Fixing a connected component in the Hilbert scheme of $\mathbb{P}^2$ is the same as fixing the Hilbert polynomial $ax+b$. Taking a primary decomposition of any subscheme in this connected component returns a curve of class $\beta$ and length zero subschemes with total length $n$. $n$ and $\beta$ depend only on $a,b$.
I was recently asked whether there was an isomorphism $$\mathrm{Hilb}^{[n]}(\mathbb{P}^2) \times \mathbb{P}^{d(d+1)/2 - 1}\rightarrow \mathrm{Hilb}_{ax+b}(\mathbb{P}^2)$$
I am skeptical of this claim but I cannot see why it is false. Certainly there is a (bi?)rational map: I think of the second productand as the moduli of degree $d$ curves. On the locus where support of curve and zero dimensional subscheme are disjoint, this is an isomorphism.
Can somebody help me understand where (and thus why) this map fails to be an isomorphism? Unless of course it is an isomorphism? One could hope for an inverse via primary decomposition; I assume this fails to preserve flat families.