I'm looking at chapter 4 of Waterhouse's *"Abelian varieties over finite fields"*; and Theorems 4.1 and 4.2 seem to use the following fact:



> 
> Suppose that $E/\mathbb{F}\_q$ is an elliptic curve over a finite field with $q=p^n$ elements and let $\pi\_E$ denote the $q$th-power Frobenius map acting on $E$. Suppose that $\pi\_E$ does not act like multiplication-by-$N$ for any integer $N$ so that $\pi\_E = [\alpha]\_E$ where $\alpha$ is a root of the polynomial
> $$x^2 - \mathrm{tr}(\pi\_E)x + q$$
> Then $E$ is ordinary if and only if $p$ is splits in $\mathbb{Q}(\alpha)$.
> 
> 
> 


Waterhouse uses some very general theory (namely a theorem of Honda/Tate and the general theory of abelian varieties) and I find his reasoning/terminology vague. Does anybody know of a more down to earth proof for this result?


Lang has a proof utilitizing the fact that $End(E)\to End(T\_p(E))$ is injective; but i'm trying to find proofs that don't use local methods


**EDIT:** Update on attempt: Here is a proof attempt; perhaps someone can help me finish it off.


We prove $E$ is supersingular if and only if $p$ is non-split in $\mathbb{Q}(\alpha)$.


First suppose that $p$ is non-split. If $p$ is inert, then since $\alpha$ has norm $q = p^n$, it follows that $p$ divides $(\alpha)$ and so $E[p]\subseteq E[\pi\_E] = \{O \}$; thus $E$ is supersingular. If $p$ ramifies then $p$ divides the discriminant of $\mathbb{Q}(\alpha)$ and so $p$ divides $\mathrm{tr}(\pi\_E)^2 - 4q$, which implies that $p$ divides $\mathrm{tr}(\pi\_E)$ and so $E$ is supersingular;


Conversely, suppose that $E$ is supersingular. Suppose that $p$ splits as $(p) = \mathfrak{p}\overline{\mathfrak{p}}$. Then $(\alpha) = \mathfrak{p}^a\overline{\mathfrak{p}}^b$ where $a+b=n$. However, since $E$ is supersingular $\alpha^N \in \mathbb{Z}$ for some positive integer $N$. Which implies that $(\alpha)^N = \mathfrak{p}^{Na}\overline{\mathfrak{p}}^{Nb}$ and so $Na=Nb$ since $(\alpha^N) = \overline{(\alpha^N)}$. Therefore $a=b$ and so $(\alpha) = \mathfrak{p}^a \overline{\mathfrak{p}}^a = (p^a)$ where $a=n/2$. This implies that $\alpha = \zeta p^{a}$ for some unit $\zeta$. This is where I am stuck.


Any help would be appreciated.



Supersingular curves over $\mathbb{F}_q$ and the splitting of $p$
Here is a solution that avoids explicit use of $\mathbf{Q}\_p$ and in particular does not require knowing that $(\operatorname{End} E) \otimes \mathbf{Q}\_p$ is a division ring. The key is to use *inseparable degree of endomorphisms*.


Identify $\alpha$ with $\pi\_E$. Let $a = \operatorname{tr} \alpha \in \mathbf{Z}$.
Let $\mathcal{O}$ be the ring of integers of $\mathbf{Q}(\alpha)$.
Let $\mathcal{O}' := (\operatorname{End} E) \cap \mathbf{Q}(\alpha)$.
For $\beta \in \mathcal{O}'$, the inseparable degree $\deg\_{\text{i}} \beta$ is multiplicative in $\beta$. The $\beta \in \mathcal{O}'$ with $\deg\_{\text{i}} \beta \ge p^m$ are the $\beta$ that factor through the $p^m$-power Frobenius morphism $F\_m \colon E \to E^{(p^m)}$; they form an additive subgroup. Thus $\beta \mapsto \log\_p \deg\_{\text{i}} \beta$ defines a valuation $v \colon \mathcal{O}' \to \mathbf{Z} \cup \{\infty\}$. Extend $v$ to the fraction field $\mathbf{Q}(\alpha)$.


**Case 1: $E$ is ordinary.** Then $p \nmid a$, so $x^2-ax+q$ mod $p$ has two distinct factors, so $p$ splits in $\mathbf{Q}(\alpha)$.


**Case 2: $E$ is supersingular.** Then $p \mid a$, so the equality $\alpha^2 - a \alpha + q = 0$ yields $\alpha^2 \in p \mathcal{O}$.


First consider the case $\mathcal{O}'=\mathcal{O}$. If $\beta \in \mathcal{O}$ satisfies $v(\beta) \ge 2n$, then $\beta$ factors through $F\_{2n} = \alpha^2 \in p \mathcal{O}$, so $\beta \in p \mathcal{O}$.
This implication for all $\beta$ shows that $v$ is the unique place above $p$.


In general, write $(\mathcal{O}:\mathcal{O}')=p^e d$ with $p \nmid d$.
If $\beta \in \mathcal{O}$ satisfies $v(\beta) \ge (e+1)2n$,
then the endomorphism $(p^e d) \beta$ is divisible (in $\mathcal{O}'$ or in $\mathcal{O}$)
by $\alpha^{2(e+1)}$ and hence by $p^{e+1}$, but $p \nmid d$,
so $\beta \in p\mathcal{O}$.
Again, this shows that $v$ is the unique place above $p$.


