$2$ and is unramified away from $2$. However, using explicit class field theory (as implemented in [@PARI2]), we may check that none of the nine fields $K$ has an extension $L/K$ with Galois group $\operatorname{GL}_2(\F_2)\cong S_3$ which is unramified away from $2$ and has quadratic subfield $K(\sqrt{\varepsilon})$. Hence $E[2]$ is not absolutely irreducible, establishing the claim for $\ell=2$. Now either $E[2]$ is reducible, in which case Corollary \[coro:2torsion-odddiscriminant\] shows that up to $2$-isogeny $v_{\id{p}}({{\mathcal{D}}}(E))=1$; or $E[2]$ is irreducible (but not absolutely irreducible) in which case Theorem \[thm:absred2\] implies that $v_{\id{p}}({{\mathcal{D}}}(E))=2$.

It remains to prove the claim for odd primes $\ell$ which divide $v_{\id{p}}({{\mathcal{D}}}(E))$. We consider first the case $\ell=3$, where as with $\ell=2$ an elementary argument establishes the claim unconditionally. If $K\ne\Q(\sqrt{-3})$, then every unit in $K$ is a cube, so the hypothesis that $3\mid v_{\id{p}}({{\mathcal{D}}}(E))$ implies that ${{\mathcal{D}}}(E)$ is a cube in $K$. In case $K=\Q(\sqrt{-3})$ we can only say that $\Delta(E)$ is a unit times a cube, and we may have $K(\sqrt[3]{{{\mathcal{D}}}(E)})=\Q(\zeta_9)$. In general, $K(\sqrt[3]{{{\mathcal{D}}}(E)})$ is the subfield of $K(E[3])$ cut out by the Sylow $2$-subgroup, which has index $3$ and order $16$, in $\operatorname{GL}_2(\F_3)$; this subgroup is the normaliser of the non-split Cartan subgroup. Hence the field 