Cuspidal cubic not coarse moduli space parametrizing one dimensional subspaces of $\mathbb{C}^2$?

Question

Let $F$ be the functor of flat families of lines through the origin in $\mathbb{C}^2$. Let $C$ be the projective curve with plane model $y^2 = x^3$, i.e. in projective space it is defined by $y^2z = x^3$. Is there a natural transformation $\Phi: F \to h_C$ making $(C, \Phi)$ a coarse moduli space for $F$?

Answer 1

Let us work with schemes over $\mathbf{C}$ (and not algebraic spaces). Let $F$ a contravariant functor from the category of schemes over $\mathbf{C}$ to the category of sets. A scheme $M$ and a transformation of functors $\phi : F \to h_M$ is called a coarse moduli scheme $M$ if (a) $\phi$ is a categorical quotient, and (b) the map $F(\text{Spec}(k)) \to h_M(\text{Spec}(k))$ is bijective for any algebraically closed extension $\mathbf{C} \subset k$.

You can find this definition in Mumford's book ``Geometric Invariant Theory'' (second edition) page 99. Yes, he is only making this definition in the setting of moduli of curves, but it is customary to extend this definition to arbitrary moduli problems. There he also defines what is a fine moduli scheme for $F$ and he tells you that a fine moduli scheme is a coarse moduli scheme (this is immediate from the definition).

Now my point is that if $(M, \phi)$ is a coarse moduli scheme, then $M$ is unique up to (unique) isomorphism by (a). This you should work out for yourself -- it is a completely categorical statement. Hence if you can show that your functor $F$ is represented (i.e., has a fine moduli scheme) by $\mathbf{P}^1_\mathbf{C}$, then any coarse moduli scheme will be isomorphic to $\mathbf{P}^1_\mathbf{C}$.

OK, so now it is your task to correctly define $F$ (which you didn't really do in the question) so that this is indeed the case.