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.



Existence of certain endomorphism of supersingular elliptic curve
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First, if you want to learn about quaternion algebras, their orders and their relation to supersingular elliptic curves, I suggest John Voight's [book](https://math.dartmouth.edu/~jvoight/quat.html).


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 $1$.


