Given an elliptic curve $E$ over a field $k$ with $j$-invariant $j$, what is the criterion on $j$ for $E$ to be connected?
I know that when an elliptic curve is in Wierstrass form $y^2=x^3+px+q$, it is connected if and only if the discriminant of the cubic in $x$ has a negative discriminant.
We have a formula for the $j$-invariant of $E$ as follows:
$$j=1728 \frac{4p^3}{4p^3+27q^2}=1728\frac{4p^3}{-\Delta}$$
So given $j$ and $p$ we can deduce whether or not it is connected. However I imagine that the choice of $p$ should not really matter as contentedness should really only depend on $j$. I therefore tried to express the $j$ invariant for the Legendre form:
$$j=256\frac{(\lambda^2-\lambda + 1)^3}{\lambda^2(\lambda - 1)^2}$$
This should give me at most $6$ choices of $\lambda$. Solving this generally however is a non-trivial task. And then converting this into Weirstrass form is not too easy, I will get a very disgusting expression for $p$.
The only reference for connectedness of $E$ I could find was in terms of the discriminant of the polynomial of the Weirstrass form. Is there a more algebro-geometric criterion for knowing connectedness?
I imagine for curves over $\Bbb C$ something like the betti homology would give the correct answer but I can't see how you can know it just from $j$.
Let us work over $\Bbb R$, as that is where connectedness is interesting.
Consider $$j-1728=1728\frac{-27q^2}{-\Delta}.$$ This shares the sign of $\Delta$. So if $j<1728$ then $E$ is connected, and if $j>1728$ is is not connected. The $j=1728$ case is CM by $\Bbb Z[i]$, and considering $y^2=x^3-x$ and $y^2=x^3+x$ we see that $E$ may or not be connected.