corresponding $\ell$-adic representation. What you are trying to do is to prove its semisimplicity, which was done by Faltings 4 years later. The presence of additive factors $G_a$ in the Zariski closure of the Galois image is not an issue as far as you are interested only in the algebraicity. In fact, since every linear nilpotent Lie algebra (in char 0) is algebraic (coincides with the Lie algebra of a certain linear algebraic group), one may replace the original representation by its semisimplification and deal with a certain semisimple $\ell$-adic representation $\rho$ of the absolute Galois group $Gal(K)$ of a number field $K$. Replacing $K$ by its suitable finite algebraic extension, one may assume that the inertia groups of places not lying over $\ell$ act as unipotent operators and $H_{\ell}$ is connected. So, if $\rho$ is semisimple abelian then it is unramified outside divisors of $\ell$. Since it is of Hodge-Tate type, the images (under $\rho$) of the inertia groups of divisors of $\ell$ are open in their Zariski closures $U_{\lambda}$ (see Serre's Abelian $\ell$-adic representations and elliptic curves), i.e., the corresponding local $\ell$-adic representations are algebraic. Now, if $U$ is is the algebraic subgroup of $H_{\ell}$ that is generated by all $U_{\lambda}$'s, then the images of all the inertia groups generate (after taking the closure) an open subgroup in $U(\mathbf Q_{\ell})$. Now it suffices to check that $H_{\ell}=U$. </p> <p>If $U$ is a proper subgroup of $H_{\ell}$ then the composition $$Gal(K)\to H_{\ell}(\mathbf Q_{\ell}) \twoheadrightarrow H_{\ell}(\mathbf Q_{\ell})/U(\mathbf Q_{\ell})$$ gives rise to a continuous surjective homomorphism from $Gal(K)$ to $\mathbf Z_{\ell}$ that corresponds to an infinite everywhere unramified abelian extension of $K$. Since such extensions do not exist, we get the desired contradiction.</p> <p>By the way, Bogomolov's Izvestiya paper is available in English <a href="http://mr.crossref.org/iPage/?doi=10.1070%2FIM1981v017n01ABEH001329" rel="nofollow">http://mr.crossref.org/iPage/?doi=10.1070%2FIM1981v017n01ABEH001329</a> .</p> http://mathoverflow.net/questions/121137/theta-group-representation/121203#121203 Answer by Yuri Zarhin for Theta group representation Yuri Zarhin 2013-02-08T14:36:11Z 2013-02-08T14:36:11Z <p>If I understand correctly, $K_1$ and $K_2$ are (mutually orthogonal maximal) isotropic subgroups of $K(L)$. Therefore they both can be lifted (non-canonically) to (finite) commutative (sic!) subgroups of the theta group. In particular, the restriction of the projective representation of $K(L)$ to $K_i$ is actually a projectivization of a certain linear representation of finite commutative $K_i$ in $H^0(X,L)$. Since a linear representation of such groups is irreducible if and only if it is one-dimensional, we get a negative answer to your question (unless $H^0(X,L)$ has dimension 1, i.e. $L$ defines a principal polarization).</p> http://mathoverflow.net/questions/119956/abelian-varieties-with-given-endomorphism-algebra/119966#119966 Answer by Yuri Zarhin for Abelian varieties with given endomorphism algebra Yuri Zarhin 2013-01-26T21:14:01Z 2013-01-26T21:14:01Z <p>Felipe, Albert is right: $r>1$. In particular, there are no complex abelian surfaces, whose endomorphism algebra is a definite quaternion algebra over the rationals. The same is true in any characteristic: see a survey article of Frans Oort entitled "Endomorphism algebras of abelian varieties" (Alg. Geom. and Commut. Algebra in Honor of M. Nagata (1987, Ed. H. Hijikata et al.), Kinokuniya Cy, Tokyo 1988; Vol. II, pp. 469-502).</p> <p>On the other hand, for any quaternion algebra $D$ over the rationals, there exists a simple $2$-dimensional complex torus $T$, whose endomorphism algebra is isomorphic to $D$. ( $D$ is definite if and only if $T$ is not algebraizable.) See my paper with Frans (Math. Ann. 303 (1995), 11--29). </p> http://mathoverflow.net/questions/115440/equation-for-simple-jacobian-of-a-genus-two-curve/115445#115445 Answer by Yuri Zarhin for Equation for simple Jacobian of a genus two curve Yuri Zarhin 2012-12-04T21:55:57Z 2012-12-04T21:55:57Z <p>In Mumford's Tata Lectures on Theta II" (Progress in Math. 43, 1984) there are explicit equations for a certain open affine (dense) subset $Z$ of the jacobian $Jac(C)$ for any hyperelliptic curve $C$; the jacobian is covered by all the translations of $Z$ by points of order $2$. (Recall that every genus 2 curve is hyperelliptic.) </p> http://mathoverflow.net/questions/111324/where-in-the-literature-does-the-anticyclotomic-mathbfz-p-extension-of-an-im/111340#111340 Answer by Yuri Zarhin for Where in the literature does the anticyclotomic $\mathbf{Z}_p$-extension of an imaginary quadratic field first appear? Yuri Zarhin 2012-11-03T03:46:14Z 2012-11-03T03:51:26Z <p>First time I heard about antycyclotomic $\Gamma$-extensions in 1972 from Pavel Kurchanov, in connection with his paper </p> <p><a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=sm&amp;paperid=3020&amp;option_lang=eng" rel="nofollow">http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=sm&amp;paperid=3020&amp;option_lang=eng</a></p> <p>Actually, his goal was to construct elliptic curves of infinite rank over $\Gamma$-extensions. (According to Mazur's conjecture, one cannot do it over the cyclotomic extensions.)</p> http://mathoverflow.net/questions/109245/finite-groups-that-admit-an-anti-automorphism-with-many-fixed-points Finite groups that admit an anti-automorphism with many fixed points Yuri Zarhin 2012-10-09T18:07:59Z 2012-10-10T04:50:11Z <blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="http://mathoverflow.net/questions/38/homomorphism-more-than-3-4-the-inverse" rel="nofollow">Homomorphism more than 3/4 the inverse</a> </p> </blockquote> <p>Let $G$ be a finite group that admits an anti-automorphism $u: G \to G$ and let $S$ be the set of fixed points of $u$. </p> <p>I am looking for references/results that deal with the structure of $G$ when $S$ is big, i.e., the ratio $$r=\#(S)/\#(G)$$ is big. For example, if $r=1$ (i.e., $S=G$) then (obviously) $G$ is abelian. One may check that if $r>1/2$ then $G$ is also abelian if we assume additionally that order of $G$ is an odd number. I've heard that if $r>3/4$ then one may conclude (without any additional assumptions) that $G$ is abelian but cannot find a reference.</p> http://mathoverflow.net/questions/107324/can-we-control-the-size-of-the-intersection-of-two-abelian-subfactors-of-an-abeli/107353#107353 Answer by Yuri Zarhin for Can we control the size of the intersection of two abelian subfactors of an abelian variety ? Yuri Zarhin 2012-09-16T23:41:46Z 2012-09-17T00:15:15Z <p>Here is a (slightly more detailed) variant of what was pointed out by grp.</p> <p>Let $E$ and $F$ be non-isogenous elliptic curves over $K$. Let $n$ be a positive integer. (If $p=char(K)>0$ and $p$ divides $n$ we assume additionally that both $E$ and $F$ are ordinary elliptic curves.) Then there are order $n$ cyclic subgroups $C_n \subset E(K)$ and $D_n \subset F(K)$. Fix a group isomorphism $\phi: C_n \cong D_n$. Let $$\Gamma(\phi)=[{(x,\phi x) \mid x \in C_n }] \subset C_n \times D_n \subset E(K) \times F(K)$$ be the graph of $\phi$; it is an order $n$ cyclic subgroup of $(E\times F)(K)$. Let us consider the quotient $A:= (E\times F)/\Gamma(\phi)$ and denote by $\pi: E\times F \to A$ the corresponding degree $n$ isogeny of abelian surfaces. Clearly, the restrictions of $\pi$ to $E \times {e_F}$ and ${e_E}\times F$ give us isomorphisms of elliptic curves $$E=E \times {e_F} \cong \pi(E \times {e_F})=: E^{\prime}\subset A,$$ $$F={e_E}\times F \cong \pi({e_E}\times F)=: F^{\prime} \subset A.$$</p> <p>(Here $e_E$ (resp. $e_F$) is the zero of group law on $E$ (resp. on $F$).) It is also clear that the intersection of $E^{\prime}$ and $F^{\prime}$ (in $A$) is a cyclic order $n$ subgroup that is the image under $\pi$ of $$[{(x,0) \mid x \in C_n }] \subset C_n \times D_n \subset E(K) \times F(K).$$ Now let $Z$ be a 1-dim'l abelian subvariety of $A$ and let $Y$ be the identity component of its preimage $\pi^{-1}(Z)$ in $E\times F$. Clearly, $Y$ is a 1-dim'l abelian subvariety of $E\times F$ and $\pi(Y)=Z$. It is also clear that (at least) one of projection maps $$Y \to E, \ Y \to F$$ is non-constant. If $Y \to E$ is non-constant then it is an isogeny of elliptic curves. Since $E$ and $F$ are non-isogenous, $Y$ is non-isogenous to $F$ and therefore $Y \to F$ is the constant map to $e_F$. It follows that $Y=E\times {e_F}$ and therefore $Z=\pi(Y)=E^{\prime}\subset A$. The same arguments prove that if $Y \to F$ is non-constant then $Z=F^{\prime} \subset A$.</p> http://mathoverflow.net/questions/106128/what-is-the-importance-of-the-conjectural-semi-simplicity-of-the-action-of-the-fr/106136#106136 Answer by Yuri Zarhin for What is the importance of the conjectural semi-simplicity of the action of the Frobenius on the etale cohomology of a variety over a finite field ? Yuri Zarhin 2012-09-01T21:03:38Z 2012-09-01T21:18:51Z <p>Here is a conjectural statement that follows from the combination of the semisimplicity conjecture and the Tate conjecture on algebraic cycles.</p> <p>Let $X$ and $Y$ be geometrically irreducible smooth projective varieties over a finite field $F$. Suppose that $X$ and $Y$ have the same number of $F^{\prime}$-points for all finite overfields $F^{\prime}$ of $F$. Then there exist a finite overfield $F_0$ of $F$ and a geometrically irreducible closed $F_0$-subvariety $Z\subset X \times Y$ such that $\dim(X)=\dim(Y)=\dim(Z)$ and both projections map $Z \to X$ and $Z \to Y$ are surjective.</p> <p>The semisimplicity is known to be true for abelian varieties (Weil). Combining the semisimplicity with Tate's theorem on homomorphisms, one may deduce that if $X$ and $Y$ are abelian varieties with the same number of $F^{\prime}$-points (for all $F^{\prime}$) then they are isogenous over $F$.</p> http://mathoverflow.net/questions/99064/weil-reciprocity-on-abelian-varieties-and-biextensions/99096#99096 Answer by Yuri Zarhin for Weil reciprocity on abelian varieties and biextensions? Yuri Zarhin 2012-06-08T08:00:54Z 2012-06-08T08:00:54Z <p>Lang's reciprocity law and Poincar\'e-Mumford biextensions come together in the context of generalized N\'eron pairings</p> <p><a href="http://iopscience.iop.org/0025-5726/6/3/A03/pdf/0025-5726_6_3_A03.pdf" rel="nofollow">http://iopscience.iop.org/0025-5726/6/3/A03/pdf/0025-5726_6_3_A03.pdf</a></p> <p>(Proof of Proposition 2 on p. 496 and p. 502).</p> http://mathoverflow.net/questions/35060/when-is-a-product-of-elliptic-curves-isogenous-to-the-jacobian-of-a-hyperelliptic/87936#87936 Answer by Yuri Zarhin for When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve? Yuri Zarhin 2012-02-08T22:03:15Z 2012-02-08T22:03:15Z <p>Over the complex numbers every product of two elliptic curves is isogenous to the jacobian of a genus 2 (hyperelliptic) curve. Indeed, the corresponding Siegel upper half-space $H_2$ is an orbit of the real symplectic group $Sp(4,R)$. Since the subgroup $Sp(4,Q)$ of its rational points is everywhere dense in $Sp(4,R)$, every $Sp(4,Q)$-orbit is everywhere dense in $H_2$ and therefore meets the (Torelli) open subset $T_2$ of $H_2$ that parametrized" the Jacobians. Now one has only to notice that if points $x,y \in H_2$ correspond to principally polarized abelian surfaces $A_x$ and $A_y$ respectively then $A_y$ is isogenous to $A_x$ if $y$ lies in the $Sp(4,Q)$-orbit of $x$. (And, of course, one should take $x$ with $A_x$ being a product of two elliptic curves.)</p> <p>Similar arguments prove that every product of three elliptic curves is isogenous to the jacobian of a genus 3 curve. (These arguments were used in Sect. 2, Remark 3 on pp. 60--61 of arXiv:0912.4325v1 [math.NT] in order to prove certain properties of the modular height.) </p> http://mathoverflow.net/questions/73879/canonical-liftings-of-endomorphisms-of-ordinary-abelian-varieties Canonical liftings of endomorphisms of ordinary abelian varieties Yuri Zarhin 2011-08-28T02:11:46Z 2011-08-29T13:01:08Z <p>I am looking for a reference to the following well known" fact.</p> <p>Let $k$ be a perfect field of prime characteristic $p$ and $W(k)$ its ring of Witt vectors. Let $A_0$ be an ordinary abelian variety over $k$ and let $A$ be an abelian scheme over $W(k)$ that is the canonical (Serre--Tate) lifting of $A_0$. Then every endomorphism $u_0$ of $A_0$ lifts to an endomorphism of $A$. In other words, the natural map $End(A) \to End(A_0)$ is bijective. </p> <p>My problem is that I need it for infinite $k$. (I know a couple of references that deal with finite $k$.) </p> http://mathoverflow.net/questions/73396/etale-endomorphisms-of-abelian-varieties-in-positive-characteristic/73874#73874 Answer by Yuri Zarhin for Etale endomorphisms of abelian varieties in positive characteristic Yuri Zarhin 2011-08-27T23:36:07Z 2011-08-28T01:45:12Z <p>If $p-1$ is divisible by $24$ then there is an explicit example of an ordinary $7$-dimensional abelian variety $X$, whose endomorphism algebra is the imaginary quadratic field $Q(\sqrt{-3})$; in particular, $X$ is absolutely simple and its $K/k$-trace is trivial. Namely, $K$ is the field of rational functions $k(t)$ and $X$ is the jacobian of the $K$-curve $y^3=x^9-x-t$. See Example 4.3 of arXiv:math/0606422 [math.NT] [MR2289628 (2007j:11077)] for details.</p> <p>If $p>2$ and $g>1$ is an odd integer then there exists a $g$-dimensional ordinary abelian variety $X$ over a suitable $K$, whose endomorphism algebra is an imaginary quadratic field; in particular, $X$ is absolutely simple and its $K/k$-trace is trivial. See Theorem 1.5(i,ii) of the same paper that is based on a construction of Oort and van der Put. (Actually, the condition $p>2$ could be dropped.)</p> http://mathoverflow.net/questions/72288/proofs-of-mordell-weil-theorem/72302#72302 Answer by Yuri Zarhin for Proofs of Mordell-Weil theorem Yuri Zarhin 2011-08-07T20:01:20Z 2011-08-07T20:01:20Z <p>Manin's proof of Mordell-Weil theorem (for abelian varieties over number fields) has appeared as an appendix to Russian translation of First edition of Mumford's Abelian varieties". Eventually it was translated into English and published as an appendix to Second and Third editions of Mumford's book.</p> http://mathoverflow.net/questions/66829/o-linear-weil-pairing-on-abelian-varieties-with-real-multiplication/66848#66848 Answer by Yuri Zarhin for O-linear Weil-pairing on abelian varieties with real multiplication Yuri Zarhin 2011-06-03T20:13:01Z 2011-06-03T21:04:25Z <p>Yes, it's just a linear algebra stuff. Probably, it's more convenient to look at the Weil pairing between the $\ell$-adic Tate modules of $A$ and its dual, taking into account that these modules are free $O\otimes Z_{\ell}$-modules of the same rank.</p> <p>From the linear algebra point of view the situation is as follows.</p> <p>Let $e: M \times N \to P$ be a perfect pairing of free $Z_{\ell}$-modules $M$ and $N$ of the same rank that takes values in a free $Z_{\ell}$-module $P$ of rank $1$. Assume also that $M$ and $N$ are free $O\otimes Z_{\ell}$-modules of the same rank. Then we get a natural $Z_{\ell}$-linear map</p> <p>$M \times N \to Hom_{Z_{\ell}}(O\otimes Z_{\ell}, P),$ $(m,n) \mapsto [a \mapsto e(ax,y)]$ for all $a \in O$. Taking into account that $$Hom_{Z_{\ell}}(O\otimes Z_{\ell}, Z_{\ell})=D^{-1}\otimes Z_{\ell}$$ (via the trace map), we get the natural pairing</p> <p>$$M \times N \to P \otimes_{Z_{\ell}}[D^{-1}\otimes Z_{\ell}].$$ Here $M$ and $N$ are the Tate modules of $A$ and its dual while $P$ is $Z_{\ell}(1)$ and $e$ is the Weil pairing.</p> http://mathoverflow.net/questions/65830/reference-for-a-theorem-of-tate-on-the-endomorphism-rings-of-avs-over-finite-fiel/65857#65857 Answer by Yuri Zarhin for Reference for a theorem of Tate on the endomorphism rings of AVs over finite fields Yuri Zarhin 2011-05-24T14:50:12Z 2011-05-24T15:39:22Z <p>I guess $p=\operatorname{char}(k)$. For another (unified) proof of Tate's theorem (that works for primes $\ell\ne p$ and $\ell=p$) see <a href="http://arxiv.org/abs/0711.1615" rel="nofollow">arXiv:0711.1615</a> [math.AG]; MR2484084 (2010a:11117). </p> http://mathoverflow.net/questions/5499/which-mathematicians-have-influenced-you-the-most/62988#62988 Answer by Yuri Zarhin for Which mathematicians have influenced you the most? Yuri Zarhin 2011-04-26T00:25:21Z 2011-04-26T00:25:21Z <p>Who: Manin, Parshin, Serre, Tate.</p> <p>When: When I was an undergraduate.</p> http://mathoverflow.net/questions/61984/rational-points-on-algebraic-curves-over-qab/62026#62026 Answer by Yuri Zarhin for rational points on algebraic curves over $Q^{ab}$ Yuri Zarhin 2011-04-17T12:15:31Z 2011-04-17T12:15:31Z <p>Actually Ken Ribet proved that if $K$ is a number field and $K(\mu_{\infty})$ is its infinite cyclotomic extension generated by all roots of unity then for every abelian variety $A$ over $K$ the torsion subgroup of $A(K(\mu_{\infty}))$ is finite: <a href="http://math.berkeley.edu/~ribet/Articles/kl.pdf" rel="nofollow">http://math.berkeley.edu/~ribet/Articles/kl.pdf</a> .</p> <p>On the other hand, Alosha Parshin conjectured that if $K_{p}$ is the extension of $K$ generated by all $p$-power roots of unity (for a given prime $p$) then the set $C(K_{p})$ is finite for every $K$-curve $C$ of genus $>1$: <a href="http://arxiv.org/abs/0912.4325" rel="nofollow">http://arxiv.org/abs/0912.4325</a> (see also <a href="http://arxiv.org/abs/1001.3424" rel="nofollow">http://arxiv.org/abs/1001.3424</a> ).</p> http://mathoverflow.net/questions/57103/do-isogenies-between-avs-over-finite-fields-separate-finite-subgroups/57112#57112 Answer by Yuri Zarhin for Do isogenies between AVs over finite fields separate finite subgroups? Yuri Zarhin 2011-03-02T13:54:06Z 2011-03-02T14:02:05Z <p>Let me describe a natural straightforward generalization of Chris Wutrich's counterexample.</p> <p>Let $B$ be a $g$-dimensional abelian variety over a field $k$ and assume that $End_k(B)$ is a principal ideal domain. Let $A$ be another abelian variety over $k$ that is not $k$-isomorphic to $B$ but $k$-isogenous to it. Then the group $Hom_k(A,B)$ becomes a free $End_k(B)$-module of rank 1. This means that there exists a (generator) isogeny $\lambda:A \to B$ such that every $k$-homomorphism $v: A\to B$ is a composition $u\lambda$ of $\lambda$ and a certain $u \in End_k(B)$. In particular, $ker(v)$ always contains $\ker(\lambda)$. Since $A$ and $B$ are not isomorphic over $k$, the isogeny $\lambda$ is not an isomorphism and therefore $H:=ker(\lambda)\subset A$ is nontrivial but is killed by every $k$-homomorphism from $A$ to $B$.</p> <p>In order to construct explicit examples (over finite fields) pick any imaginary quadratic field $K$ of discriminant 1 amd let $O$ be the ring of integers in $K$, which is PID. Then for a half" of the primes $p$ there exist a finite field $k$ of characteristic $p$ and an ordinary elliptic curve $B$ over $k$ with $End_k(B)=End(B)=O$. Enlarging (if necessary) $k$, we may find an elliptic curve $A$ over $K$ that is not $k$-isomorphic to $A$ but $k$-isogenous to it. For example, if a prime $\ell$ is different from $p$ and inert in $O$ then we may pick a cyclic order $\ell$ subgroup $C$ in $B(k)$ (enlarging $k$ if necessary) and put $A=B/C$. Then the cyclic order $\ell$ subgroup $H=B_{\ell}/C\subset A$ is killed by every homomorphism $A \to B$.</p> http://mathoverflow.net/questions/56884/does-there-exist-a-family-of-curves-or-abelian-varieties-on-the-punctured-line/56899#56899 Answer by Yuri Zarhin for Does there exist a family of curves (or abelian varieties) on the punctured line with specified monodromy on H^1? Yuri Zarhin 2011-02-28T14:06:46Z 2011-02-28T14:06:46Z <p>There are several restrictions. First, the existence of potential semistable reduction (Grothendieck-Mumford) implies that your representative $\sigma$ at every puncture must be quasi-unipotent of level 2, i.e, there exists a positive integer $N$ such that $(\sigma^N-1)^2=0$. Second, the Zariski closure $G\subset Sp_{2g,Q}$ of the global monodromy group $\Gamma \subset Sp(2g,Z)$ must satisfy the following properties (Deligne).</p> <ol> <li><p>Its identity component $G^0$ is semisimple.</p></li> <li><p>All absolutely simple quotients of $G^0$ (over the field $C$ of complex numbers) are classical algebraic groups (A_r,B_r,C_r,D_r) and their" natural nontrivial irreducible subrepresentations in $C^{2g}$ are (fundamental) minuscule, i.e., the corresponding set of weights is an orbit of the Weyl group.</p></li> </ol> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56546#56546 Answer by Yuri Zarhin for Groups acting on Riemann Surfaces Yuri Zarhin 2011-02-24T18:36:26Z 2011-02-24T18:36:26Z <p>A counterexample to Q3 is provided by the genus 2 compact Riemann surface $X$ of $y^2=x^5-1$. Indeed, the order 10 cyclic group $C_{10}$ acts on $X$ (by changing sign of $y$ and multiplying $x$ by $5$th roots of unity). It is known that the jacobian of $X$ has endomorphism ring $Z[\zeta_5]$ - the $5$th cyclotomic ring of integers and any finite multiplicative subgroup of $Z[\zeta_5]$ is a subgroup of $\mu_{10}\cong C_{10}$. This implies that $Aut(X)=C_{10}$. On the other hand, the dihedral group $D_{10}$ of order $10$ has the same Sylow subgroups as $C_{10}$ but is not isomorphic to it. In other words, there is no faithful action of $D_{10}$ on $X$ while its Sylow subgroups $C_5$ and $C_2$ act faithfully on $X$.</p> <p>If $Y$ is a compact Riemann surface of genus $g$ and its jacobian $J$ has no nontrivial automorphisms (i.e., $End(J)$ is the ring of integers $Z$) then either $Y$ is non-hyperelliptic and $Aut(Y)={1}$ or $Y$ is hyperelliptic and $Aut(Y)=C_2$. For example, if $g>1$ and $Y_g$ is the hyperelliptic Riemann surface $y^2=x^{2g+1}-x-1$ then its jacobian $J_g$ has no nontrivial endomorphisms (Math. Research Letters 7 (2000), 123--132) and therefore $Aut(Y_g)=C_2$. If $p$ is an odd prime then for each integer $n \ge 5$ the automorphism group of the compact Riemann surface $y^p=x^n-x-1$ is the cyclic group $C_{p}$. Indeed, the endomorphism ring of the jacobian is the $p$th cyclotomic ring $Z[\zeta_p]$ (Math. Proc. Cambridge Philos. Soc. 136 (2004), 257--267) and one may easily check, using the differentials of the first kind that the curve is non-hyperelliptic.</p> <p>Using Del Pezzo surfaces of degree 2, one may construct non-hyperelliptic genus 3 curves $Y$, whose jacobian has no nontrivial endomorphisms (AMS Translations Series 2, vol. 218 (2006), 67--75; MR2279305, 2007k:14060) and therefore $Aut(Y)={1}$. For the genus 4 case see a paper of Anthony Várilly-Alvaradoa and David Zywina (LMS Journal of Computation and Mathematics (2009), 12: 144-165); their approach makes use of Del Pezzo surfaces of degree 1 (see also Math. Ann. 340 (2008), 407--435).</p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55814#55814 Answer by Yuri Zarhin for faithful unipotent representations of (finite) $p$-groups Yuri Zarhin 2011-02-18T02:44:44Z 2011-02-18T02:44:44Z <p>Here is a slightly different approach. If $G$ is a $p$-group (infinite or finite) and $k$ is a field of characteristic $p$ then in the group algebra $k[G]$ we have $(x-1)^{p^r}=x^{p^r}-1$ for all $x \in k[G]$. (Actually, if elements $a$ and $b$ of any $k$-algebra do commute then $(a-b)^{p^r}=a^{p^r} - b^{p^r}$, thanks to divisibility properties of binomial coefficients.) Applying it to $x=g$ where $g$ is an element of $G$ of order $p^r$, we conclude that $(g-1)^{p^r}=g^{p^r}-1=1-1=0$ in $k[G]$. Since every representation space $V$ of $G$ over $k$ is a module over $k[G]$, we conclude that $g-1$ acts on $V$ as a nilpotent operator.</p> <p>An example of a faithful representation of $G$ is provided by the regular representation where $V$ is the space of all $k$-valued functions on $G$. Another example is provided by its $G$-invariant space $V_0$ of all functions $f: G \to k$ with finite support (i.e., vanishing at all but finitely many points of $G$). Notice that for each $g \in G$ the space $V_0$ splits into a direct sum of $g$-invariant finite-dimensional subspaces (that correspond to finite left cosets of the cyclic group generated by $g$).</p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55687#55687 Answer by Yuri Zarhin for faithful unipotent representations of (finite) $p$-groups Yuri Zarhin 2011-02-17T03:13:33Z 2011-02-17T03:25:32Z <p>I guess you mean finite $p$-groups. Then every finite-dimensional representation of a finite $p$-group $G$ is unipotent in characteristic $p$. (Indeed, all eigenvalues of every element of $G$ are $p$-power roots of unity and therefore must be equal to $1$. This means that every element of $G$ acts as a unipotent operator.) An example of a faithful unipotent representation of $G$ is provided by its regular representation over any field of characteristic $p$. In particular, the regular representation of $G$ over the prime field $F_p$ is faithful and unipotent.</p> <p>(The same construction works for infinite $p$-groups $G$ as well. Of course, in this case the regular" representation space of functions on $G$ with finite support will be infinite-dimensional.)</p> http://mathoverflow.net/questions/17899/lower-bound-for-torsion-of-abelian-varieties/54443#54443 Answer by Yuri Zarhin for lower bound for torsion of abelian varieties Yuri Zarhin 2011-02-05T18:54:53Z 2011-02-05T18:54:53Z <p>Now let me address your last question. For the sake of simplicity, let us assume that $K$ is a global field of characteristic $p>2$ and the ring $End(A)$ of all endomorphisms of $A$ (over an algebraic closure of $K$) is the ring $Z$ of integers. Then my old results (Math. Notes: 21 (1978), 415--419 and 22 (1978), 493--498) imply that for all but finitely many primes $\ell$ the Galois module $A[\ell]$ is absolutely simple and the Galois group $Gal(K(A[\ell])/K)$ is noncommutative. </p> <p>I claim that for all but finitely many primes $\ell$ the order of $Gal(K(A[\ell])/K)$ is divisible by $\ell$ and therefore $[K(A[\ell]):K] > \ell$.</p> <p>Indeed, suppose that for a given $\ell$ the order of $Gal(K(A[\ell])/K)$ is not divisible by $\ell$. Let us call such $\ell$ exceptional. Then the natural representation of $Gal(K(A[\ell])/K)$ in $A[\ell]$ can be lifted to characteristic zero, i.e., $Gal(K(A[\ell])/K)$ is isomorphic to a (finite) subgroup of $GL(2\dim(A),C)$ where $C$ is the field of complex numbers. Now, by a theorem of Jordan, there exists a positive integer $N$ that depends only on $\dim(A)$ anf such that $Gal(K(A[\ell])/K)$ contains a normal commutative subgroup, whose index does not exceed $N$. This means that $K(A[\ell])/K$ contains a Galois subextension $K_{0,\ell}/K$ such that $[K_{0,\ell}:K] \le N$ and the field extension $K(A[\ell])/K_{0,\ell}$ is abelian.</p> <p>Let $S$ be the (finite) set of places of bad reduction for $A$. The field extension $K(A[\ell])/K$ is unramified outside $S$; the same is true for the subextension $K_{0,\ell}/K$. (In the number field case one should add the divisors of $\ell$ but we live in characteristic $p$!). Recall that the Galois extensions of global $K$ with degree $\le N$ and ramification only at $S$ constitute a finite set. Let $L/K$ be the compositum of all such extensions. Clearly, $L$ is also a global field while $L/K$ is a finite Galois extension that contains $K_{0,\ell}$. In addition, the Galois group $Gal(L(A[\ell])/L)$ is commutative. Now considering $A$ as an abelian variety over $L$ and applying previously mentioned results, we obtain that the set of exceptional primes $\ell$ is finite.</p> http://mathoverflow.net/questions/17899/lower-bound-for-torsion-of-abelian-varieties/54422#54422 Answer by Yuri Zarhin for lower bound for torsion of abelian varieties Yuri Zarhin 2011-02-05T17:49:00Z 2011-02-05T17:49:00Z <p>Let $E$ be a supersingular elliptic curve over a finite field $K$ of characteristic $p$. If $K$ is sufficiently large then the generator (Frobenius automorphism) of the absolute Galois group of $K$ acts on all $E[\ell]$ as multiplication by the square root $q'$ of $q$ where $q$ is the cardinality of $K$. Let $d=d(\ell)$ be the smallest positive integer such that $(q')^d-1$ is divisible by $\ell$. Then $[K(E[\ell]):K]=d$. </p> <p>Now suppose that $p=2$, $q$ is, at least, $4$ and let $r$ be a prime such that $\ell=2^r-1$ is a (Mersenne) prime. Then $[K(E[\ell]):K]=d(\ell)&lt;\log(\ell+1)$. So, if the set of Mersenne primes is infinite then we get a negative answer to your first question. Similarly, if the set of Fermat primes is infinite then we also get a negative answer to your first question. Of course, it would be good enough to know that infinitely many Mersenne (or Fermat) numbers have sufficiently large" prime factors.</p> http://mathoverflow.net/questions/53937/possible-cm-types-of-abelian-varieties/54036#54036 Answer by Yuri Zarhin for possible CM-types of abelian varieties Yuri Zarhin 2011-02-01T23:44:35Z 2011-02-01T23:44:35Z <p>There are certain additional conditions on multiplicities" that are spelled out in the original Shimura's paper. Let me discuss a couple of interesting cases.</p> <p>The case of CM type, i.e. when $n=g$. Let $\Phi$ be the corresponding CM type, which is a $n$-element set of embeddings of $K$ into the field $C$ of complex numbers. For each pair of complex-conjugate embeddings $K \to C$ exactly one of them lies in $\Phi$. If $K$ does not contain a proper CM subfield then everything is fine, i.e., the endomorphism algebra of $A$ is $K$. However, if $K$ contains a proper CM subfield $L$ then a bad" choice of $\Phi$ would imply that the endomorphism algebra of $A$ is a matrix algebra over $L$ rather than $K$. In order to guarantee that the endomorphism algebra is $K$, one should require that if $\tau_1, \tau_2 \in \Phi$ and their restrictions to the maximal totally real subfield $L_0$ of $L$ coincide then their restrictions to $L$ also coincide. (Of course, one should require it for all CM subfields.) For example, if $n=g=2$ and a quartic CM field $K$ contains an imaginary quadratic subfield then $K$ must contain another imaginary quadratic subfield as well! In this case one may check that the endomorphism algebra of $A$ is always bigger than $K$ (and $A$ is isogenous to a square of a CM elliptic curve).</p> <p>The case of an abelian surface $A$ and an imaginary quadratic field $K$, i.e., $g=1, n=2$. It is known that the endomorphism algebra of an abelian surface is never an imaginary quadratic field. (Actually, as was pointed out by Frans Oort, this assertion remains true in characteristic p.) More precisely, suppose we are given an embedding of $K$ in the endomorphism algebra of $A$. Then either $A$ is isogenous to a square of an elliptic curve with multiplication by $K$ or the endomorphism algebra of $A$ is an indefinite quaternion algebra over the rationals.</p> http://mathoverflow.net/questions/53816/fundamental-group-of-a-product-of-two-curves/53849#53849 Answer by Yuri Zarhin for Fundamental group of a product of two curves Yuri Zarhin 2011-01-31T02:02:31Z 2011-01-31T02:02:31Z <p>All two-dimensional complex tori $T$ have the same fundamental group, because such a torus is homeomorphic to a product of four copies of the unit circle $S^1$. Among them there are all the products of $E_1 \times E_2$ of elliptic curves. Since each elliptic curve is homeomorphic to a product of two copies of $S^1$, the fundamental groups of $T$ and $E_1 \times E_2$ are isomorphic (for all $T, E_1,E_2$). However, almost all two-dimensional complex tori are not biholomorphically isomorphic to a product of elliptic curves.</p> <p>Shafarevich's Basic algebraic geometry" contains examples of two-dimensional complex tori that do not contain complete complex curves at all and therefore are not the products of two curves. One may also get an explicit example of such a torus (without curves), starting with a totally complex quartic number field $F$ that does not contain an imaginary quadratic subfield, choosing a rank 4 discrete lattice $\Gamma$ in the realification $F_R$ of $F$ and putting $T=F_R/\Gamma$ (Math. Ann. 303 (1995), 11--29).</p> <p>As for complex abelian surfaces $A$ (i.e., algebraizable two-dimensional complex tori), almost all of them are also not isomorphic to a product of elliptic curves. An explicit example is provided by the jacobian $J(C)$ of the genus 2 curve $C:y^2=x^5-x-1$. Actually, it is known (arXiv:math/9909052 [math.AG]) that $J(C)$ has no nontrivial endomorphisms and therefore is not isomorphic to a product of elliptic curves.</p> http://mathoverflow.net/questions/45984/complete-extensions-of-valuations-from-q-to-r/45998#45998 Answer by Yuri Zarhin for Complete extensions of valuations from Q to R. Yuri Zarhin 2010-11-14T00:40:51Z 2010-11-14T04:36:53Z <p>Any nontrivial non-archimedean valuation on the field of rational numbers $Q$ is essentially $p$-adic for some prime $p$, i.e., there exists a prime $p$ such that $|p|&lt;1$ and one may check that such $p$ is unique: $|q|=1$ for all other primes $q$. (This is a classical theorem of Ostrowsky, see first pages of Koblitz's p-adic numbers, p-adic analysis and zeta-functions.) So, if R is complete wrt some extension of the valuation then it must contain the $p$-adic completion of $Q$, i.e., the field $Q_p$ of $p$-adic numbers. More precisely, the field of real numbers $R$ must contain a subfield isomorphic to $Q_p$. In particular, all rational numbers that are squares in $Q_p$ are squares in $R$. But this is not the case: for example, $1-p^3$ is a square in $Q_p$ but not in $R$, since it is negative. This proves that $R$ does not contain a subfield isomorphic to $Q_p$. This implies that in the case of usual valuations the answer to your question is positive in the following sense: any extension of of a nontrivial non-archimedean valuation on $Q$ to $R$ fails to make $R$ a complete field. </p> <p>Of course, if the original non-archimedean valuation on $Q$ is trivial. i.e., every nonzero rational number has norm 1 then one may extend it to the trivial valuation on $R$; every field is complete wrt the trivial valuation, including $R$.</p> http://mathoverflow.net/questions/124502/does-the-manin-drinfeld-theorem-hold-over-number-fields/124507#124507 Comment by Yuri Zarhin Yuri Zarhin 2013-03-14T20:24:12Z 2013-03-14T20:24:12Z You are welcome. Actually, since the end of 1960th all major Russian mathematical journals (Izvestija, MatSbornik, Uspekhi, Functional Analysis, MatZametki, . . .) are translated into English. http://mathoverflow.net/questions/121137/theta-group-representation/121203#121203 Comment by Yuri Zarhin Yuri Zarhin 2013-02-15T17:07:19Z 2013-02-15T17:07:19Z The answer is the same: iff the polarization is principal. If it is not principal then one may take the projectivization $P(W)$ of any proper nonzero $K_i$-invariant vector subspace $W$ of $H^0(X,L)$; this $P(W)$ is $K_i$-stable. http://mathoverflow.net/questions/116941/tate-conjecture-on-decomposition-of-motives/116944#116944 Comment by Yuri Zarhin Yuri Zarhin 2012-12-22T01:07:44Z 2012-12-22T01:07:44Z In general, one should choose a maximal commutative semisimple $Q$-subalgebra $E$ of $End_Q(X)\otimes Q$ and consider the $E\otimes_Q \bar{Q}$-module $H^1(X(C),\bar{Q})$. We have a splitting $E\otimes_Q \bar{Q}=\oplus E_j$ where each $E_j$ is the field $\bar{Q}$. Now one has put $M_j =E_j \cdot H^1(X(C),\bar{Q})$. Tensoring $M_j$ by $\bar{Q}_{\ell}$ over $\bar{Q}$, we obtain an irreducible representation of $G_Q$ (it follows from Faltings' theorem that describes the centralizer of $G_Q$ in $End_{Q_{\ell}}(H^1(X(C),Q_{\ell})$ and the semisimplicity of the action of $G_Q$ on $H^1$). http://mathoverflow.net/questions/116941/tate-conjecture-on-decomposition-of-motives/116944#116944 Comment by Yuri Zarhin Yuri Zarhin 2012-12-22T00:56:20Z 2012-12-22T00:56:20Z @mtb, You are welcome. So, we deal with $H^1$ and replacing $X$ by its Albanese variety, assume that $X$ is an abelian variety of positive dimension over the field $Q$ of rational numbers. Let me restrict myself to Part 1 of Conjecture 1.2. If $End_{Q}X$ is the ring of integers $Z$ then by Faltings' theorem $H^1(X(C),Q_{\ell}$ is an absolutely irreducible representation of the absolute Galois group $G_Q$ of $Q$ and we put $M_1=H^1(X(C),\bar{Q})$. Clearly, tensoring $M_1$ by $\bar{Q}_{\ell}}$, we obtain an irreducible representation $H^1(X(C),\bar{Q}_{\ell}$ of $G_Q$. http://mathoverflow.net/questions/116941/tate-conjecture-on-decomposition-of-motives/116944#116944 Comment by Yuri Zarhin Yuri Zarhin 2012-12-21T03:26:08Z 2012-12-21T03:26:08Z The Faltings' paper was translated into English, see &quot;Arithmetic Geometry&quot; by Cornell and Silverman, Springer-Verlag, 1984. See also &quot;Rational Points&quot; by Faltings, Wuestholz et al. (Vieweg, several rditions), Proceedings of Szpiro S\'eminaire (Ast\'erisque, 127 (1985) and my survey paper with Parshin, arXiv:0912.4325. http://mathoverflow.net/questions/107324/can-we-control-the-size-of-the-intersection-of-two-abelian-subfactors-of-an-abeli/107353#107353 Comment by Yuri Zarhin Yuri Zarhin 2012-09-17T19:01:33Z 2012-09-17T19:01:33Z You are welcome. http://mathoverflow.net/questions/106128/what-is-the-importance-of-the-conjectural-semi-simplicity-of-the-action-of-the-fr/106136#106136 Comment by Yuri Zarhin Yuri Zarhin 2012-09-01T21:20:51Z 2012-09-01T21:20:51Z Thanks for pointing out: I also claim (please see the corrected version) that $\dim(Z)=\dim(X)=\dim(Y)$. http://mathoverflow.net/questions/99064/weil-reciprocity-on-abelian-varieties-and-biextensions/99096#99096 Comment by Yuri Zarhin Yuri Zarhin 2012-06-08T10:06:51Z 2012-06-08T10:06:51Z Sorry to hear that. However, the original Russian version is available for free (see pp. 502 and 508--509 of) <a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=im&amp;paperid=2308&amp;option_lang=eng" rel="nofollow">mathnet.ru/php/&hellip;</a> <a href="http://www.mathnet.ru/links/07870c75a71989615a88f79957d8d509/im2308.pdf" rel="nofollow">mathnet.ru/links/07870c75a71989615a88f79957d8d509/&hellip;</a> http://mathoverflow.net/questions/35060/when-is-a-product-of-elliptic-curves-isogenous-to-the-jacobian-of-a-hyperelliptic/87936#87936 Comment by Yuri Zarhin Yuri Zarhin 2012-02-08T22:57:52Z 2012-02-08T22:57:52Z @Dror Speiser You are welcome. Alas, I don't know anything non-trivial when the product is of more curves except some negative results that tell that jacobians of certain superelliptic curves are not isogenous to products of elliptic curves. http://mathoverflow.net/questions/73879/canonical-liftings-of-endomorphisms-of-ordinary-abelian-varieties Comment by Yuri Zarhin Yuri Zarhin 2011-08-28T18:10:52Z 2011-08-28T18:10:52Z @ roy smith: Yes - he gave me several references that deal with the case of finite $k$. http://mathoverflow.net/questions/73396/etale-endomorphisms-of-abelian-varieties-in-positive-characteristic/73874#73874 Comment by Yuri Zarhin Yuri Zarhin 2011-08-28T16:58:53Z 2011-08-28T16:58:53Z In fact, it seems that in my first example, neither r nor t are &#233;tale: one has to look at their actions on the differentials of the first kind on the curve. http://mathoverflow.net/questions/73396/etale-endomorphisms-of-abelian-varieties-in-positive-characteristic/73874#73874 Comment by Yuri Zarhin Yuri Zarhin 2011-08-28T16:32:21Z 2011-08-28T16:32:21Z @ Damian R&#246;ssle: You are welcome. Actually, it seems that I've answered a different question inspired by your remark about ordinary variety. Indeed, if $X$ is not defined over a finite field then the Frobenius and Verschiebung morphisms (or their powers) are not endomorphisms of $X$ over any field. My apologies! http://mathoverflow.net/questions/61984/rational-points-on-algebraic-curves-over-qab/62014#62014 Comment by Yuri Zarhin Yuri Zarhin 2011-04-17T20:27:40Z 2011-04-17T20:27:40Z OOPS! Sorry. I've mistook the underline as bar over K. http://mathoverflow.net/questions/61984/rational-points-on-algebraic-curves-over-qab/62026#62026 Comment by Yuri Zarhin Yuri Zarhin 2011-04-17T14:54:34Z 2011-04-17T14:54:34Z You are welcome. http://mathoverflow.net/questions/61984/rational-points-on-algebraic-curves-over-qab/62014#62014 Comment by Yuri Zarhin Yuri Zarhin 2011-04-17T14:54:02Z 2011-04-17T14:54:02Z As for a simple abelian variety $A$ over a number field $K$, the torsion subgroup of $A(K^{ab})$ is finite if and only if $A$ is not of CM-type over&quot; $K$. (in other words, $\bar{K}$ on the last line of your comment should be replaced by $K$.)