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Suzet
Suzet
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Let $E/\mathbb Q$ be a quadratic totally imaginary number field with ring of integers $\mathcal O_E$. Let $p > 2$ be an odd prime number which ramifies in $\mathcal O_E$. Let $\mathcal P$ be the place of $\mathcal O_E$ above $p$, so that $p\mathcal O_E = \mathcal P^2$. The $\mathcal P$-adic completion of $E$ is a ramified quadratic extension of $\mathbb Q_p$. Up to isomorphism, there are only two such quadratic ramified extension of $\mathbb Q_p$. I make the assumption that $E_{\mathcal P} \simeq \mathbb Q_p[\sqrt{\epsilon p}]$ where $-\epsilon \in \mathbb Z_p^{\times}$ is not a square.

In other words, $E \simeq \mathbb Q[\sqrt{-\delta}]$ for some squarefree positive integer $\delta$ such that $p|\delta$ and $\frac{\delta}{p}$ is not a square in $\mathbb Z_p^{\times}$.

Is there a supersingular abelian variety $A$ over $\mathbb F_p$ with a non-trivial $\mathcal O_E$-action, ie. equipped with an injective map $\iota_A:\mathcal O_E \otimes \mathbb Z_{(p)} \hookrightarrow \mathrm{End}_{\mathbb F_p}(A) \otimes \mathbb Z_{(p)}$? If so, may we require $A$ to be superspecial as well?

Here, recall that an abelian variety $A$ of dimension $g$ over a field $k$ of characteristic $p$ is said to be supersingular (resp. superspecial) if, for some (or for all) algebraically closed field $K \supset k$, the base change $A \otimes_k K$ is isogeneous (resp. isomorphic) to $\mathcal E^g$ for some supersingular elliptic curve $\mathcal E$ over $K$. Besides, an elliptic curve $\mathcal E$ over $k$ is supersingular if $\mathcal E[p](K) = 0$ for some (or for all) $K$ as above. Equivalently, $A$ is supersingular if and only if $\mathrm{End}_{K}(A\otimes K)$ has rank $(2g)^2$ over $\mathbb Z$, or again if and only if the Dieudonné module of the $p$-divisible group of $A\otimes K$ has only slope $1/2$ in the isoclinic decomposition.

My question arises from the observation that it seemingly fails for elliptic curve. Indeed, any supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ satisfies $\mathrm{End}_{\mathbb F_p}(\mathcal E) \simeq \mathbb Q[\pi]$, where $\pi = \sqrt{-p}$ is the Frobenius action. In particular, if I had a supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ with $\mathcal O_E$-action, taking $p$-adic completion I would have an embedding $\mathbb Z_p[\sqrt{\epsilon p}] \subset \mathbb Z_p[\sqrt{-p}]$, which is certainly not possible. Note that, however, $\mathcal E \otimes \mathbb F_{p^2}$ does have $\mathcal O_E$-action

Let $E/\mathbb Q$ be a quadratic totally imaginary number field with ring of integers $\mathcal O_E$. Let $p > 2$ be an odd prime number which ramifies in $\mathcal O_E$. Let $\mathcal P$ be the place of $\mathcal O_E$ above $p$, so that $p\mathcal O_E = \mathcal P^2$. The $\mathcal P$-adic completion of $E$ is a ramified quadratic extension of $\mathbb Q_p$. Up to isomorphism, there are only two such quadratic ramified extension. I make the assumption that $E_{\mathcal P} \simeq \mathbb Q_p[\sqrt{\epsilon p}]$ where $-\epsilon \in \mathbb Z_p^{\times}$ is not a square.

In other words, $E \simeq \mathbb Q[\sqrt{-\delta}]$ for some squarefree positive integer $\delta$ such that $p|\delta$ and $\frac{\delta}{p}$ is not a square in $\mathbb Z_p^{\times}$.

Is there a supersingular abelian variety $A$ over $\mathbb F_p$ with a non-trivial $\mathcal O_E$-action, ie. equipped with an injective map $\iota_A:\mathcal O_E \otimes \mathbb Z_{(p)} \hookrightarrow \mathrm{End}_{\mathbb F_p}(A) \otimes \mathbb Z_{(p)}$? If so, may we require $A$ to be superspecial as well?

Here, recall that an abelian variety $A$ of dimension $g$ over a field $k$ of characteristic $p$ is said to be supersingular (resp. superspecial) if, for some (or for all) algebraically closed field $K \supset k$, the base change $A \otimes_k K$ is isogeneous (resp. isomorphic) to $\mathcal E^g$ for some supersingular elliptic curve $\mathcal E$ over $K$. Besides, an elliptic curve $\mathcal E$ over $k$ is supersingular if $\mathcal E[p](K) = 0$ for some (or for all) $K$ as above. Equivalently, $A$ is supersingular if and only if $\mathrm{End}_{K}(A\otimes K)$ has rank $(2g)^2$ over $\mathbb Z$, or again if and only if the Dieudonné module of the $p$-divisible group of $A\otimes K$ has only slope $1/2$ in the isoclinic decomposition.

My question arises from the observation that it seemingly fails for elliptic curve. Indeed, any supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ satisfies $\mathrm{End}_{\mathbb F_p}(\mathcal E) \simeq \mathbb Q[\pi]$, where $\pi = \sqrt{-p}$ is the Frobenius action. In particular, if I had a supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ with $\mathcal O_E$-action, taking $p$-adic completion I would have an embedding $\mathbb Z_p[\sqrt{\epsilon p}] \subset \mathbb Z_p[\sqrt{-p}]$, which is certainly not possible. Note that, however, $\mathcal E \otimes \mathbb F_{p^2}$ does have $\mathcal O_E$-action

Let $E/\mathbb Q$ be a quadratic totally imaginary number field with ring of integers $\mathcal O_E$. Let $p > 2$ be an odd prime number which ramifies in $\mathcal O_E$. Let $\mathcal P$ be the place of $\mathcal O_E$ above $p$, so that $p\mathcal O_E = \mathcal P^2$. The $\mathcal P$-adic completion of $E$ is a ramified quadratic extension of $\mathbb Q_p$. Up to isomorphism, there are only two such quadratic ramified extension of $\mathbb Q_p$. I make the assumption that $E_{\mathcal P} \simeq \mathbb Q_p[\sqrt{\epsilon p}]$ where $-\epsilon \in \mathbb Z_p^{\times}$ is not a square.

In other words, $E \simeq \mathbb Q[\sqrt{-\delta}]$ for some squarefree positive integer $\delta$ such that $p|\delta$ and $\frac{\delta}{p}$ is not a square in $\mathbb Z_p^{\times}$.

Is there a supersingular abelian variety $A$ over $\mathbb F_p$ with a non-trivial $\mathcal O_E$-action, ie. equipped with an injective map $\iota_A:\mathcal O_E \otimes \mathbb Z_{(p)} \hookrightarrow \mathrm{End}_{\mathbb F_p}(A) \otimes \mathbb Z_{(p)}$? If so, may we require $A$ to be superspecial as well?

Here, recall that an abelian variety $A$ of dimension $g$ over a field $k$ of characteristic $p$ is said to be supersingular (resp. superspecial) if, for some (or for all) algebraically closed field $K \supset k$, the base change $A \otimes_k K$ is isogeneous (resp. isomorphic) to $\mathcal E^g$ for some supersingular elliptic curve $\mathcal E$ over $K$. Besides, an elliptic curve $\mathcal E$ over $k$ is supersingular if $\mathcal E[p](K) = 0$ for some (or for all) $K$ as above. Equivalently, $A$ is supersingular if and only if $\mathrm{End}_{K}(A\otimes K)$ has rank $(2g)^2$ over $\mathbb Z$, or again if and only if the Dieudonné module of the $p$-divisible group of $A\otimes K$ has only slope $1/2$ in the isoclinic decomposition.

My question arises from the observation that it seemingly fails for elliptic curve. Indeed, any supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ satisfies $\mathrm{End}_{\mathbb F_p}(\mathcal E) \simeq \mathbb Q[\pi]$, where $\pi = \sqrt{-p}$ is the Frobenius action. In particular, if I had a supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ with $\mathcal O_E$-action, taking $p$-adic completion I would have an embedding $\mathbb Z_p[\sqrt{\epsilon p}] \subset \mathbb Z_p[\sqrt{-p}]$, which is certainly not possible. Note that, however, $\mathcal E \otimes \mathbb F_{p^2}$ does have $\mathcal O_E$-action

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Suzet
Suzet
  • 769
  • 3
  • 13

Existence of supersingular abelian variety over $\mathbb F_p$ with $\mathcal O_E$-action where $E/\mathbb Q$ is quadratic imaginary ramified at $p$

Let $E/\mathbb Q$ be a quadratic totally imaginary number field with ring of integers $\mathcal O_E$. Let $p > 2$ be an odd prime number which ramifies in $\mathcal O_E$. Let $\mathcal P$ be the place of $\mathcal O_E$ above $p$, so that $p\mathcal O_E = \mathcal P^2$. The $\mathcal P$-adic completion of $E$ is a ramified quadratic extension of $\mathbb Q_p$. Up to isomorphism, there are only two such quadratic ramified extension. I make the assumption that $E_{\mathcal P} \simeq \mathbb Q_p[\sqrt{\epsilon p}]$ where $-\epsilon \in \mathbb Z_p^{\times}$ is not a square.

In other words, $E \simeq \mathbb Q[\sqrt{-\delta}]$ for some squarefree positive integer $\delta$ such that $p|\delta$ and $\frac{\delta}{p}$ is not a square in $\mathbb Z_p^{\times}$.

Is there a supersingular abelian variety $A$ over $\mathbb F_p$ with a non-trivial $\mathcal O_E$-action, ie. equipped with an injective map $\iota_A:\mathcal O_E \otimes \mathbb Z_{(p)} \hookrightarrow \mathrm{End}_{\mathbb F_p}(A) \otimes \mathbb Z_{(p)}$? If so, may we require $A$ to be superspecial as well?

Here, recall that an abelian variety $A$ of dimension $g$ over a field $k$ of characteristic $p$ is said to be supersingular (resp. superspecial) if, for some (or for all) algebraically closed field $K \supset k$, the base change $A \otimes_k K$ is isogeneous (resp. isomorphic) to $\mathcal E^g$ for some supersingular elliptic curve $\mathcal E$ over $K$. Besides, an elliptic curve $\mathcal E$ over $k$ is supersingular if $\mathcal E[p](K) = 0$ for some (or for all) $K$ as above. Equivalently, $A$ is supersingular if and only if $\mathrm{End}_{K}(A\otimes K)$ has rank $(2g)^2$ over $\mathbb Z$, or again if and only if the Dieudonné module of the $p$-divisible group of $A\otimes K$ has only slope $1/2$ in the isoclinic decomposition.

My question arises from the observation that it seemingly fails for elliptic curve. Indeed, any supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ satisfies $\mathrm{End}_{\mathbb F_p}(\mathcal E) \simeq \mathbb Q[\pi]$, where $\pi = \sqrt{-p}$ is the Frobenius action. In particular, if I had a supersingular elliptic curve $\mathcal E$ over $\mathbb F_p$ with $\mathcal O_E$-action, taking $p$-adic completion I would have an embedding $\mathbb Z_p[\sqrt{\epsilon p}] \subset \mathbb Z_p[\sqrt{-p}]$, which is certainly not possible. Note that, however, $\mathcal E \otimes \mathbb F_{p^2}$ does have $\mathcal O_E$-action