# $\ell$-adic Representations of Elliptic Curves If you're somehow seeing this right now, look away. It's not finished, and I'm not sure when/if it will be.

The title of this post is a bit of a misnomer. We won’t be talking about representations of elliptic curves, but about representations attached to (associated with?) elliptic curves. In particular, the goal of this post is to prove that given an elliptic curve $E/\Q$ defined over the rationals, and a prime $\l$, the $\l$-adic representation

is irreducible where $G_{\Q}=\Gal(\Qbar/\Q)$ is the absolute Galois group of $\Q$ and $T_\l(E)=\invlim E[\l^n]$ is $E$’s $\l$-adic Tate module. If you don’t know what some of these words mean, don’t worry; I’ll explain. 1

It’ll be quite a while before we get into defining and proving irreducibility of these representations since doing so requires a lot of ideas I have not introduced on this blog before. To begin, we’ll introduce some of the basics of the general theory of algebraic curves before focussing specifically on elliptic curves. Since I’ve said the word $\spec$ here before, I could set things up in those terms 2, but I won’t; instead, I’ll take an approach that’s more concrete. 3 With that said, let’s get started$\dots$

Note: I’m writing this post like a madman (in a rushed manner manner split over several days without much of a game plan ahead of time), so it likely contains more mistakes than usual (e.g. there may be some circular arguments here). Some time after finishing it and putting it online, I’ll take another look at it and try to resolve all these issues. Until then, recovering a coherent post from what’s below is left as an exercise to the reader 4.

# Algebraic Varieties

This section is mostly definitions 5, and so can be skipped and referred back to whenever you see something but don’t know what it means.

Fix a field $k$ with algebraic closure $\bar k$. Let $\bar k[X]=\bar k[x_1,\dots,x_n]$ be a polynomial ring in $n$ variables. Let $\A^n=\A^n(\bar k)=\bracks{P=(p_1,\dots,p_n):p_i\in\bar k}$ denote affine $n$-space (over $\bar k$).

Let $I\subset\bar k[X]$ be an ideal. Its vanishing set is $$V(I)=\bracks{P\in\A^n:f(P)=0\text{ for all }f\in I}.$$ Any set $C$ of the form $C=V(I)$ for some $I$ is called an affine algebraic set.
Given any set $C\subset\A^n$, we associate to it the ideal $$I(C)=\bracks{f\in\bar k[X]:f(P)=0\text{ for all }P\in C}.$$ When $C$ is an algebraic set, we see that $C$ is defined over $k$, denoted $C/k$, exactly when $I(C)$ can be generated by polynomials in $k[X]$; equivalently, exactly when $I(C)=I(C/K)\bar k[X]$ where $I(C/K)=I(C)\cap k[X]$.
Let $G=G_{\bar k/k}=\Gal(\bar k/k)$ be the absolute Galois group of $k$. Then, $G$ naturally acts on $\A^n$, and because $f(P^\sigma)=f(P)^\sigma$ for any $f\in\bar k[X]$ and $\sigma\in G$, we see that $G$ even acts on algebraic sets defined over $k$.

Note that Hilbert’s basis theorem tells us that $\bar k[X]$ is noetherian, so any algebraic set is given by the vanishing of only finitely many polynomials.

When $C$ is defined over $k$, we write $$C(K)=C\cap\A^n(K)=\bracks{P\in C:P^\sigma=P\text{ for all }\sigma\in G_{\bar k/k}}$$ for the set of $k$-rational points on $C$.
An affine algebraic set $C/k$ is called an affine variety if $I(C)$ is a prime ideal in $\bar k[X]$. In this case, its affine coordinate ring is $k[C]=k[X]/I(C/k)$, and its function field is $k(C)=\Frac k[C]$.
Note that an element $f\in\bar k[C]$ induces a well-defined function $f:C\to\bar K$ since we only killed functions vanishing along $C$. Furthermore, the action of $G_{\bar k/k}$ on $k[X]$ descends to one on $\bar k[V]$ (and $\bar k(V)$). As with points, we denote the action of $\sigma\in G_{\bar k/k}$ on $f$ by $f\mapsto f^\sigma$, so we see that $$(f(P))^\sigma=f^\sigma(P^\sigma)$$ always.

That was a lot of definitions, and there are only more to come. If you haven’t seen material like this before, this may be a good place to mention that my previous post on algebraic geometry might be helpful for understanding why some of the definitions here are reasonable. For example,

Let $V$ be a variety. Then, its dimension is $\dim(V)=\trdeg_{\bar k}\bar k(V)$.
If $V=\A^n$, then $I(V)=(0)$, $\bar k[V]=\bar k[X]$, and $\bar k(V)=\bar k(X)=\bar k(x_1,\dots,x_n)$, so $\dim V=n$.
If $f\in\bar k[X]$ and $V=V(f)$ is the vanishing set of a single (irreducible) nonconstant polynomial, then $\dim(V)=n-1$.