TITLE: Abelian variety with prescribed endomorphism ring
QUESTION [11 upvotes]: Consider the cyclotomic field $L={{\mathbb{Q}}}(\zeta_8)={{\mathbb{Q}}}(\sqrt{2},i)$, where $\zeta_8$ is a primitive 8-th  root of unity. Let $\Lambda={{\mathbb{Z}}}[\zeta_8]$ denote the ring of integers of $L$.

Question 1. Does there exist an abelian variety $A$  over ${{\mathbb{C}}}$ of dimension two  with
  ${\mathrm{End\,}} A\simeq \Lambda$ ?

I expect the answer "no". Note that I want the endomorphism ring to be exactly $\Lambda$, not just contain $\Lambda$. In particular, my abelian variety should be simple.

Question 2. Does there exist an abelian variety $A$  over ${{\mathbb{C}}}$ of dimension four  with
  ${\mathrm{End\,}} A\simeq \Lambda$ ?

I expect  the answer "yes".

Question 3. Is it possible to find an abelian variety $A$  over ${{\mathbb{C}}}$ of dimension 2 or 4  with
  ${\mathrm{End\,}} A\simeq \Lambda$  and such that $A$ is definable (without endomorphisms) over a small number field,
  say over ${{\mathbb{Q}}}(i)$ or even over ${{\mathbb{Q}}}$?

REPLY [9 votes]: The answer to your Question 3 is YES with the ground field $\mathbb{Q}$.
Here is a sketch of the proof. For each positive integer $q$ and a "parameter" $t$ (in char 0) consider the smooth projective model $C_{q,t}$ of an affine curve $y^q=x^3-x-t$. Let $P_{8,t}$ be the Prym variety of the double cover
$$C_{8,t}\to C_{4,t}, (x,y)\mapsto (x,y^2).$$
Then $P_{8,t}$ is an abelian fourfold provided with an embedding 
$$\mathbb{Z}[\zeta_8]\hookrightarrow End(P_{8,t}).$$
One may deduce from Theorem 1.5 of  arXiv:math/0601072 [math.AG] that if $t$ is  a transcendental number then $P_{8,t}$ does not contain positive-dimensional abelian subvarieties of CM type. It follows from  Th. 1.1 of a paper by Jiangwei Xue and Chia-Fu Yu, arXiv:1304.6202 [math.NT] that for such $t$ the endomorphism ring $End(P_{8,t})$ coincides with $\mathbb{Z}[\zeta_8]$. Now, by Masser's specialization theorem
one may choose a rational number $c$ such that
$End(P_{8,c})$ equals $End(P_{8,t})$
and therefore coincides with  $\mathbb{Z}[\zeta_8]$.