most recent 30 from http://mathoverflow.net 2013-05-24T02:21:37Z http://mathoverflow.net/feeds/question/119956 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119956/abelian-varieties-with-given-endomorphism-algebra Felipe Voloch 2013-01-26T18:45:30Z 2013-01-26T21:14:01Z

<p>I am confused by a statement in the very classical paper of A. A. Albert "On the construction of Riemman matrices II", Ann. Math. 1935, Thm 16. If I understand what he saying, the theorem says that given a quaternion algebra $D$ over a totally real field $F$ of degree $t$ over $\mathbb{Q}$, which is not split at any of the infinite places of $F$, there exists an abelian variety (over $\mathbb{C}$) of dimension $n$ with endomorphism ring (tensored with $\mathbb{Q}$) equal to $D$ if and only if $n = 2tr, r>1$. (I changed the notation slightly).</p> <p>Now, I thought that such things existed even with $r=1$. For example, if $t=1, F = \mathbb{Q}$ one can have an abelian surface ($n=2$) with endomorphism ring $D$. Another place where these things are alluded to (for arbitrary $t$) is a paper of Y. Morita "on potential good reduction of abelian varieties" J. Fac. Sci. Univ. Tokyo, 1975, remark 3.1.</p> <p>I guess my questions are, first is whether Albert is right, wrong or misunderstood by me. In case Albert is wrong, is there a place where examples with $r=1$ are constructed? Can every such $D$ occur , when $r=1$?</p>

http://mathoverflow.net/questions/119956/abelian-varieties-with-given-endomorphism-algebra/119966#119966 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>