Hilbert Scheme of Points of Riemann Sphere

The Hilbert scheme of $n$ points on $\mathbb P^1$ is the space of unordered families of $n$ non-necessarily distinct points of $\mathbb P^1$, and we will see that it is isomorphic to $\mathbb P^n$.
To understand this concretely it is best to interpret $\mathbb P^n$ as the projective space associated to the vector space $\mathbb C[x,y]_n$ of homogeneous polynomials of degree $n$ in two variables .
In other words, the polynomial $a_0x^n+a_1x^{n-1}y+\cdots+a_ny^n$ is seen as the point $[a_0:a_1:\cdots:a_n]\in \mathbb P^n$
The required isomorphism then comes from the morphism $F:(\mathbb P^1)^n\to \mathbb P^n$ sending an $n$-tuple $([u_i:v_i])_{i=1}^n$ to the point corresponding to the polynomial $\Pi_{i=1}^n (v_i x-u_iy)\in \mathbb C[x,y]_n$.
Dividing out by the action of the symmetric group $S_n$ on $(\mathbb P^1)^n$ we get the required isomorphism $$f:\operatorname {Hilb}^n(\mathbb P^1)=(\mathbb P^1)^n/S_n=\operatorname {Sym}^n(\mathbb P^1)\stackrel {\cong}{\to} \mathbb P^n$$

$Hilb^n(\mathbb{P}^1) \cong \mathbb{P}^n$.

EDIT (an explanation). A subscheme $Z$ of length $n$ is determined by its ideal $I_Z$ and its embedding $I_Z \hookrightarrow \mathcal{O}$. Every ideal on a smooth curve is invertible, hence $I_Z \cong \mathcal{O}(-n)$, and its embedding is determined by an element of $Hom(\mathcal{O}(-n),\mathcal{O}) = H^0(\mathbb{P}^1,\mathcal{O}(n)) = k^{n+1}$. Thus $Hilb^n(\mathbb{P}^1) \cong \mathbb{P}(k^{n+1}) \cong \mathbb{P}^n$.

Here is another, extremely geometric, vision of the Hilbert scheme $\mathcal H=\mathcal H^{[n]}$ of $n$ points in the projective line $\mathbb P^1_{x:y}$, in other words the Hilbert scheme of the subschemes $S\subset \mathbb P^1$ with Hilbert polynomial the constant polynomial $n\in \mathbb Q[T]$.
Like all hypersurfaces of any projective space, each $S=S_a\subset \mathbb P^1_{x:y}$ has an equation, namely $f(a;x,y)= a_0x^n+a_{n-1}x^{n-1}y+\cdots+a_ny^n=0$ and the Hilbert scheme is the hypersurface $$ \mathcal H=V(f(a;x,y)) \subset \mathbb P^n_a\times \mathbb P^1_{x:y} $$ endowed with its projection $$p:\mathcal H\to \mathbb P^n_a:([a_0:a_1:\cdots:a_n],[x:y])\mapsto [a_0:a_1:\cdots:a_n]$$ so that $p^{-1}(a)=S_a\subset \mathbb P^1$, the subscheme with equation $a_0x^n+a_{n-1}x^{n-1}y+\cdots+a_ny^n=0$.
The Hilbert scheme $\mathcal H$ is smooth and connected but its fibres over $\mathbb P^n$ are not: the non-smooth fibres $S_a$ of $p$ correspond to those $a\in \mathbb P^n$ such that the discriminant $$\operatorname {discr} \: (a_0x^n+a_{n-1}x^{n-1}y+\cdots+a_ny^n)\in \mathbb C[a_0,\cdots,a_n]$$ is zero.
This is in line with the fact that $\mathbb P^n$ being simply connected cannot have an étale connected covering of degree two and thus $p:\mathcal H\to \mathbb P^n$ must have a non-empty ramification locus.

Remarks
1) The construction shows that $\mathcal H$ is projective so that all morphisms to $\mathbb A^1_\mathbb C$ are constant, which answers one of Hamed's questions.
2) The connectednes of $\mathcal H$ is an extremely special case of Hartshorne's brilliant Ph.D: he proved that every Hilbert scheme (corresponding to some fixed Hilbert polunomial) is connected
3) The smoothness of $\mathcal H$ is a low-tech calculation using only the implicit function theorem of advanced calculus.
These calculations can easily be generalized to the case of the Hilbert scheme of hypersurfaces of $\mathbb P^N$.
Unfortunately these calculations cannot be found in the literature (as far as I know), and they could be the theme of another question...