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.
Form of minimal integral Weierstrass equation for elliptic curve over $Q$ with good reduction at $2$ and $3$.
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0Good point. I foolishly overlooked that fact about reduction at $2$. – 2012-03-20
1 Answers
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.
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0@Hoffden, you are welcome! – 2012-03-21