For any elliptic curve $E/\mathbb{Q}$, there are infinitely many primes $p$ such that $E$ is ordinary.

Question

Let $E$ be an elliptic curve defined over $\mathbb{Q}$, and fix a Weierstrass equation for $E$ having coefficients in $\mathbb{Z}$. How do I see that there are infinitely many primes $p \in \mathbb{Z}$ such that the reduced curve $E/\mathbb{F}_p$ has Hasse invariant $1$?

Joe Silverman writes the following here.

Elkies and Serre have independent proofs that for every $\epsilon>0$ there is a constant $C_\epsilon>0$ such that $$ {\#\{p\le X : E/\mathbb{F}_p \text{ is ordinary}\}} \ge C_\epsilon X^{3/4-\epsilon}. $$

However, I just want to know that there are infinitely many ordinary primes, which should be much easier to see.

Answer 1

Just to get this off the unanswered list, let me provide a solution to the exercise that Silverman mentions. $\newcommand \Q{\mathbb{Q}}$ $\newcommand \Z{\mathbb{Z}}$ $\newcommand \F{\mathbb{F}}$ $\newcommand \E{\mathcal{E}}$ $\newcommand \p{\mathfrak{p}}$ $\newcommand \h{\mathcal{O}}$

So, let $E/\Q$ be an elliptic curve and let $\E/\Z$ be a minimal Weierstrass model (which exists since $\Q$ has class number $1$). Fix an ancillary prime $\ell$. We claim that if $p$ is a prime satisfying the following:

• $p\geqslant 5$
• $p\nmid\text{Cond}(E)$ (where $\text{Cond}(E)$ is the conductor of $E$)
• $p$ splits completely in $\mathcal{O}_K$ where $K:=\mathbb{Q}(E[\ell](\overline{\mathbb{Q}}))$

then $E$ has good ordinary reduction at $p$. This proves the claim since there are infinitely many $p$ satisfying these conditions (there are infinitely many primes that split in $K$ by, applying a hammer to a fly, Cebotarev density and there are only finitely many primes not satisfying the first two conditions).

To see this, let $\p$ be any prime $\mathcal{O}_K$ dividing $p$. Note that $E_K$ has good reduction at $\p$ with, in fact, a model over $(\h_K)_\p$ given by $\E_{(\h_K)_\p}$. Now, note that basic theory (smooth proper base change if we're applying hammers to flies again :P) says that we have an isomorphism $T_\ell E_{K_\p}\cong T_\ell \E_{k(\p)}$ (with $k(\p):= \h_K/\p$) intertwining the surjection $G_{K_\p}\to G_{k(\p)}$. But, note that since $p$ split in $K$ we have that $k(\p)=\F_p$. So, let $\sigma\in G_{K_\p}$ lift $\text{Frob}_p\in G_{\F_p}$. Note then that by our isomorphism if Tate modules we have that $\text{tr}(\sigma)=\text{tr}(\text{Frob}_p)$ as elements of $\mathbb{Z}$. But, note that

$$\text{tr}(\sigma)\mod \ell=\text{tr}(\sigma \mod \ell)=\text{tr}(\sigma\mid_{E[\ell]})=\text{tr}(\text{id})=2$$ and thus, evidently, we may conclude that $\text{tr}(\text{Frob}_p)\ne 0$. Since $p\geqslant 5$ this implies that $\E_{\F_p}$ is ordinary as claimed.