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. See Figure 10.1.3 for an example, in which $(0,2)$ and $(1,0)$ add to $(3,4)$ in the group law.

Figure 10.1.3: The Group Law: $(1,0)+(0,2)=(3,4)$ on $y^2=x^3-5x+4$

That the above condition defines an abelian group structure on $E(K)$ is not obvious (the trickiest part is seeing that the operation is associative). The best way to understand the group operation on $E(K)$ is to view $E(K)$ as a class group, very similar to class groups of number fields. Let $\Div (E/K)$ be the free abelian group on the points of $E$, which is analogous to the group of fractional ideals of a number field. 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 associted 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:

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

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.

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.