are equal up to squares, so that the fields $K(P)$ and $K(P')$ are the same.

Let $A$ be the surface $E\times E'$ and let $\lambda$ be the product principal polarization on $A$. Let $\alpha$ be the automorphism of $A$ that sends a point $(U,V)$ to $(U, V+\psi(U))$. Let $G$ be the $K$-defined subgroup $$G = \{(O,O), (P, P'), (Q, Q'), (R, R')\}$$ of $A[2]$ and let $G' = \alpha(G)$, so that $G'$ is the $K$-defined subgroup $$G'= \{(O,O), (P, R'), (Q, Q'), (R, P')\}.$$ Let $(B,\mu)$ and $(B,\mu')$ be the principally-polarized surfaces obtained from $G$ and $G'$ as in Section \[S-curves\]. The polarizations $\mu$ and $\mu'$ will be indecomposable except in unusual circumstances, so there will usually be curves $C$ and $C'$ whose polarized Jacobians are isomorphic to $(B,\mu)$ and $(B,\mu')$, respectively. If $E$ and $E'$ are given to us by explicit equations, then $C$ and $C'$ can also be given by explicit equations — see [@HLP §3.2], where the unusual circumstances are also explained.

To make this outline explicit and to prove Theorem \[T-nonsimple2\] we must start with an explicit $2$-isogeny $\psi:E\to E'$, where the discriminants of $E$ and $E'$ are equal up to squares. Let $t$ be an 