Proving that an elliptic curve does not have CM

$$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} = \Bbb{Q} \quad\text{or}\quad \Bbb{Q}(\sqrt{-d})\;(\mathrm{CM})$$

$\operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} \hookrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \hookleftarrow \Bbb{Q}(\operatorname{Frob}_p)$.
$a_p := \operatorname{Tr}(\operatorname{Frob})$
$p\nmid a_p \Longleftrightarrow \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}}$ is a quadratic field
If $E$ has CM by $\Bbb{Q}(\sqrt{-d})$, then
$$\begin{aligned} a_p \equiv 0 \bmod p &\Longleftrightarrow p\text{ inert or ramified in }\Bbb{Q}(\sqrt{-d}) \\ &\Longleftrightarrow \operatorname{End}_{\Bbb{Q}} E^{\mathrm{al}} \not\simeq \operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \end{aligned}$$
If $E$ is non-CM, then $\operatorname{End}_{\Bbb{Q}} E_p^{\mathrm{al}} \cap \operatorname{End}_{\Bbb{Q}} E_q^{\mathrm{al}} \simeq \Bbb{Q}$ with prob. 1;
and we expect $\operatorname{Prob}(a_p \equiv 0 \bmod p) \sim 1/\sqrt{p}$

All primes considered are primes of good reduction. Here $a_p=p+1-\#E(\Bbb{F}_p)$ and $c_p(t)=t^2-a_pt+p$.

At an ordinary prime, $\operatorname{End}_{\Bbb{Q}}E_p^{\mathrm{al}}\simeq\Bbb{Q}[t]/(c_p(t))$; at a supersingular prime it is a quaternion algebra.

There are many ways to distinguish the two cases; for example, we could compute the $j$-invariant.

By looking at the number of points mod $p$, we may deduce the endomorphism algebra of an elliptic curve.

$\operatorname{End}_{\Bbb{Q}}$ means endomorphisms tensored with $\Bbb{Q}$. If two ordinary good reductions have non-isomorphic quadratic endomorphism fields, then the geometric endomorphism algebra in characteristic zero is $\Bbb{Q}$. For a fixed non-CM curve, Lang-Trotter predicts a count $C_E\sqrt B/\log B$ for good primes with $a_p=0$; the constant depends on the curve.