Suppose $A\_0$ is an abelian variety over $\mathbb{C}$, $E$ is a CM field ,denote $A= A\_0\otimes\_Q E$, is there an isomorphism $ H\_1(A\_0\otimes\_Q E,Q)=H\_1(A\_0,Q)\otimes\_Q E$?how it comes?



homology of abelian variety ?
If $A$ over $\mathbb{C}$ is an abelian variety of dimension $g$, then for all $i \in \mathbb{N}$, $H\_i(A,\mathbb{Q}) \cong H^i(A,\mathbb{Q}) \cong \mathbb{Q}^{ {2g \choose i}}$. 


Taking $i = 1$, the conclusion you are asking about is true if and only if $\operatorname{dim} A\_0 \otimes\_\mathbb{Q} E = [E:\mathbb{Q}] \operatorname{dim} A\_0$. 


Which it probably is, if $A\_0 \otimes\_{\mathbb{Q}} E$ means the Serre tensor construction. I can't quite remember how this goes at the moment (and I'll wait for you to confirm your notation before trying). 


If you tell/remind us exactly what $A\_0 \otimes\_{\mathbb{Q}} E$ means, we could probably give you a natural isomorphism between these two homology groups. 


**Addendum**: I found a nice online treatment of Serre's tensor construction [here](http://math.stanford.edu/~conrad/vigregroup/vigre04/serre_constr.pdf).


