Question

Let $K$ be a field and $E$ be an elliptic curve defined over $K$. It well understood the points on $E$ forms an abelian group. What is the structure of this group?(depending on char($K$)?) Is it a direct sum of some well known abelian groups such as $\mathbb{Z}/m\mathbb{Z}$?

Answer 1

First case: Complex number. Over $\mathbb C$ the structure as an abstract group is $\mathbb S^1 \oplus \mathbb S^1$ where $\mathbb S^1$ is the circle, i.e., $\mathbb R/\mathbb Z$. This follows as Robin Chapman denotes below, i.e., it is a complex torus in the form $\mathbb C/\Lambda$ where $\Lambda$ is a lattice in $\mathbb C$.

Let $K$ be an algebraically closed field of char $p$. If $n$ is prime to $p$, the the $n$-torsion is $\mathbb{Z}/n\mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$. The $p^e$-torsion could be either trivial for all $e$, or $\mathbb{Z}/p^e\mathbb{Z}$ for all $e$.

Over a non-algebraically closed-field, this is going to be much more complicated. I will try to give just an introduction.

Over $\mathbb Q$ and number fields: Over $\mathbb Q$ or a number field, it is finitely generated by the Mordell-Weil theorem. So it has a torsion part and free part. The free part could be arbitrarily large.

The Mordell-Weil theorem is in fact true for arbitrary finitely generated fields. This is due to Néron.

Over $\mathbb Q$, The torsion part has exactly $15$ possibilities according to the theorem of Mazur. Over number fields, this had been generalized that the torsion part is uniformly bounded.

Over finite fields, the torsion group would be $\mathbb{Z}/n\mathbb{Z} \oplus \mathbb{Z}/m\mathbb{Z}$ where $n |m$. There is no free part.

And it could go on like this. Please have a look at Silverman's "Advanced topics in the Arithmetic of Elliptic Curves" for elliptic curves over real numbers, $p$-adic numbers, function fields, etc..

Answer 2

If you don't specify more about the structure of the field $K$, then we can't say much about the structure of the group $E(K)$. There are special cases (described in the Wikipedia article):

1. If $K$ is a number field, then the Mordell-Weil theorem implies the group is finitely generated (and this has been generalized, as Anweshi mentioned). In fact, for each number field, there is a global bound on the size of the torsion of any elliptic curve over that field. In particular, if $K = \mathbf{Q}$, then it is a direct sum of a free abelian group of finite rank with a torsion group that is one of 15 types.
2. If $K$ is finite of order $q$, then (by a theorem of Hasse) the group is finite of order about $q+1$ with error bounded by $2\sqrt{q}$. It is a sum of two cyclic groups.

If $K$ is larger than that, then $E(K)$ can be quite large. For example, if $K$ is separably closed, then $E(K)$ is divisible. In this case, if $K$ has characteristic zero, then $E(K) \cong (\mathbf{Q}/\mathbf{Z})^2 \oplus \bigoplus \mathbf{Q}$. If $K$ has characteristic $p>0$, then $E(K) \cong \bigoplus_{\ell \neq p} (\mathbf{Q}_\ell/\mathbf{Z}_\ell)^2 \oplus (\mathbf{Q}_p/\mathbf{Z}_p)^h \oplus \bigoplus \mathbf{Q}$. Here, $h$ is zero or one depending on whether the curve is supersingular or ordinary, and the $\bigoplus \mathbf{Q}$ is a vector space whose dimension is:

1. zero if K is an algebraic closure of a finite field.
2. countably infinite if $K$ is countable and not an algebraic closure of a finite field.
3. equal to the cardinality of $K$ otherwise.

Away from the separably closed case, you get a subgroup of one of these groups, but you can have very complicated subgroups of $\mathbf{Q}$ as summands, and very complicated torsion subgroups.