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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?

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Doesn't make sense to tensor "over Q" against ab. var. (Q doesn't act on ab. var., except in isogeny-category sense that is linguistics). So first use an order in $E$ (& then pass to isog. category): for finite flat $\mathbf{Z}$-alg. $R$, the functor $S \mapsto A_0(S) \otimes_ {\mathbf{Z}} R$ on analytic spaces is rep'td by ab. var. (call it $A_0 \otimes R$), and map $A_0 \rightarrow A_0 \otimes R$ induces ${\rm{H}}_1(A_0) \rightarrow {\rm{H}}_1(A_0 \otimes R)$ whose linearization $R \otimes {\rm{H}}_1(A_0) \rightarrow {\rm{H}}_1(A_0 \otimes R)$ is an isom. Nice exercise with uniformization. – BCnrd May 12 '10 at 15:58
you are right ,it is in the isogeny-categoty,which I forget to say. – TOM May 12 '10 at 16:02
The role of $R$ above only matters as $\mathbf{Z}$-module: for finite free $\mathbf{Z}$-module $M$ can similarly define/characterize $A_0 \otimes M$, and to any $m \in M$ we associate $A_0 \rightarrow A_0 \otimes M$ defined by $a \mapsto a \otimes m$ (via def'n of right side). The existence/uniqueness/isom. problems all make sense for any $M$ (recovering above when $M$ is ring), and behavior for direct sum via products reduces us to the case $M = \mathbf{Z}$. Can improve arguments to work with $\mathbf{Z}$ replaced by other (assoc.) rings and $M$ a finite projective (left/right?) module. – BCnrd May 12 '10 at 16:48
@TOM: Note that the tensor construction is not a case of "base change" in any sense that I know: in particular you start and end with an abelian variety over $\mathbb{C}$. You might want to retitle your question accordingly. – Pete L. Clark May 12 '10 at 18:08
up vote 3 down vote accepted

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.

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Well, BCnrd's terse comment is enlightening as usual. Let me know if you need any help fleshing it out. – Pete L. Clark May 12 '10 at 16:13
I find this statement in a paper,which does not tell us the exactly meaning of $A_0\otimes_Q E$,and I have thought that it is just base change ,maybe I am wrong.And the paper has another statement$H^1(A_0\otimes_Q E)=H^1(A_0)\otimes_Q E$ – TOM May 12 '10 at 16:17
Can you tell me the "Serre tensor construction",or any reference ? – TOM May 12 '10 at 16:26
And the condtion that dim $A_0\otimes_Q E=[E:Q]$dim$A_0$is true in the paper – TOM May 12 '10 at 16:30
Perhaps its the paper that includes: "Recall that the category of abelian varieties up to isogeny is obtained from the category of abelian varieties by taking the same class of objects but replacing $Hom(A,B)$ with $Hom(A;B)\otimes\mathbb{Q}$. We shall always regard an abelian variety as an object in the category of abelian varieties up to isogeny: thus $Hom(A,B)$ is a vector space over $\mathbb{Q}$." If so, $A\otimes E$ means $A\otimes_{\mathbb{Z}}\mathcal{O}_{E}$, which is explained by Torsten. – JS Milne May 12 '10 at 17:04

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.

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