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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.

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    $\begingroup$ 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. $\endgroup$ – user18119 Mar 20 '12 at 21:08
  • $\begingroup$ Good point. I foolishly overlooked that fact about reduction at $2$. $\endgroup$ – Hoffden Mar 20 '12 at 23:19
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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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