Let $S^n\mathbb{P}^1$ be $n$-th symmetric power of $\mathbb{P}^1$ over $k$, this is just given by the quotient by the action of the permutation group $\Sigma_n$ on the $n$-th cartesian product of $\mathbb{P}^1$ . I want to prove that $S^n\mathbb{P}^1 = \mathbb{P}^n$, but first I'd like to understand in detail the simpler case when $n=2$.
We know that $$ k[\mathbb{P}^1 \times \mathbb{P}^1] \cong \bigoplus_{d=0}^{\infty} k[x_0,x_1]_d \otimes k[x_0,x_1]_d $$ where $k[x_0,x_1]_d$ is the vector space generated by the monomials of degree $d$ in $x_0, x_1$. Now we need to conclude $k[\mathbb{P}^1 \times \mathbb{P}^1]^{\Sigma_2} \cong k[\mathbb{P}^2]$ i.e. the ring of regular functions invariant to the action of $\Sigma_2$ is isomorphic to $k[\mathbb{P}^2]$. My idea consists in simply defining isomorphisms between the gradings $$ (k[x_0,x_1]_d \otimes k[x_0,x_1]_d)^{\Sigma_2} \to k[x_0,x_1,x_2]_d $$ We also know that $\mathrm{dim}_k \> k[x_0,x_1,x_2]_d = \binom{d+2}{d} = \frac{(d+2)(d+1)}{2}$.
Is this reasoning correct so far?
And if so, how can one find a basis for $(k[x_0,x_1]_d \otimes k[x_0,x_1]_d)^{\Sigma_2}$ or at least determine its dimension?