# Tate models and subgroups of type$(m,m)$ - re Silverman, Hindry 88

I need help in understanding a passage in a paper by Hindry and Silverman, "The Canonical Height and Integral Points on Elliptic Curves". (re. page 439)

Let $E(K)$ be an elliptic curve with multiplicative reduction at a valuation $\nu$, and K a function field of zero characteristic. The torsion points of this curve are supposed to have the form $E(K)_\mathrm{tors}$ $\cong$ $\dfrac{\mathbb{Z}}{m\mathbb{Z}} \times \dfrac{\mathbb{Z}}{n\mathbb{Z}}$.

They select a point P so that it kills the right side of the product, i.e., $\dfrac{E(K)_\mathrm{tors}}{\langle P\rangle} \cong \dfrac{\mathbb{Z}}{m\mathbb{Z}} \times \dfrac{\mathbb{Z}}{m\mathbb{Z}}$.

Then, they use the isogeny theorem to construct another elliptic curve $E'$ such that $E$ is isogenous with $E'$.

Next, they construct the Tate models for these curves: $E(K_\nu) \cong \dfrac{K_\nu^{*}}{\langle q\rangle }$ and $E'(K_\nu) \cong \dfrac{K_\nu^{*}}{\langle q'\rangle}$, where $K_\nu$ is the completion at $\nu$, $q$ and $q'$ have the property that $\nu(q) > 1$ and $\nu(q') > 1$.

Up to this point everything is fine. However, in the next line they find that there are elements $q_1$ and $q'_1$ such that $q = q_1^m$ and $q' = q'_1^m$; the reason being that there is a subgroup of type $(m,m)$ in both curves.

At first, I thought that one could see this by elementary group theory, but that only gives that $q^{k} = q_1^m$ for some power $k$ of $q$, and I could not continue from this point. Am I missing something? Can anyone help me understand this?

cheers

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TeX is not so crude that you need to write $<p>$. I changed it to $\langle p\rangle$. –  Michael Hardy Aug 16 '12 at 3:18
Fixed a choice of $q^{1/m}$ and a primitive $m$-th root $\xi_m$ in an algebraic closure $\bar{K}_\nu$ of $K_\nu$.
When you have a parametrization $E(K_ν)≅K_ν^*/⟨q⟩$, the $m$-torsion points in $E({K}_\nu)$ are given by $z\in K_\nu$ such that $z^m=q^k$ for some $k\in \mathbb Z$. So $z=(q^{1/m}){}^k \xi_m^r$ for some $k, r\in\mathbb Z$. There are at most $m^2$ solutions. It this bound is reached, then $z\in K_\nu$ for all $k,r$, so $q^{1/m}\in K_\nu$.