What is the quadratic twist of an elliptic curve in Legendre Form? How do you show an elliptic curve and its quadratic twist is isomorphic when they are in Legendre Form?

2$\begingroup$ Is "legendre form" just y^2=x(x1)(xlambda)? A typical quadratic twist of this is dy^2=x(x1)(xlambda). They're isomorphic over C via the change of variables y>y.sqrt(d). I'm not sure I can make any more sense of your question other than these remarks. $\endgroup$ – Kevin Buzzard Feb 21 '10 at 9:14

$\begingroup$ I think James is asking how to write the quadratic twist itself in Legendre form. $\endgroup$ – Qiaochu Yuan Feb 21 '10 at 22:20

1$\begingroup$ Qiaochu: can you make that question make sense? Over a general field why should a random elliptic curve be able to be put into Legendre form? See Victor's answer for a proof that over Q this can almost never be done, for example. And over e.g. an alg closed field of char not 2, the Legendre form always exists but the Legendre form of the twist is the same as of the original curve, because you can't see twists over an alg closed field. As I said before, I can't make sense of the question. Can you? $\endgroup$ – Kevin Buzzard Feb 21 '10 at 22:31

$\begingroup$ Yes like Qiaochu Yuan said, I was asking how to write the quadratic twist in Legendre Form. But Victor cleared things up for me. $\endgroup$ – James Feb 22 '10 at 0:51
To amplify on what Kevin Buzzard said: the Legendre form of an elliptic curve is
$E_{\lambda} : y^2 = x(x1)(x\lambda)$
Since it has only one parameter, it has much less freedom than $y^2 = x^3 + ax + b$. Strictly speaking, an elliptic curve over a field $K$ is a twist of another elliptic curve over $K$, if they are not isomorphic over $K$ but they are over an extension $L/K$. So it doesn't make sense to ask when the quadratic twist (meaning that $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt{d})$ is a quadratic extension) is isomorphic to the curve. However, I think that you mean the curve
$E_{\lambda,d}: d y^2 = x(x1)(x\lambda)$ rewritten into Legendre form. However, we do have
$j = 256 \frac{(\lambda^2  \lambda + 1)^3}{\lambda^2 (\lambda1)^2}$
and there are six "versions" of $\lambda$ which yield isomorphic curves in Legendre form:
$\lambda,1\lambda,1/\lambda,1/(1\lambda),\lambda/(\lambda1),(\lambda1)/\lambda$.
This seriously restricts the possible $d$'s for which $E_{\lambda,d}$ can be written in Legendre form to a at most six values, namely if $\pm d \lambda, \pm d, \pm d (1\lambda)$ are squares.
[Added later]: It might make more sense to think of $Y^2 = X(Xa)(Xb)$ as a "generalized Legendre form". In that line of thought, the "standard" Weierstrass form $Y^2 = X^3 + a X + b$ is associated with the moduli space $X(1)$, and $a$ is a modular form for weight 4, and b is a modular form of weight 6 for $\Gamma(1)$ (suitably normalized Eisenstein forms). The "generalized Legendre form" is associated with $X(2)$ which parametrizes elliptic curves along with the full 2torsion subgroup (satisfying a normalization of the Weil Pairing). In that case $a$ and $b$ are modular form of weight 2 for $\Gamma(2)$

$\begingroup$ To amplify, if you have an elliptic curve in the form $Y^2=X(Xa)(Xb)$ where $a,b \in K$ some field, then it is isomorphic over $K$ to a curve of them form $Y^2=X(X1)(X\lambda)$, for $\lambda \in K$ if and only if one of $\pm a, \pm b,(1+a)/b,(1+b)/a$ is a square in $K$. For your "twist" you have $a=d, b=d \lambda$. $\endgroup$ – Victor Miller Feb 21 '10 at 22:49