The Torsion Subgroup of $E(\mathbb{Q})$ and the Rank

For any abelian group $G$ , let $G_{\tor }$ be the subgroup of elements of finite order. If $E$ is an elliptic curve over $\mathbb {Q}$ , then $E(\mathbb{Q})_{\tor }$ is a subgroup of $E(\mathbb{Q})$ , which must be finite because of Theorem 6.5.1 (see Exercise 6.6). One can also prove that $E(\mathbb{Q})_{\tor }$ is finite by showing that there is a prime $p$ and an injective reduction homomorphism $E(\mathbb{Q})_{\tor } \hookrightarrow E(\mathbb{Z}/p\mathbb{Z})$ , then noting that $E(\mathbb{Z}/p\mathbb{Z})$ is finite. For example, if $E$ is $y^2=x^3-5x+4$ , then $E(\mathbb{Q})_{\tor } = \{\O, (1,0)\} \cong \mathbb{Z}/2\mathbb{Z}{}.$

The possibilities for $E(\mathbb{Q})_{\tor }$ are known.

Theorem 6.5 (Mazur, 1976)   Let $E$ be an elliptic curve over  $\mathbb {Q}$ . Then $E(\mathbb{Q})_{\tor }$ is isomorphic to one of the following 15 groups:

$$\mathbb{Z}/n\mathbb{Z}{} n\leq 10 n=12,$$

$$\mathbb{Z}/2\times \mathbb{Z}/2n n \leq 4.$$

The quotient $E(\mathbb{Q})/E(\mathbb{Q})_{\tor }$ is a finitely generated free abelian group, so it is isomorphism to $\mathbb{Z}^r$ for some integer $r$ , called the rank of $E(\mathbb{Q})$ . For example, using descent one finds that if $E$ is $y^2=x^3-5x+4$ , then $E(\mathbb{Q})/E(\mathbb{Q})_{\tor }$ is generated by the point $(0,2)$ . Thus $E(\mathbb{Q}) \cong \mathbb{Z}\times (\mathbb{Z}/2\mathbb{Z})$ .

The following is a folklore conjecture, not associated to any particular mathematician:

Conjecture 6.5   There are elliptic curves over  $\mathbb {Q}$ of arbitrarily large rank.

The world record is the following curve, whose rank is at least $28$ :

$$y^2 + xy + y = x^3 - x^2 -$$

$$20067762415575526585033208209338542750930230312178956502x +$$

$$344816117950305564670329856903907203748559443593191803612\ldots$$

$$\ldots 66008296291939448732243429$$

It was discovered in May 2006 by Noam Elkies of Harvard University.