Timeline for Endomorphisms ring of complex abelian variety under isogenies
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2022 at 7:16 | comment | added | Martina Monti | Thank everyone for the answers, now it is more clear! | |
| Jan 30, 2022 at 20:21 | comment | added | François Brunault | 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. | |
| Jan 30, 2022 at 17:23 | comment | added | Mikhail Borovoi | 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. | |
| Jan 30, 2022 at 17:17 | comment | added | Chris | 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 $\mathcal{O}_K$. | |
| Jan 30, 2022 at 17:15 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
Typos corrected
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| S Jan 30, 2022 at 16:13 | review | First questions | |||
| Jan 30, 2022 at 17:03 | |||||
| S Jan 30, 2022 at 16:13 | history | asked | Martina Monti | CC BY-SA 4.0 |