Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
To me, the $j$-function means Klein's modular function ( 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
@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
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 :) ]

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.