Let $B_1 := \mathbb{C}[p_1, q_1], B_2 := \mathbb{C}[p_2, q_2]$ , and $(B_1 \otimes B_2)^{\mathcal{S}_2} = \mathbb{C}[p_1, q_1, p_2, q_2]^{\mathcal{S}_2}$. Here, the action of $\mathcal{S}_2$ is $(p_1, q_1) \leftrightarrow (p_2, q_2)$.
My question is: How to compute $(B_1 \otimes B_2)^{\mathcal{S}_2}$ in this case ? Or, what is a minimal basis ?
I think, if the group action is $\mathcal{S}_2 \times \mathcal{S}_2$; $p_1 \leftrightarrow p_2$, and $q_1 \leftrightarrow q_2$, then
$\mathbb{C}[p_1, p_2, q_1, q_2]^{\mathcal{S}_2 \times \mathcal{S}_2} \cong \mathbb{C}[p_1 + p_2, p_1p_2, q_1+q_2, q_1q_2]$.
But, I don't know how to compute $(B_1 \otimes B_2)^{\mathcal{S}_2}$.
The ring of invariants is generated by $$ p_1 + p_2,\ q_1 + q_2,\ (p_1 - p_2)^2,\ (p_1 - p_2)(q_1 - q_2),\ (q_1 - q_2)^2 $$ with a unique (evident) quadratic relation.
Indeed, denote $$ p = p_1 + p_2,\ q = q_1 + q_2,\ r = p_1 - p_2,\ s = q_1 - q_2. $$ Then $\mathbb{C}[p_1,p_2,q_1,q_2] = \mathbb{C}[p,q][r,s]$, and the group preserves $p$ and $q$ and acts by negation on $r$ and $s$. Now it is easy to see that the ring of invariants is generated over $\mathbb{C}[p,q]$ by $r^2$, $rs$, and $s^2$.