What is the argument used to dsitinguish the cases (a) and (b)

Question

We know from [B. Mazur, Modular curves and the Eisenstein ideal, Publ. math. IHES 47 (1977), 33-186] that if $C$ is an elliptic curve of the form ($C:y²=x³+ax+b$ with $a,b∈ℤ$), then $C(ℚ)^{tors}$ (the subgroup of elements of finite order in $C(ℚ)$ and $C(ℚ)$ is the Mordell-Weil group) is one of the following $15$ groups:

(a) $ℤ/nℤ$, with $1≤n≤10$ or $n=12$, or

(b) $ℤ/2mℤ× ℤ/2ℤ$, $1≤m≤4$.

Note that each of these groups occurs infinitely often as the torsion subgroup of an elliptic curve over $ℚ$. My question is: .What is the argument (justification) used to dsitinguish the two cases (a) and (b) (given $C/Q$, how do we tell if we are in one case of (a) or one case of (b))?

Could you clarify your question... what do you mean by "distinguish the two cases"? Do you mean, given $E/\mathbb{Q}$, how do we tell if we are in one case of (a) or one case of (b)? — Álvaro Lozano-Robledo, Apr 29 '14 at 16:57
@ÁlvaroLozano-Robledo: In you answer you mean ...$E(Q)[2]≅Z/2mZ⊕Z/2Z$ is completely defined over $Q$. — DER, May 04 '14 at 19:23
No, I don't. The two torsion is a subgroup of $\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$, and I mean the case when the $2$ torsion is "full", i.e., isomorphic to $\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$. — Álvaro Lozano-Robledo, May 04 '14 at 19:53
@ÁlvaroLozano-Robledo: So in (b) we have $Z/2mZ×Z/2Z$ is isomorphe to $Z/2mZ⊕Z/2Z$. — DER, May 04 '14 at 20:11
@ÁlvaroLozano-Robledo: there is an isomorphism between the tow groups. — DER, May 04 '14 at 20:15

score 2 · Accepted Answer · answered Apr 29 '14 at 18:25

2

Let $E/\mathbb{Q}: y^2=x^3+ax+b$ be an elliptic curve. Then, by Mazur's theorem, $E(\mathbb{Q})_\text{tors}$ falls in case (b) if and only if the two torsion $E(\mathbb{Q})[2]\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ is completely defined over $\mathbb{Q}$, and this occurs if and only if the polynomial $x^3+ax+b$ factors completely into (non-repeated) linear factors over $\mathbb{Q}$. Notice that in case (a) the two torsion over $\mathbb{Q}$ is either trivial of $E(\mathbb{Q})[2]\cong \mathbb{Z}/2\mathbb{Z}$.

answered Apr 29 '14 at 18:25

Álvaro Lozano-Robledo

15,390

1

To make Alvaro Lozano-Robledo's answer more clear for the less experts, note that the $2$-torsion points of an elliptic curve in Weierstrass form are the point at infinity (which is actually the zero element of the group) and the three intersections with the $x$-axis, because by simmetry these are the points with tangent meeting the point at infinity of the curve. – Andrea Mori May 08 '14 at 16:02

What is the argument used to dsitinguish the cases (a) and (b)

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