In Hartshorne, there's a formula for the j-invariant in terms of $\lambda$. It says that $$ j = 2^8 \frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}.$$ Can one reverse this formula? That is, can we express $\lambda$ in terms of $j$?
Of course, there will be six different solutions because we are dealing with a polynomial in $\lambda$ of degree six.
If you are interested in, see here.