# Wednesday April 8th

Exercise
:   If $\ch K \neq 2,3$ and $E_1, E_2/ K$ are elliptic curves, set $L = K(\sqrt{d})$.
    If $E_1 \not\cong_K E_2$ but $E_1 \cong_{K(\QQ)} E_2$, then $E_2 \cong E_1^d$ (so $E_2$ is a quadratic twist of $E_1$).

> Note that we take the twist with respect to the inversion map.

This is not generally true for quadratic twists of $(V, \iota)$.
It depends on $\aut(V)$.
If $\aut(E) \theset{\pm 1}$, this holds for $j =  0 ,1728$.
Then for $E/K$ an elliptic curve, $J(E) = \theset{ E'_K \suchthat E' \cong_{K(\sqrt d)} E}$, which is in correspondence with $H^1_{\text{gal}} (\Aut(K(\sqrt d) / K), \aut(E) )$.
This is given by $H^1(\theset{1, \sigma}, I)$, which reduces to a hom set.

Last time we justified the argument that every ramified quadratic twist of a Tate curve, resulting in an elliptic curve with additive reduction.

Question:

Given $E_1, E_2/K$, how we can decide if $E_1 \cong_K E_2$ if $j\neq 0,1728$?

1. Check if they are isomorphic over $\bar K \iff j(E_1) = j(E_2)$.
    If not, stop, otherwise if so, they are quadratic twists of each other.
2. Use an invariant (see Silverman) $\gamma(E)$; then $E_1 \cong E_2 \iff j(E_1) = j(E_2)$ and $\gamma(E_1) = \gamma(E_2)$.

Proposition (An Application)
:   Let $F$ be a field, $F \injects \CC$, $E/F$ an elliptic curve with complex multiplication.
    Then $\endo(E) = \OO$ (taken over the algebraic closure) is some order in an imaginary quadratic field $K  = \QQ(\sqrt{\Delta})$, and $j(E) \in \bar{\ZZ}$, the algebraic integers.
    Then $\QQ(j(E))$ is a number field $L$, and for all $\mfp \normal \ZZ_\ell$, $E$ has potential good reduction at $\mfp$ (i.e. the $j\dash$invariant is $p\dash$adically integral).

> We more or less need to show that complex multiplication and "being a Tate curve" are mutually exclusive.

Proof (Sketch)
:   Let $K = \endo(E) \tensor_\ZZ \QQ$ and $L\definedas K(j(E))$.
    Then there are two cases:

    1. If $j(E)$ is algebraic then $L$ is a number field.
    2. Otherwise $j(E)$ is transcendental, and thinking of $j$ as an indeterminate we have $L \cong K(j)$, a rational function field.

    **Case 1:**
    Take $\mfp \in \ZZ_L$ and $E/L_\mfp$ a $p\dash$adic field.
    We want to show that $v_\mfp (j(E)) \geq 0$.
    Assume the proposition doesn't hold, toward a contradiction.
    Then after a quadratic base extension $M_\mfp / L_\mfp$, the curve $E/M_\mfp$ is a Tate curve.

    \

    For $\ell \gg 0$, define $N_\ell \definedas M_\mfp (E[\ell])$ and there is a map $\theta_\ell: g_{M_\mfp} \to \GL(2, \ZZ/\ell\ZZ)$.
    Note that $N_\ell / M_\mfp$ is Galois, and $\theta_\ell \definedas \Aut(N_\ell / M_\mfp) \cong \rho_\ell(g_{M_\mfp})$.

    \

    Because $E / M_\mfp$ has $\mu_\mfp\dash$ central (?) complex multiplication, $G_\ell$ is commutative and $\rho_\ell: g_{M_\mfp} \to \GL(1, \OO/\ell\OO)$ which is commutative.
    Since it's a Tate curve, we have an explicit description of the $\ell\dash$torsion coming from Kummer theory.
    This buys us the fact that $\mu_\mfp(E[\ell]) = M_\mfp(\mu_\ell, q^{1/\ell})$, which is just the splitting field of $t^\ell - q$ over $M_\mfp$ by "non-abelian Kummer Theory".

    > Qual alert: computing with these splitting fields shows up often on Algebra quals!

    Choosing $\ell \gg 0, \ell > 2$, and $\ell$ not dividing $v_\mfp(q)$ makes this splitting field irreducible.
    Thus the galois group is given by $G_\ell \cong \ZZ/\ell \ZZ \semidirect \chi_\ell(g_{M_\mfp})$ where the RHS is the cyclotomic character sitting inside of $\qty{\ZZ/\ell\ZZ}\units$, with the action given by multiplication.
    (See Lang's Algebra or Pete's Field Theory notes.)

    \

    What is the abelianization?
    We can compute $\#[G_\mfp: G_\ell] = \# \chi_\rho(g_{M_\mfp})$, and eventually this reduces to the case of a finite field $\FF_q$  and considering $[\FF(?): \FF_q]$.

    **Case 2**:
    Suppose $j(E)$ is transcendental over $K$, then $K(j)$ is a rational function field equal to $ff(K[j])$ with $v_\infty \definedas -\deg(P(j))$ and $v_\infty(j) = -1$.
    Letting $K_\infty$ be the completion of $K(j)$ with respect to $v_\infty$, then $K_\infty \cong K((1/j))$ with residue field $K = \QQ(\Delta)$.
    Now $E/K_\infty$ has complex multiplication, $v_\infty(j(E)) < 0$, and we again have a quadratic extension and $E$ is a Tate curve with complex multiplication.
    Using the fact that a number field only contains finitely many roots of unity, the same argument works.

There is a higher-genus notion of a Mumford curve which are analogously $p\dash$adically uniformized (much in the same way $\HH$ uniformizes) by lattices in $\GG_m^g$.