Are all elliptic curves from $w^3 = \text{cubic}(z)$ isomorphic? I've been playing around with Riemann surfaces of cubics, and it seems to me that all surfaces obtained as coverings of the Riemann sphere from equations of the form
$w^3 = q(z)$, where $q(z)$ is a cubic with three distinct roots, must be isomorphic.
Argument:
we have a critical point of multiplicity 3 at each of the roots of $q(z)$. Monodromy around a small counterclockwise circuit about any of these points multiplies $w$ by the same cube root of unity ($\neq 1$). So monodromy around a circuit enclosing all three roots of $q(z)$ leaves $w$ unchanged, so no branch points over $\infty$.
Now we can move the three roots of $q(z)$ to any other positions using a Möbius transformation, so by the previous paragraph, we should be able to establish an analytic isomorphism between the surfaces via continuation.
Is this correct? If so, what is the common J-invariant? And since we can choose $q(z)=z^3-1$, this means that all these curves have CM, right?
References appreciated.
 A: This is a great observation!
Let $E/\mathbb{Q}$ be a curve given by a model $v^3=q(u)$, for some cubic polynomial $q\in\mathbb{Q}[u]$, and assume that the projective closure of this model, i.e., $V^3=W^3q(U/W)$, is smooth, and it has a rational point $P$. Then, $(E,P)$ is an elliptic curve defined over $\mathbb{Q}$. Moreover, $E$ admits an endomorphism
$$[\rho] : E\to E$$
that sends $[U,V,W]$ to $[U,\rho V,W]$, where $\rho$ is a primitive third root of unity. Hence, $\operatorname{End}(E)$ is strictly larger than $\mathbb{Z}$, and so $E/\mathbb{Q}$ is a curve with complex multiplication. Further, 
$$\operatorname{End}(E)\otimes \mathbb{Q} \cong \mathbb{Q}(\rho),$$
and therefore the complex multiplication is by an order in the ring of integers of the quadratic imaginary field $\mathbb{Q}(\rho)=\mathbb{Q}(\sqrt{-3})$. It is a fact that follows from the theory of complex multiplication that all the elliptic curves with complex multiplication by $\mathbb{Q}(\sqrt{-3})$ have $j$-invariant equal to $0$. It follows that all the elliptic curves of the form $V^3=W^3q(U/W)$ are isomorphic over $\mathbb{C}$ (as the $j$-invariant classifies elliptic curves up to isomorphism over $\mathbb{C}$).
For example, consider $E: V^3=U^3-W^3$ (or $v^3=u^3-1$ in affine coordinates), with $P=[1,1,0]$. Then, $E$ is isomorphic to 
$$E': Y^2Z-9YZ^2=X^3-27Z^3$$
or $y^2 - 9y = x^3 - 27$ in affine coordinates, via the map
$\phi:E'\to E$ that sends
$$\phi([X,Y,Z])=(Y-9Z,Y,3X),$$
which sends $[0,1,0]$ to $[1,1,0]$. The curve $E'$ is now easily checked to have $j=0$ via the usual formulas, and therefore it has complex multiplication by $\mathbb{Q}(\sqrt{-3})$.
