Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $E$ is an elliptic curve over $\mathbb{Q}$ which has good reduction at $2$ and $3$, is it always possible to find a minimal integral Weierstrass equation for $E$ of the form $y^2 = x^3 + Ax + B$ ($A,B \in \mathbb{Z}$)? Minimal here means that the absolute value of the discriminant $\Delta = -16(4A^3 + 27B^2)$ of the equation is minimal.

share|improve this question
1  
If $E$ has good reduction at $2$, its minimal discriminant must be prime to $2$. But as you observed, an equation as you want has discriminant divisible by $16$, so it can't have good reduction at $2$. However, it is perfectly possible to have good reduction at $3$. But not every $E$ having good reduction at $3$ can have such an equation. –  user18119 Mar 20 '12 at 21:08
    
Good point. I foolishly overlooked that fact about reduction at $2$. –  Hoffden Mar 20 '12 at 23:19

1 Answer 1

up vote 4 down vote accepted

The answer is no. Consider the elliptic curve $E$ with Cremona label "11a1" given by the Weierstrass equation $$E:y^2 + y = x^3 - x^2 - 10x - 20.$$ The conductor of $E$ is $11$, the discriminant of this model for $E$ is $-11^5$, so this model is minimal, and $E$ has good reduction at $2$ and $3$. Now consider the model $$E':y^2 = x^3 - 13392x - 1080432.$$ The curves $E$ and $E'$ are isomorphic over $\mathbb{Q}$, with a change of variables $\varphi:E\to E'$ that sends $$(x,y)\mapsto (36x - 12,\ 216y + 108).$$ However, the discriminant of $E'$ is $-2^{12}3^{12}11^5$. Now, if $E''$ was another model for $E$ of the form $$E'':y^2=x^3-Ax-B$$ with $A,B\in\mathbb{Z}$, and minimal discriminant $-11^5$, then there is a change of variables from $E''$ to $E'$ that sends $(x,y)\mapsto (x/u^2,y/u^3)$, for some $u\in\mathbb{Q}$ and therefore $13392=Au^4$ and $1080432=Bu^6$, and $\Delta(E')=\Delta(E'')\cdot u^{12}$. The equation relating discriminants says that $u=6$. However, $$13392=2^4\cdot 3^3\cdot 31, \quad \text{ and } 1080432=2^4\cdot 3^3\cdot 41\cdot 61,$$ and such $A$ and $B\in \mathbb{Z}$ cannot exist.

share|improve this answer
    
Thank you very much. –  Hoffden Mar 20 '12 at 23:19
    
@Hoffden, you are welcome! –  Álvaro Lozano-Robledo Mar 21 '12 at 0:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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