# Discriminants and Weierstrass form of elliptic curves

I'm confused by what appears to be contradictory information.

In this post, the claim is made that

"Every elliptic curve over $\mathbb{Q}$ can be written in the form $y^{2}= x^{3}+ax+b$ where $a,b∈ \mathbb{Z}$ with discriminant $Δ=−16(4a^{3}+27b^{2})≠0$. So the number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a,b)∈ \mathbb{Z}^{2}$ such that $D=−16(4a^{3}+27b^{2})$."

Is this true? this post gives an example of an elliptic curve with integer coefficients that is not isomorphic to any curve $y^2 = x^3 + Ax + B$ for $A, B$ integers and discriminant equal to the original. Isn't this in contradiction to the original claim? That is, the curve in the second link would not be counted as a curve with its discriminant, right?

In fact, it seems to me (after plugging through the relevant changes of variables) that most curves of the form $$E: y^2 + a_{1} xy + a_{3} y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6},$$ with discriminant $\Delta$

cannot be written in the form $$E' : y^{2} = x^{3} + Ax + B$$ with discriminant also $\Delta$ for some $A, B \in \mathbb{Z}$.

Any help?

## 1 Answer

The statement:

"The number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a,b)∈ \mathbb{Z}^{2}$ such that $D=−16(4a^{3}+27b^{2})$."

is certainly false as stated. For example, the discriminant of the model for the curve $$E:y^2 + y = x^3 - x^2 - 10x - 20$$ is $-11^5$. Any model of the form $y^2=x^3+Ax+B$ with integers $A,B$ has a discriminant of the form $D=-16(4a^3+27b^2)$. But the equation $$-11^5 = -16(4a^3+27b^2)$$ is impossible in integers $a,b$ as the left hand side is odd and the right hand side is even.

One could salvage the statement above, however, and write something in a similar spirit. Going from a generic Weierstrass model to a short Weierstrass model means that the discriminant is multiplied, at worst, by $2^{12}3^{12}$. So one could say

"The number of elliptic curves of discriminant $D$ is bounded above $n_1+n_2+n_3+n_4$, where

• $n_1$ is the number of non-trivial pairs $(a,b)\in\mathbb{Z}^2$ such that $D=-16(4a^3+27b^2)$,

• $n_2$ is the number of non-trivial pairs $(a,b)\in\mathbb{Z}^2$ such that $2^{12}D=-16(4a^3+27b^2)$,

• $n_3$ is the number of non-trivial pairs $(a,b)\in\mathbb{Z}^2$ such that $3^{12}D=-16(4a^3+27b^2)$, and

• $n_4$ is the number of non-trivial pairs $(a,b)\in\mathbb{Z}^2$ such that $6^{12}D=-16(4a^3+27b^2)$."

and I believe this is a true statement.

• Regarding "Going from a generic Weierstrass model to a short Weierstrass model means that the discriminant is multiplied, at worst, by $2^{12} 3^{12}$" Is this true? I've only seen that going from one "short" Weierstrass model to another results in multiplying the discriminant by a $12$th power of $2, 3$ or $6$. In fact in Silverman (pages $42-43$ he gives that $y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}$ is isomorphic to $y^{2} = x^{3} - 27c_{4}x - 54c_{6}$ but these two discriminants don't necessarily (and indeed it seems very rarely) differ by a twelfth power. – 112358 Feb 25 '14 at 5:10
• If you go from an equation of the form $y^2+Axy+By=f(x)$ to one of the form $Y^2=f(x)$, you need to do a change of the form $Y=y+(Ax+B)/2$. This may introduce a $2^{12}$ in the discriminant. If now you go from $y^2=x^3+Ax^2+Bx+C$ to $y^2=X^3+A'x+B'$, you need a change of the form $X=x-A/3$, and this may introduce a $3^{12}$ in the discriminant. – Álvaro Lozano-Robledo Feb 25 '14 at 14:10
• Okay, I had miscalculated something. I now agree that what you wrote is true. Thanks – 112358 Feb 25 '14 at 18:03