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Let $\Lambda(\lambda) := \left\{\lambda, 1 - \lambda, \frac{1}{\lambda}, \frac{1}{1-\lambda}, \frac{\lambda - 1}{\lambda}, \frac{\lambda}{\lambda - 1} \right\}$, and consider the $j$-function

$$j(\lambda) = 256\frac{(\lambda ^2 - \lambda + 1)^3}{\lambda ^2 (\lambda - 1)^2}.$$

I'm trying to prove that $\Lambda(\lambda) = \Lambda(\lambda ')$ iff $j(\lambda) = j(\lambda ')$.

That the values of the $j$-function coincide if the sets are the same is fairly obvious by a straightforward calculation, but I'm stuck trying to prove the converse.

I'd welcome any hints.

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To me, the $j$-function means Klein's modular function (en.wikipedia.org/wiki/J-invariant). Is your $j$-function related to Klein's? If not, then I think the title of your question is seriously misleading and in need of editing. –  Gerry Myerson Nov 24 '11 at 22:39
    
@GerryMyerson This is the $j$-invariant of the elliptic curve whose Legendre form is specified by $\lambda$, I think. –  Dylan Moreland Nov 25 '11 at 0:06
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@DylanMoreland, OK, thanks. Even so, it's really just a question about rational functions, innit? No knowledge of elliptic curves (or algebraic geometry) needed to understand and/or answer it? Maybe title and tag should be edited. –  Gerry Myerson Nov 25 '11 at 0:18

1 Answer 1

up vote 2 down vote accepted

$j(\lambda)$ is a rational function of degree $6$, so for a generic $z$ there a $6$ $\lambda$'s such that $j(\lambda)=z$, and for the critical values there are less than $6$ $\lambda$'s. If $\lambda$ is such that $\Lambda(\lambda)$ has $6$ elements and if $j(\lambda)=j(\lambda')$, it follows that $\lambda'\in\Lambda(\lambda)$. If $\Lambda(\lambda)$ has less than $6$ elements, you can check it manually (there are only two possible values of $j$ in this case) [there is a better explanation, but perhaps this one will do :) ]

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