Consider the elliptic curve given by $y^2=x^3-ax/p^2+b/p^3$ for $p \geq 5$, where $a,b \in \mathbb{Z}_p^{\times}$ are such that $p|4a^3+27b^2$ and $4a^3+27b^2 \neq 0$. For instance, $a=-3,b=2+p$.
Emerton’s answer is, I think, on a less elementary level than that: it’s addressing the case of what $p$-torsion becomes mod $p$. This case is rather simpler: the kernel of the $n$-torsion extends to a smooth finite scheme $G$ over $\mathbb{Z}_{(p)}$. By base changing to the ring of integers of a local field $K$ (with residue field $k$), we know that $G$ is finite smooth free of rank $m$ [in this case $m=n^2$] coprime to $p$ over $O_K$.
Let $C \subset G$ be a closed open subset and let $A=O(C)$ (free $O_K$-algebra of finite rank, with $A \otimes k$ reduced), we want to show that $A$, $A \otimes K$, and $A \otimes k$ have the same idempotents (which will imply the result – as I’ll explain afterwards).
The bijection from the idempotents of $A$ to the idempotents of $A \otimes k$ is well-known (it’s basically Hensel). The inclusion from the idempotents of $A$ to those of $A \otimes K$ is clearly well-defined and injective. But if $e \in A\otimes K$ is idempotent, then for some $a \in O_K$ (a power of the uniformizer, chosen to be minimal – suppose for the sake of contradiction that $a$ is noninvertible) $e’=ae \in A$ and $e’^2=ae’$. But this means that $e’^2$ vanishes in $A \otimes k$ so $e’$ was already divisible in $A$ by the uniformizer, so we could have chosen $a$ with a smaller valuation!
Why is this enough? Well, consider the residue disk $C$ of the unit point in the special fiber of $G$: $G$ is finite as a topological space, so it is the finite reunion of closures of points on the generic fiber, so $C$ is closed. But $G$ is finite as a topological space, so $C$ is constructible. Since $C$ is of course stable under generization, so $C$ is open.
So $C$ is a closed open subscheme of $G$ with its special fiber reduced to a point, so by the above its generic fiber must be connected. But $C$ is smooth over $O_K$ so its generic fiber is the spectrum of a finite reduced, connected $K$-algebra with a $K$-point (the unit), so it must be $K$, ie reduced to a point, so we are done.
Best Answer
An elliptic curve $E$ over $\overline{\mathbb{Q}}$ can be defined over $\mathbb{Q}$ if and only if its j-invariant $j(E)$ is a rational number. More precisely, $E$ can be defined over a number field $K\subset \overline{\mathbb{Q}}$ if and only if $j(E)\in K$.
As Mathmo123 points out above, you can find many elliptic curves over $\mathbb{Q}(i)$ which can not be defined over $\mathbb{Q}$:
https://www.lmfdb.org/EllipticCurve/2.0.4.1/?field=2.0.4.1&Qcurves=non-Q-curve
But, it is actually quite easy to find examples yourself. Consider the elliptic curve $y^2 = x^3 + Ax + 1$. For which $A$ is the $j$-invariant a rational number?
Additional remark: If $\lambda\in \overline{\mathbb{Q}}\setminus \{0,1]\}$, then the Legendre elliptic curve $E_{\lambda}$ is defind by $y^2 = x(x-1)(x-\lambda)$. If $\lambda$ lies in $K$, then it is obviously defined over $K$.
The $j$-invariant of $E_{\lambda}$ is given by
$$ j(E_{\lambda}) = \frac{256(1-\lambda(1-\lambda))^3}{(\lambda(1-\lambda))^2} = \frac{256(1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}. $$ So, $E_{\lambda}$ is defined over $\mathbb{Q}$ if and only if there is a rational number $q$ such that $$q\lambda^2(1-\lambda)^2 - 256(1-\lambda+\lambda^2)^3 =0. $$ So, choosing $\lambda $ (for example) to be an eleventh root of $2$ obviously gives a non-rational $j$-invariant. (Because this $\lambda$ does not satisfy a polynomial relation over $\mathbb{Q}$ of degree smaller than $11$.) So, this is a relatively easy way of writing down elliptic curves which can not be defined over the rationals.