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Let $E$ be a supersingular elliptic curve over an algebraically closed field $K$ of characteristic $p$. Let $R = \operatorname{End}(E)$ be its ring of endomorphisms. Then, it is known $R \otimes_{\mathbb Z} \mathbb Q$ is an order in a quaternion algebra $D$. In particular, $D$ has a multiplicative norm function $N\colon D^* \rightarrow \mathbb Q$, which can be defined as $N(\phi) = \phi \hat \phi$, where $\hat\phi$ denotes the dual of the endomorphism $\phi$, or equivalently, by saying that $N(\phi)$ is the degree of an endomorphism $\phi \in R$.

My question is: Does there exist an element $\phi \in R$ such that $N(\phi)$ is singly divisible by $p$, i.e. such that the $p$-adic valuation of $N(\phi)$ is 1?

If $E$ is defined over $\mathbb F_p$, then Frobenius is an endomorphism, and has degree $p$, so that gives an affirmative answer, but some supersingular elliptic curves are only defined over $\mathbb F_{p^2}$. In that case, Frobenius is not an endomorphism of $E$, but an isogeny $E \rightarrow E^{(p)}$ where $E^{(p)}$ is the Frobenius twist of $E$. The question is equivalent to: Does there exists a separable isogeny $E^{(p)} \rightarrow E$?

Also, I don't know much about quaternion algebras over $\mathbb Q$, but I have read that $D$ is ramified at $p$, and if I understand the consequences of that correctly, that implies that there exists an element $\phi$ of $D$ such that $N(\phi)$ has $p$-adic valuation equal to 1. If that's true, then, the question is whether such an element can be chosen to be in $R$, and not just in $D$. I have also read that $R$ is a maximal order in $D$, which seems like it should help, but I'm not sure how to make use of that fact in a non-commutative setting.

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$\newcommand{\Z}{\mathbb{Z}}$ First, if you want to learn about quaternion algebras, their orders and their relation to supersingular elliptic curves, I suggest John Voight's book.

The answer to your question is yes. To see this, you can use the fact that (because $R$ is maximal) $R/p^2R$ contains an element $\pi$ such that $\pi^2 = p$, and therefore satisfying $N(\pi)\equiv \pm p\bmod{p^2}$. Let $\phi\in R$ be such that $\phi\equiv \pi \bmod{p^2R}$, then the valuation at $p$ of $N(\phi)$ is $1$.

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  • $\begingroup$ Thanks @Aurel. Voight's book is helpful and I'm using it to fill in the details. In particular, the results in Chapter 13 show that the p-adic valuation of the norm on D is a valuation, and Thm. 13.3.10(c) implies that this valuation is surjective onto the integers, which shows that my claim in the last paragraph is true. Then, because begin a maximal order localizes (Lemma 10.4.2) and is then compatible with completion (Lem. 9.4.6), we know that $R \otimes_{\mathbb Z} \mathbb Z_p$ is a maximal order, and and then Prop. 13.3.4 shows that it consists of the integral elements, such as $\pi$. $\endgroup$ – Dustin Cartwright Aug 29 '18 at 17:46
  • $\begingroup$ However, Voight also has a proof that the endomorphism ring of E is a maximal order, which itself uses that maximality is a local property (Thm. 42.1.9), and so shows directly that $R \otimes \mathbb Z_p$ is a maximal order. Adapting that to my question: once we know that the p-adic valuation of the norm on $D$ is surjective onto the integers, then there is a $\phi\in R$ such that the $p$-adic valuation of $N(\phi)$ is odd, say $2k+1$. Then, we claim that $\phi$ is divisible by $p$, for example because its purely inseparable part is the $2k+1$ power of Frobenius, and $p$ is the $2k$th power. $\endgroup$ – Dustin Cartwright Aug 29 '18 at 18:03

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