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I’m trying to understand if over $\mathbb{C}$ two abelian varieties have the same complex multiplication if and only if they are isogenous. Is it true?

If it is true this means that if I consider $A$ abelian variety over $\mathbb{C}$ and $p$ a $n$-torsion point, let $t_p$ the translation on $A$ by the point $p$ which defines an action of finite cyclic group of $A$, let $B=A/t_p$: then we have that $A$ and $B$ have the same endomorphisms ring in any case? Because if $A$ has no CM then $B$ has no CM, while if $A$ has CM then $B$ has the same CM because they are isogenous?

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    $\begingroup$ Welcome to MathOverflow! What do you mean by "have the same complex multiplication"? If you mean "they have the same endomorphism ring", this is false: take for example two CM elliptic curves $\mathbb{C}/\mathcal{O}_K$ and $\mathbb{C}/\mathcal{O}$, where $\mathcal{O}_K$ is the ring of integers of a quadratic imaginary number field $K$, and $\mathcal{O}\subset \mathcal{O}_K$ is an order strictly contained in $\mathcal{O}_K$. $\endgroup$
    – Chris
    Commented Jan 30, 2022 at 17:17
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    $\begingroup$ No. Let $A$ be an elliptic curve over $\mathbb C$ with complex multiplication, and let $B$ and $C$ be non-isogenous elliptic curves over $\mathbb C$ without complex multiplication. Then the abelian surfaces $A\times B$ and $A\times C$ have isomorphic endomorphism algebras (and even isomorphic endomorphism rings), but they are not isogenous. $\endgroup$
    – Mikhail Borovoi
    Commented Jan 30, 2022 at 17:23
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    $\begingroup$ On the other hand, two isogenous abelian varieties $A,B$ over $\mathbb{C}$ have the same endomorphism algebra, that is $\mathrm{End}(A) \otimes \mathbb{Q} \cong \mathrm{End}(B) \otimes \mathbb{Q}$. This is because an isogeny $\phi : A \to B$ has an inverse after tensoring with $\mathbb{Q}$, in other words in $\mathrm{Hom}(B,A) \otimes \mathbb{Q}$, using the dual isogeny. $\endgroup$
    – François Brunault
    Commented Jan 30, 2022 at 20:21
  • $\begingroup$ Thank everyone for the answers, now it is more clear! $\endgroup$
    – Martina Monti
    Commented Jan 31, 2022 at 7:16

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