Galois module $E[\ell]$ is unramified away from $\ell$, and also that its determinant is the $\ell$th cyclotomic character. We claim that the hypotheses of the theorem imply that for every prime number $\ell$ dividing $v_{\id{p}}({{\mathcal{D}}}(E))$, the module *$E[\ell]$ is not absolutely irreducible*.

Assume the claim. For odd $\ell$, Theorem \[thm:Kraus\] below implies that either $E$, or an $\ell$-isogenous curve, has a rational torsion point of order $\ell$. In Section \[smallimage\] all curves of odd prime conductor with an $\ell$-torsion point are computed (see Table \[table:oddtorsioncurves\]) with the exception of curves with a rational $3$-torsion point over $\Q(\sqrt{-3})$. It is easy to verify from Table \[table:oddtorsioncurves\] that all curves of prime conductor listed there have (up to an $\ell$-isogeny) discriminant with valuation prime to $\ell$. Finally, Theorem \[thm:3-torsionspecialcase\] proves that any curve with a rational $3$-torsion point over $\Q(\sqrt{-3})$ either has discriminant valuation prime to $3$, or a $3$-isogenous curve does. Hence, up to isogeny, the claim (for all odd $\ell$) implies that $v_{\id{p}}({{\mathcal{D}}}(E))$ has no odd prime factors.

Suppose that $v_{\id{p}}({{\mathcal{D}}}(E))$ is even. The hypothesis implies that ${{\mathcal{D}}}(E)=\varepsilon\pi^2$ for some $\varepsilon\in{{\mathcal{O}}}_K^*$ and $\pi\in{{\mathcal{O}}}_K$. If $E[2]$ is absolutely irreducible, then ${{\mathcal{D}}}(E)$ is not a square, and the extension $K(\sqrt{\varepsilon})/K$ has degree 