# Monday March 30th

We'll begin by recalling some properties that Tate curves have.

## Tate Curves

Let $(K, \abs{\wait})$ be a complete nonarchimedean normal field and $q\in K$ with $0 < \abs{q} < 1$ so $\abs{j(E_q)} = {1 \over \abs q} > 1$.
Let $R$ be its valuation ring.
Then

**Remark (a)**:

There exists an elliptic curve $E_1/K$ and a $K\dash$analytic group isomorphism
\begin{align*}
\phi_K: K\units/\gens{q} &\mapsvia{\cong} E_q(K)
,\end{align*}
a *multiplicative uniformization*.

For all finite $L/K$, the norm extends uniquely over $L$, and there is a compatibility in the following sense

\begin{tikzcd}
K\units / \gens{q} \ar[d, hook] \ar[r, "\cong \phi_K"] & E_q(K) \ar[d, hook] \\
K\units/ \gens{q} \ar[r, "\cong \phi_L"]               & E_q(L)
\end{tikzcd}


And taking the $\directlim$ over separable $L/K$ yields
\begin{align*}
\phi: (K\sep)\units/\gens{q} \mapsvia{\cong} E_q(K\sep)
,\end{align*}
which is an isomorphism of $g_K\dash$modules where $g_K = \aut(K\sep/K)$.

Note that in this case, $E_q(K)$ will not have potentially good reduction, and in fact we precisely get all curves with split multiplicative reduction.

**Remark (b)**:

Suppose $*K, \cdot)$ is discretely valued.
Then $E_1$ has split multiplicative reduction and the component group is cyclic of order $v(q)$.

Example
: If $\pi$ is a uniformizing element for $K$ and $q = \pi^n$ then $E_q(K)[\tors] \cong K\units / (\pi^n)$ and the class of $\pi$ has order $n$.
  So the torsion can be arbitrary large, but we can understand this quotient using Kummer theory.

Suppose that $K$ is a complete discrete valuation field with perfect residue field $k$.

- If $\abs{j(E)} \leq 1$ then $E$ has potentially good reduction.
  - Note that we just need to pass to an extension of degree 2 or 4.

- If $\abs{j(E)} > 1$ then $E$ has **no** potentially good reduction and is a unique quadratic twist of a *unique* Tate curve.

The latter is a strong statement: we'll attached a unique $q$ and $E_q$ to $E$ such that $j(E) = j(E_q)$.
Thus these $E$ will have explicit identifications, and since we understand the former case of potentially good reduction somewhat well, we'll be in good shape.

Theorem
:   If $(K, v)$ is a complete discrete valuation field with perfect residue field $k$.
    Let $E/K$ be an elliptic curve suchj that $\abs{j(E)} < 0$ (so it's a quadratic twist of a Tate curve).
    Let $\ell \geq 3$ where $\ell$ does not divide $\ch K v(j(E))$.

    > If $\ch K = 0$ and $\ch k = p$, then $\ell = p$ is allowed.

    Let $I_K$ be the inertia group of $K$ (noting that we're over a local field so $g_K$ is local), a subgroup of the absolute Galois group of $K$, i.e.
    $1 \to I_K \to g_K \to g_k \to 1$.

    Then there exists a basis $e_1, e_2$ of $E[\ell](K\sep)$ and $\sigma \in I_K$ such that
    \begin{align*}
    \phi_\ell(\sigma) =
    \begin{bmatrix}
    1 & 1 \
    0 & 1
    \end{bmatrix}
    ,\end{align*}
    i.e. $e_1 \mapsto e_1$ and $e_2 \mapsto e_1 + e_2$.

Note that this matrix is in Jordan form.
It is unipotent, and every unipotent matrix in a field of $\ch = \ell$ has order $\ell$.
Thus we have a very particular element of order $\ell$ in the Galois group.


### Proof

Proof
:   We can replace $K$ with any algebraic extension $L/K$ such that $\phi(L/K)$ is finite and prime to $\ell$.
    In particular, we can assume that $L$ contains the maximal unramified extension $L \supset K^\text{unr}$, which occurs iff $k = \bar k$ is algebraically closed.
    Thus we can take Galois groups instead of inertia groups.

    We may also assume that $E$ is isomorphic to a Tate curve over $L$, i.e. $E \cong_L E_q$ by replacing $L$ with a quadratic extension, and that $L$ has a primitive $\ell$th root of unit $\zeta_\ell$.
    This is because the ramification index is equal to the degree $\ell-1$ which is still prime to $\ell$.
    Thus we can use the Tate parametrization and Kummer theory.

    Let $Q \in L\sep$ such that $Q^\ell = q$, i.e. take an $\ell$th root of the Tate parameter.
    By classical Kummer theory, $L(Q) / L$ is cyclic of degree $\ell$, so let $\sigma \in \aut(L(Q) / K)$ be a generator.
    Then $\zeta \definedas \sigma(Q) / Q$ is nontrivial and thus a primitive $\ell$th root of unity.

    Note that $E_1(L\sep) \cong (L\sep)\units / \gens{q}$.
    So we can construct a basis by taking $e_1 = \zeta$ and $e_2 = Q$.

Corollary
:   For $K, E/K, \ell$ as in the theorem, so $\abs{j(E)} > 1$ and $\ell$ does not divide $\ch K v(j(E))$
    Then $E/K$ has an **unique** $K\dash$rational $\ell\dash$isogeny.

Proof
:   There exist a quadratic twist $E_q$ of $E$ that is a Tate curve since the reduction is not potentially good, so $j\neq 0, 1728$, so the only twists are quadratic.
    Thus WLOG we can assume $E \cong E_q$.

    > Next time: to each quadratic twist we'll associate a quadratic character $\chi$, to which we associate a Galois representation.

    If we take $H \definedas \theset{\mu_\ell \gens{q}} < \GG_m/\gens{q}$ the group of roots of $\ell$th roots of unity, which is a $g_K\dash$stable subgroup.
    We also have $\sigma \in I_K$ with
    $?\phi(\sigma) = \begin{bmatrix}
    1 & 1 \
    0 & 1
    \end{bmatrix}$
    and the cyclic subgroups of $E[\ell]$  are generated by $[1, 0]$ and $[x, 1]$ where $x\in ?$.