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I have a couple of questions about this answer by Noam D. Elkies showing that there exist no elliptic curve $E$ over $\mathbb{C}[t, t^{-1}]$ having nonconstant $j_E$-invariant.

The strategy is to consider $G:={\rm Gal}\bigl(\overline{{\mathbb C}(t)}\big/{\mathbb C}(t)\bigr)$ on the group $E[p]$-action on $p$-torsion points $E[p]$ of such putative elliptic curve $E / {\mathbb C}(t)$ that has good reduction at all $t \neq 0, \infty$.

What I not understand is the following statement:

[...]On the other hand, once $p$ exceeds the order of a pole of the $j$-invariant $j_E^{\phantom.}$, the image of Galois includes a $p$-cycle ramified above that pole. [...]

Why that's the case? Could somebody elaborate this argument on precise connection of pole orders of $j_E$ at certain points and existence of such $p$-cycle in the image under the induced $G \to \text{Aut}_{\mathbb C(t)}(E[p])$ & its ramification behavior with details?

The flavour of the statement reminds me loosely on Neron-Ogg's, but here the $j$-invariant is not integral due to presence of poles, so I not know which result is here going to be invoked.

Going a step back, what does it mean that mean that a $p$-cycle (...more generally a subgroup) inside the Aut group ramifies about some point/prime ideal? I know what it means to be "(un)ramified" as Galois representation, or as a morphism of schemes, but what does it mean that a $p$- cycle (or a subgroup) inside Aut group ramifies as stated above?
Or does Elkies consider here $ \text{Aut}_{\mathbb C(t)}(E[p])$ & resp its subgroups "en passant" as schemes over $\Bbb C[t]$ interpreting the discussed $p$-cycle to ramified over a prime/ point in this sense?

And if that's the case, how it this related to the poles of $j$- function? The way Noam Elkies stated this, suggests that there is a for specialists a well known "big" result relating poles of $j$-function with ramification behavior of subfields of the $p$-torsion field extension, from which the quoted claim follows immediately. But which is it?

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Let me expand that sentence, which may hint at the confusion in this and the subsequent question.

Let $K=\mathbb{C}(t)$ and see the ring $R=\mathbb{C}[t]$ as a Dedekind ring. Let $v$ be a place of $R$ such that the $j$-invariant has a pole at $v$. Suppose $p$ is a prime such that $p>-v(j)$.

For a place $w$ in $L=K\bigl(E[p]\bigr)$ above $v$, there is the decomposition group $D_{w/v}$ inside the Galois group of $L/K$. The curve $E$ over the completion $K_v$ has multiplicative reduction and the curve is isomorphic to a Tate curve $E(L_w) \cong L_w^{\times}/q^{\mathbb{Z}}$ for $q=\tfrac{1}{j}+\cdots\in K_w$ of valuation $-v(j)$. The $p$-torsion points are generated by a primitive $p$-th root of unity $\zeta\in\mathbb{C}^\times \subset L_w^{\times}$ and a $p$-th root of $q$. Since $p>v(q)$, this prime $p$ cannot divide $v(q)$ and hence $L_w$ is simply $K_v\bigl(\sqrt[p]{q}\bigr)$ which is a degree $p$ extension and $D_{w/v}$ is cyclic of order $p$. In terms of the Galois representation, the image of $D_{w/v}$ in $\operatorname{GL}_2(\mathbb{F}_p)$ is $\bigl(\begin{smallmatrix} 1 & * \\ 0 & 1 \end{smallmatrix}\bigr)$. That is your element of order $p$. The extension is clearly totally ramified, i.e. $D_{w/v} = I_{w/v}$ is the inertia group.

I recommend appendix A.1 in Serre's "Abelian $\ell$-adic representations and elliptic curve" and Silverman II. There $K$ is a number field and $K_v$ is a $p$-adic field, the above works just as well here.

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  • $\begingroup$ just to check if I understood the logic of the argumentation (even though this might be a triviality): Isn't $L_w$ by construction already $K_v\bigl(\sqrt[p]{q}\bigr)$ independently of additional condition $p >v(q)$? Or does the latter condition only serve to assure that that the extension is non trivial? $\endgroup$
    – user267839
    Commented Jun 14 at 19:32
  • $\begingroup$ Sure, but if $p$ divides $v(q)$, we could have that $q$ is a $p$-th power in $K_v$ in which case $K_v=L_w$ is a trivial extension. The important part is that the extension is non-trivial of degree $p$. $\endgroup$ Commented Jun 14 at 20:50

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