$\F_{\ell^2}$) or absolutely irreducible. The hypothesis in the theorem excludes only the absolutely irreducible case. Over $\Q$, or any number field with at least one real place, the second case cannot occur, due to the action of complex conjugation: any invariant line over $\F_{\ell^2}$ must be defined over $\F_{\ell}$. In our situation, we need to consider both the reducible and the absolutely reducible cases separately.

$\bullet$ [**Reducible case:**]{} without loss of generality, the residual representation is of the form $$\label{eq:reducibledecomposition}
      \overline{\rho_{E,\ell}} \simeq \left(\begin{smallmatrix} \theta_1 & *\\ 0
    & \theta_2\end{smallmatrix} \right),$$ for $\theta_i$ characters of $\operatorname{Gal}(\overline{K}/K)$ such that $\theta_1 \theta_2 = {\chi_{\ell}}$ (the cyclotomic character). Since $E$ is semistable, the conductors of the characters $\theta_i$ are supported in $\ell$ (see [@Kraus Lemme 1]). Also since $K$ has class number $1$, the only unramified character is the trivial character. Hence we must show that at least one of the characters is unramified, since $\theta_1$ trivial implies that $E$ has a point of order $\ell$ while if $\theta_2$ is trivial then the isogenous curve has such a point. If $\ell$ is unramified in $K$, then the result follows from the proof of [@Kraus Corollaire 1] (pages 249-250), as we now explain.

If $\ell$ 