Lecture 23: Finishing up
In this final lecture, we complete the classification of torsion groups of elliptic curves over the raionals. Actually, we only give an overview of the proof, as it is a rather lengthy casebycase analysis.
Where we are
Our goal in this course has been to prove the following theorem:
Theorem (Main theorem). Let $G$ be a finite abelian group. Then $G=E(\bQ)_{\tors}$ for some elliptic curve $E/\bQ$ if and only if $G$ is one of the following groups:

$\bZ/N\bZ$ for $1 \le N \le 10$ or $N=12$.

$\bZ/2\bZ \times \bZ/N\bZ$ with $N \in \{2,4,6,8\}$.
We have proven the following:
Theorem (Mazur, MazurTate). Let $N$ be a prime $\gt 7$. Then no elliptic curve over $\bQ$ has a rational point of order $N$.
To complete the proof of the theorem, then, it remains to show the following:

$E(\bQ)$ does not have any $N$torsion for $N \in \{14, 15, 16, 18, 20, 21, 24, 25, 27, 35, 49 \}$.

$E(\bQ)$ does not have a subgroup of the form $\bZ/2 \bZ \times \bZ/N\bZ$ for $N \in \{10, 12\}$.

The 15 groups that should occur as torsion subgroups actually do.
Most of the work was carried out by Kubert in MR0434947.
The groups that should occur do
Let $N$ be an integer with $1 \le N \le 10$ or $N=12$. The curve $X_1(N)$ then has genus 0, and, in fact, these are the only values of $N$ for which this is true. Furthermore, $X_1(N)$ has rational points, namely, its rational cusps. It follows that $X_1(N)$ is isomorphic to $\bP^1$ over $\bQ$, and therefore has infinitely many rational points. In particular, $Y_1(N)$ has rational points. For $N \ge 4$, the scheme $Y_1(N)$ represents the moduli problem of elliptic curves with $N$torsion, and so there exist elliptic curves over $\bQ$ with a rational point of order $N$. For $N=2,3$, the same conclusion holds, and is easy to see directly. (Any curve of the form $y^2+axy+by=x^3$ admits $(0,0)$ as a 3torsion point, and any curve of the form $y^2=f(x)$ where $f$ has a rational root has a 2torsion point.)
Now let $N \in \{4,6,8\}$. Let $Y$ be the moduli scheme parametrizing elliptic curves equipped with an injection from $\bZ/2\bZ \times \bZ/N\bZ$ (this does exist as a scheme), and let $X$ be the corresponding compactification. Then, again, $X$ has genus and a rational point, and is therefore isomorphic to $\bP^1$ over $\bQ$. It follows that $Y(\bQ)$ is nonempty, and so $\bZ/2\bZ \times \bZ/N\bZ$ does occur as a subgroup of the MordellWeil group of an elliptic curve over $\bQ$. It is easy to see that this conclusion continues to hold for $N=2$: consider $y^2=f(x)$ where $f$ has only rational roots.
We have thus shown that each of the 15 groups $G$ there exists an elliptic curve $E/\bQ$ such that $E(\bQ)$ contains a subgroup isomorphic to $G$. In fact, one can find $E$ for which $E(\bQ)_{\tors}$ is isomorphic to $G$. We just explain the idea in one case. Let's show that $\bZ/5 \bZ$ occurs exactly as the torsion subgroup. If it didn't, that'd mean that every elliptic curve over $\bQ$ with a rational 5torsion point would also have an additional rational torsion point, which necessarily would be 2torsion by Ogg's conjecture. In other words, the map $X_1(10) \to X_1(5)$ would induce a surjection on rational points. But this is a degree 3 map of rational curves, so that cannot happen: in fact, most points in the target will not be in the image.
Remark. For $N \ge 4$ it is possible to explicitly write down the universal elliptic curve with a point of order $N$ (or a subgroup of the form $\bZ/2\bZ \times \bZ/N\bZ$). For example, the universal curve with a point of order 4 is given by
$$ y^2 + xy  ty = x^3  t x^2. $$The 4torsion point is $(0,0)$.
When X_0(N) has genus 1
Excluding certain torsion orders
Proposition. Let $N$ belong to $\{11, 14, 15, 17, 19, 20, 21, 24, 27, 32, 36, 49\}$. Then no elliptic curve over $\bQ$ has a point of order $N$.
Proof. The curve $X_0(N)$ has genus 1. Choosing one of the cusps as the origin, this gives $X_0(N)$ the structure of an elliptic curve. One can then use standard methods to study the rational points on $X_0(N)$, and one finds that it has rank 0. Determining the torsion subgroup is straightforward, and so one can explicitly write down all the rational points on $X_0(N)$. If all the rational points on $X_0(N)$ are cuspdial then we are done. If not there is more work to do.
Suppose there are noncuspidal rational points on $X_0(N)$. Enumerate them as $x_1, \ldots, x_n$. Each can be represented by a pair $(E_i, G_i)$ where $E_i/\bQ$ is an elliptic curve and $G_i$ is a cyclic subgroup of order $N$. If $y$ is a rational point of $X_1(N)$, corresponding to $(E, P)$, then $y$ maps to some $x_i$, and so $E$ is isomorphic to some twisted form of $E_i$. (Twisted form because $X_0(N)$ is only a coarse moduli space.) To prove the proposition, it therefore suffices to check that no twisted form of any $E_i$ has a point of order $N$. To do this, consider $E_i[N]$ as a Galois representation. Since $E_i$ admits a cyclic isogeny of degree $N$, we have a short exact sequence
$$ 0 \to (\bZ/N\bZ)(\alpha_i) \to E_i[N] \to (\bZ/N\bZ)(\beta_i) \to 0 $$where $\alpha_i$ and $\beta_i$ are homomorphisms $\Gal(\ol{\bQ}/\bQ) \to (\bZ/N\bZ)^{\times}$. Suppose $E_i$ is nonCM, and let $E_i^{(d)}$ be the quadratic twist of $E_i$ by $d$. We then have an exact sequence
$$ 0 \to (\bZ/N\bZ)(\alpha_i \chi_d) \to E_i^{(d)}[N] \to (\bZ/N\bZ)(\beta_i \chi_d) \to 0 $$where $\chi_d$ is the quadratic character associated to $d$. Thus $E_i^{(d)}$ has a rational $N$torsion point if and only if $\alpha_i$ is a quadratic character, or $E_i[N]$ is split and $\beta_i$ is a quadratic character. (When $E_i$ is CM a similar statement holds.) Thus, to prove the proposition, it suffices to check this is not the case for each $i$, which is a straightforward finite computation. ◾
Example. When $N=21$ there are four noncuspidal rational points on $X_0(N)$. The $E_i$ are given by:
$$ \begin{aligned} y^2 &= x^3 + 45 x  18 \\ y^2 &= x^3  75 x  262 \\ y^2 &= x^3  1515 x  46106 \\ y^2 &= x^3  17235 x  870894 \end{aligned} $$None of these curves are CM, and so it suffices to show that no quadratic twist of any of them has a 21torsion point.
Example. When $N=27$ there is one noncuspidal rational point on $X_0(N)$. It is given by
$$ y^2 + y = x^3  30x  5 $$This curve has CM by $\sqrt{27}$.
Remark. There are rational noncuspdial points on $X_0(N)$ if and only if $N \in \{ 11, 14, 15, 17, 19, 21, 27 \}$.
Remark. When $N \in \{11,14,15\}$, the curve $X_1(N)$ has genus 1. Since $X_1(N) \to X_0(N)$ is an isogeny, it follows that $X_1(N)$ has rank 0. One can therefore easily compute all of its rational points, and one finds that they are all cusps. This is easier than the argument given above.
Remark. Combined with the previous remark and what we already know (or will prove), the above proposition is really only needed for $N \in \{20,21,24,27,49\}$. There are noncuspidal rational points only for $N \in \{21,27\}$, so only in these cases does the proof require the additional computation.
Excluding certain product groups
Lemma. Let $E/\bQ$ be an elliptic curve such that $E(\bQ)$ contains a subgroup of the form $\bZ/2\bZ \times \bZ/2\bZ$ (resp. $\bZ/2\bZ \times \bZ/4\bZ$). Then there exists a degree 2 isogeny $E \to E'$ defined over $\bQ$ such that $E'$ admits a rational cyclic isogeny of degree 4 (resp. 8).
Proof. Let $P$ be the point of order 2 and $Q$ an independent point of order 2 (resp. 4). Let $E_1=E/P$ and $E_2=E/Q$ and let $f \colon E \to E_1$ and $g \colon E \to E_2$ be the natural isogenies. Consider the dual isogeny $f^{\vee} \colon E_1 \to E$. The image of $E_1[2]$ under $f^{\vee}$ is just the point $P$. This point is not in the kernel of $g$, and so $gf^{\vee} \colon E_1 \to E_2$ is a cyclic isogeny of degree 4 (resp. 8). ◾
Proposition. Let $E/\bQ$ be an elliptic curve. Then $E(\bQ)$ does not contain a subgroup isomorphic to $\bZ/2\bZ \times \bZ/10\bZ$ or $\bZ/2\bZ \times \bZ/12\bZ$.
Proof. Suppose $E$ did contain such a subgroup. Then there would be a 2isogeny $E \to E'$ such that $E'$ admits a cyclic 20 or 24 isogeny, and thus defines a rational point on $X_0(20)$ or $X_0(24)$. By the above results, the only such points are cuspidal, which is a contradiction. ◾
The case of 16 torsion
This was dealt with by Lind in his 1940 thesis. The proof is supposed to be easy, but I haven't found it written down anywhere (and can't get the thesis). The curve $X_1(16)$ has genus 2, while $X_0(16)$ has genus 0.
The cases of 18 and 25 torsion
These cases are similar to the case of 13 torsion in that $X_0(N)$ has genus 0. This means that there's no map from $X_1(N)$ to an elliptic curve, and so one cannot reduce to the study of an elliptic curve. Kubert carries out a fairly direct generalization of the MazurTate argument in these two cases.
The curve $X_1(18)$ has genus 2. The torsion in $J_1(18)$ is $\bZ/21\bZ$, and meets $X_1(18)$ only at the cusps. It therefore suffices to show that $J_1(18)$ has rank 0. The automorphism $\gamma_5$ of $X_1(18)$ satisfies $\gamma_5^2+\gamma_5+1=0$ on $J_1(18)$, and so $J_1(18)$ contains the ring $\bZ[x]/(x^2+x+1)$ in its endomorphism algebra. The prime 7 factors as $\pi \ol{\pi}$ in this ring, and Kubert shows that multiplication by $\pi$ is surjective on the MordellWeil group of $J_1(18)$.
The case of 25torsion is more complicated.
The case of 35 torsion
At this point, it remains only to exclude 35 torsion. In fact, Kubert shows that $X_0(35)$ has no noncuspidal rational points. The idea is to consider the quotient $E=X_0(35)/w_5$, where $w_5$ is the AtkinLehner involution at 5. This curve has genus 1. One shows that $E(\bQ)=\bZ/3\bZ$, and computes the preimages of these three rational points in $X_0(35)$. They turn out to be cusps.