Abelian Groups Attached to Elliptic Curves

If $E$ is an elliptic curve over $K$, then we give the set $E(K)$ of all $K$-rational points on $E$ the structure of abelian group with identity element $\O$. If we embed $E$ in the projective plane, then this group is determined by the condition that three points sum to the zero element $\O$ if and only if they lie on a common line.

For example on the curve $y^2=x^3-5x+4$, we have $(0,2) + (1,0) = (3,4)$. This is because $(0,2)$, $(1,0)$, and $(3,-4)$ are on a common line (so sum to zero):

$$(0,2) + (1,0) + (3,-4) = \O$$

and $(3,4)$, $(3,-4)$, and $\O$ (the point at infinity on the curve) are also on a common line, so $(3,4)=-(3,-4)$. See the illustration below:

Iterating the group operation often leads quickly to very complicated points:

That the above condition--three points on a line sum to zero--defines an abelian group structure on $E(K)$ is not obvious. Depending on your perspective, the trickiest part is seeing that the operation satisfies the associative axiom. The best way to understand the group operation on $E(K)$ is to view $E(K)$ as being related to a class group. As a first observation, note that the ring

$$R = K[x,y]/(y^2 +a_1 xy + a_3 y - (x^3 + a_2 x^2 + a_4 x + a_6))$$

is a Dedekind domain, so $\Cl (R)$ is defined, and every nonzero fractional ideal can be written uniquely in terms of prime ideals. When $K$ is a perfect field, the prime ideals correspond to the Galois orbits of affine points of $E(\overline{K})$.

Let $\Div (E/K)$ be the free abelian group on the Galois orbits of points of  $E(\overline{K})$, which as explained above is analogous to the group of fractional ideals of a number field (here we do include the point at infinity). We call the elements of $\Div (E/K)$ divisors. Let $\Pic (E/K)$ be the quotient of $\Div (E/K)$ by the principal divisors, i.e., the divisors associated to rational functions $f\in K(E)^*$ via

$$f \mapsto (f) = \sum_{P} \ord _P(f) [P].$$

Note that the principal divisor associated to $f$ is analogous to the principal fractional ideal associated to a nonzero element of a number field. The definition of $\ord _P(f)$ is analogous to the ``power of $P$ that divides the principal ideal generated by $f$''. Define the class group $\Pic (E/K)$ to be the quotient of the divisors by the principal divisors, so we have an exact sequence:

$$1\to K(E)^*/K^* \to \Div (E/K) \to \Pic (E/K) \to 0.$$

A key difference between elliptic curves and algebraic number fields is that the principal divisors in the context of elliptic curves all have degree 0, i.e., the sum of the coefficients of the divisor $(f)$ is always 0. This might be a familiar fact to you: the number of zeros of a nonzero rational function on a projective curve equals the number of poles, counted with multiplicity. If we let $\Div ^0(E/K)$ denote the subgroup of divisors of degree 0, then we have an exact sequence

$$0\to K(E)^*/K^* \to \Div ^0(E/K) \to \Pic ^0(E/K) \to 0.$$

To connect this with the group law on $E(K)$, note that there is a natural map

$$E(K) \to \Pic ^0(E/K), \qquad P \mapsto [P-\O].$$

Using the Riemann-Roch theorem, one can prove that this map is a bijection, which is moreover an isomorphism of abelian groups. Thus really when we discuss the group of $K$-rational points on an $E$, we are talking about the class group $\Pic ^0(E/K)$.

Recall that we proved (Theorem 7.1.2) that the class group $\Cl (\O_K)$ of a number field is finite. The group $\Pic ^0(E/K) =E(K)$ of an elliptic curve can be either finite (e.g., for $y^2 + y = x^3 - x + 1$) or infinite (e.g., for $y^2 + y = x^3 - x$), and determining which is the case for any particular curve is one of the central unsolved problems in number theory.

The Mordell-Weil theorem (see Chapter 12) asserts that if $E$ is an elliptic curve over a number field $K$, then there is a nonnegative integer $r$ such that

$$E(\mathbf{Q}) \approx \mathbf{Z}^r \oplus T,$$ (10.2)

where $T$ is a finite group. This is similar to Dirichlet's unit theorem, which gives the structure of the unit group of the ring of integers of a number field. The main difference is that $T$ need not be cyclic, and computing $r$ appears to be much more difficult than just finding the number of real and complex roots of a polynomial!

Also, if $L/K$ is an arbitrary extension of fields, and $E$ is an elliptic curve over $K$, then there is a natural inclusion homomorphism $E(K)\hookrightarrow E(L)$. Thus instead of just obtaining one group attached to an elliptic curve, we obtain a whole collection, one for each extension of $L$. Even more generally, if $S/K$ is an arbitrary scheme, then $E(S)$ is a group, and the association $S\mapsto E(S)$ defines a functor from the category of schemes to the category of groups. Thus each elliptic curve gives rise to map:

$$\left\{\text{Schemes over $K$}\right\} \longrightarrow \left\{\text{Abelian Groups}\right\}$$