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?

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. 


Serre's construction (or, I believe, a version of it which is enough here) takes a commutative ring $R$ and an $R$category $C$ for which all idempotents have kernels. (An $R$category is a category enriched in $R$modules and which has direct sums.) Then for every finitely generated projected $R$module $P$ and every object $A$ of $C$ we can define $A\bigotimes_RP$ characterised by the adjunction equality $\mathrm{Hom}_R(P,\mathrm{Hom}_C(A,B))=\mathrm{Hom}_A(A\bigotimes_RP,B)$. For $P=R^n$ we clearly can put $A\bigotimes_RP=A^n$ and for a general $P$ we write it as a summand of some $R^n$ and use the fact that idempotents have kernels in $C$. It is purely formal that this tensor product commutes with additive functors and $H_1(,\mathbb Q)$ as a functor on the isogeny category has that property. If I remember correctly Serre's construction is a version of this where $R$ is an arbitrary ring and we have a fixed ring homomorphism $R \rightarrow \mathrm{End}(A)$ and a right projective finitely generated $R$module $P$ (and $C$ is an arbitrary additive category whose idempotents have kernels). We can then define $P\bigotimes_RA$ by the property that $\mathrm{Hom}_R(P,\mathrm{Hom}(A,B))=\mathrm{Hom}(P\bigotimes_RA,B)$. The proof of existence is almost identical to the one above. 

