Let p be a prime and let K be a field containing the p'th roots of unity. Let E be an elliptic curve over K. We consider the the moduli problem $Y_E(p)$, which sends L to set of elliptic curves F/L, and symplectic isomorphisms $\phi:E[p] \rightarrow F[p]$. We know that this moduli problem is representable by a curve over $K$, and we let the compactification of this curve be $X_E(p)$. We know $X_E(p)$ is a twist of $X(p)$. Similarly, we can construct $X_E(p^2)$, and we see that $X_E(p^2)$ is a normal cover of $X_E(p)$. I think the Galois group of $X_E(p^2)/X_E(p)$ is $(Z/pZ)^3$. If that is the case, then given any K point of $X_E(p)$, we can look at fiber over this point. This fiber is defined over K, hence it will define a field extension of K, with Galois group a subset of $(Z/pZ)^3$.

This means, if we have E and F defined over K, with $E[p] \equiv F[p]$, then we should be able to construct a cyclic extension of K of order p. What is that extension?


The answer is that in fact this construction does not produce cyclic extensions! The problem is that $X_E(p^2) \to X_E(p)$ is not generically Galois; it is so only after extension of the ground field.

Here is a more detailed explanation. Assume that $p \ge 3$ and that $p \nmid \operatorname{char} K$. Then $Y_E(p^2) \to Y_E(p)$ is a torsor not under $(\mathbb{Z}/p\mathbb{Z})^3$, but under the étale group scheme $G$ corresponding to the Galois module of $\mathbb{F}_p$-linear endomorphisms $g \colon E[p] \to E[p]$ of trace zero. Explicitly, $g \in G(\overline{K})$ maps $(F,\Phi) \in Y_E(p^2)(\overline{K})$ to $(F,\Phi')$ where $\Phi'(x):=\Phi(x+g(px))$. (Any $\Phi'\colon E[p^2] \to F[p^2]$ with the same restriction to $E[p]$ as $\Phi$ arises from some $g \in \operatorname{End} E[p]$ in this way, and the trace-zero condition is what guarantees that $\Phi'$ is symplectic.) Finally, this Galois module is typically irreducible, in which case it does not have $\mathbb{Z}/p\mathbb{Z}$ as a quotient.

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