I found a reference which has the required result with proof (which I've copied here). I claim no originality.

**Lemma (38.9.9):** Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $[d]:A\to A$ is étale if and only if $d$ is invertible in $k$.

**Proof:** Observe that $[d](x+y)=[d](x)+[d](y)$. Since translation by a point is an automorphism of $A$, we see that the set of points where $[d]:A\to A$ is étale is either empty or equal to $A$ (some details omitted). Thus it suffices to check whether $[d]$ is étale at the unit $e\in A(k)$. Since we know that $[d]$ is finite locally free (Lemma 38.9.8) to see that it is étale at $e$ is equivalent to proving that $d[d]:T_{A/k,e}\to T_{A/k,e}$ is injective. See Varieties, Lemma 32.14.8 and Morphisms, Lemma 28.36.16. By Lemma 38.6.4 we see that $d[d]$ is given by multiplication by $d$ on $T_{A/k,e}$.

Also, here (proposition 3.7) shows that for abelian varieties, an isogeny being étale is equivalent to it being separable is equivalent to it having étale kernel (meaning, a kernel which is étale as a subgroup scheme). So, it turns out that inseparability *does* have bearing on being étale, at least in the case of isogenies between abelian varieties.

An elliptic curve over $\mathbf Q$ cannot have complex multiplication (defined over $\mathbf Q$). It's possible for a rational elliptic curve to have extra endomorphisms, but these will only be defined over a finite extension.

But let's instead take an elliptic curve $E$ over a number field $K$ with complex multiplication. Then the associated Galois representation is *reducible**!

Indeed, if
$$\rho_{E,\ell}:G_K\to \mathrm{GL}_2(\overline{\mathbf Q}_\ell)$$
is the associated $\ell$-adic representation, then its not too hard to check that
$$\mathrm{End}(E)\otimes_\mathbf Z\overline{\mathbf Q}_\ell\hookrightarrow\mathrm{End}(\rho_{E,\ell}).$$

In particular, if $\mathrm{End}(E)\ne \mathbf Z$, then $\mathrm{End}(\rho_{E,\ell})$ is not a field, so $\rho_{E,\ell}$ is reducible. Its subrepresentations are one dimensional Galois representations, which by class field theory, correspond to the Grossencharacters of $E$.

In fact the above map is an isomorphism (by Faltings' theorem). So if $\mathrm{End}(E) = \mathbf Z$, then $\mathrm{End}(\rho_{E,\ell})$ is a field, so $\rho_{E,\ell}$ is irreducible.

If $E$ does not have complex multiplcation over $K$, but obtains extra endomorphisms over a finite extension, then the above argument shows that $\rho_{E, \ell}$ is irreducible. However, $\rho_{E, \ell}$ will not be surjective. By Mackey theory, since $\rho_{E, \ell}$ is irreducible, but $\rho_{E, \ell}|_{G_L}$ is reducible for some $L$, we find that $\rho_{E, \ell}$ is induced from a character of a quadratic extension. In particular, its image cannot be $\mathrm{GL}_2(\mathbf Z_\ell)$.

*By reducible, I mean that it becomes reducible over the algebraic closure $\overline{\mathbf Q}_\ell$. It may still be irreducible over $\mathbf Z_\ell$.

## Best Answer

Okay it is possible to deduce the trace bound from just what we have here. For any endomorphism $\alpha$, we see that the subring generated by $1$ and $\alpha$ is an order in some quadratic extension of $\mathbb{Q}$, since it is a commutative subring of endomorphisms of the Tate module generated by one element. Then the degree equals determinant result gives us that every element of this subring has positive norm, so this is in fact an order in an imaginary quadratic extension of $\mathbb{Q}$. This then gives our result that any polynomial has complex conjugate roots, and we’re done.