Supersingular curves over $\mathbb{F}_q$ and the splitting of $p$ 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.
 A: 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 $p$.
A: Here is a proof that does not use the Honda-Tate theory.
Notice that the quadratic field
$K=Q(\alpha)$ may be viewed as the subfield of the endomorphism algebra $End^0(E)=End(E)\otimes Q$ with the same identity element. Here $End(E)$ is the algebra of ALL endomorphisms of $E$ over an algebraic closure of $F_q$.

*

*Suppose that $E$ is ordinary and $T_p(E)$ is its physical $p$-adic Tate module, which is a free $Z_p$-module of rank 1. Let us consider the corresponding $Q_p$-vector space $V_p(E)=T_p(E)\otimes_{Z_p}Q_p$, which is a one-dimensional vector space over the field $Q_p$ of $p$-adic numbers.

By functoriality, there is the natural $Q_p$-algebra homomomorphism
$$K_p=K\otimes_Q Q_p \to End_{Q_p}(V_p(E))=Q_p,$$
which sends $1$ to $1$ and therefore is not zero.  Since $K_p$ has $Q_p$-dimension $2>1$, this homomorphism is not injective and therefore $K_p$ is NOT a field, i.e., $p$ splits in $K=Q(\alpha)$.


*Suppose that $E$ is supersingular. Then
$$End(E)\otimes Q_p=End^0(E)\otimes_Q Q_p$$
is a division algebra over $Q_p$ of dimension 4 that contains
$K_p=K\otimes_Q Q_p$ as a $Q_p$-subalgebra. Since $End^0(E)\otimes_Q Q_p$ has no zero divisors, $K_p$ also has no zero divisors, i.e., $p$ does NOT split in $K=Q(\alpha)$.

