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.