1
$\begingroup$

This is an exercise comment made by Dale Husemoller's Elliptic curve on Chpt 1, sect 3, Exercise 1.

$y^2=x^3+1$. It follows from (3.5) that this is all of group $E(Q)$ where $E(Q)$ denotes the rational points of $E$ over $Q$.

(3.5) Let $E$ be an elliptic curve defined by $y^2=x^3+a$ with $a$ being 6-th power free. Then torsion $\operatorname{Tor}(E(Q))=Z_6$ if $a=1$.

$\textbf{Q:}$ How does this follow that $E(Q)$ has no more rational points. In other words, why I do not have infinite order elements? I do not know whether he is assuming some other background information. The most probable one would be $Q(\zeta_3)$'s integral closure being PID which relates to class number information.

Ref: D.Husemoller Elliptic Curves Chpt 1, Sec 3.

$\endgroup$
  • $\begingroup$ Need more context. It’s certainly true that this $E$ has no non-torsion points, and that what you’ve quoted does not by itself show this. Maybe if you tell us what’s going on before, in 1.3 ? $\endgroup$ – Lubin Aug 30 '18 at 23:03
  • $\begingroup$ @Lubin The section starts with the theorem for 2-torsions. Originally, the book says, "1. Determine the five points in the (x, y)-plane which together with 0 form the group of six torsion points on y2 = x3 + 1. Note that from (3.5)[Should be 3.3] this is all of the group E(Q). Which of these points are generators of this cyclic group of order 6?" Sec 1.2 is computing group law for elliptic curves by direct computation. $\endgroup$ – user45765 Aug 30 '18 at 23:08
  • $\begingroup$ Well, this says nothing, but the only proof I know involves far more than that. I do not think that what you’ve just quoted is justified by what has gone before. Others, more skilled than I, can gainsay me. $\endgroup$ – Lubin Aug 31 '18 at 1:20
  • $\begingroup$ @Lubin Would you mind pointing me the reference where I can find the proof that this equation involves no torsion points? Thanks. $\endgroup$ – user45765 Aug 31 '18 at 1:25
  • $\begingroup$ Oy. You’re asking the wrong guy to supply a reference! Have you looked in Joe Silverman’s book? (Best is if Joe himself sees this and gives you an idea.) $\endgroup$ – Lubin Aug 31 '18 at 3:18
1
$\begingroup$

I doubt that from $ (3.5) $ one can deduce that the elliptic curve $E:\space y^2=x^3+1$ has no more than six rational points and I think there's a misprint in between. You need, I guess, "other background information" as you have written. Prove your curve has rank $0$ could be difficult without help of another strong enough property. Am I wrong in this point of view?

We have the following six rational points: $$A=(2,3),B=(0,1),C=(-1,0),-A=(2,-3),-B=(0,-1)$$ and the point at infinity taken as the zero of the group as usual.

Since $\dfrac{3-1}{2-0}=\dfrac{0-1}{-1-0}$, lines $AB$ and $BC$ are the same so $A+B+C=0$ by definition of the group law and obviously $-A-B-C=0$

enter image description here

We prove that $A$ is a generator of the torsion group $$\{A,B,C,-A,-B,O\}$$ using the formulas of sum $-(A+B)$ and $-2A$ (note the attached figure shows that $2A=B$).

The formulas for $y^2=x^3+ax+b$ with $a=0$ and $b=1$ are $$►(x_a,y_a)+(x_b,y_b)=-(p^2-x_a-x_b,\space y_a+p(p^2-2x_a-x_b))$$ where $p=\dfrac{y_a-y_b}{x_a-x_b}$.

$$►2(x,y)=(x,y)+(x,y)=-(p^2-2x,-y+p(p^2-3x))$$ where $p=\dfrac{3x^2}{2y}$.

Then easy calculations give $$\begin{cases}A=A\\2A=B\\3A=B+A=C\\4A=C+A=-B\\5A=3A+B=C+B=-A\\6A=2C=0\end{cases}$$ Besides $2B=4A=-B$ and $3B=-B+B=0$ and $2C=0$ i.e. $A$ has order $6$,$\space B$ has order $3$ and $C$ has order $2$ and it is so with $-A,-B$ and $-C$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.