Let us have $\mathbb{P}^n\times\cdots\times \mathbb{P}^n$ ($d$ copies of $\mathbb{P}^n$), and we have the symmetric group $S_d$ act by permuting the factors.
Is the quotient projective? I have the feeling that we can embed the quotient into some large projective space via the invariant polynomials, but the symmetric polynomials have different degrees, so I am confused how to proceed.
Thank you for any help or direction
Send $(v_1, \ldots, v_d)$, where $v_1, \ldots, v_d$ are non-zero elements in $\mathbb{C}^{n+1}$ to their symmetric tensor product $v_1 \odot \cdots \odot v_d$ (over $\mathbb{C}$). This induces a holomorphic map from the product of $d$ copies of $\mathbf{P}^n$, quotiented out by $S_d$, into $\mathbf{P}^N$, where $N = \binom{n+d}{d} - 1$. It remains to check that this map is an embedding (which I think it is, but please double check). This is a kind of symmetric version of the Segre embedding, and I am sure it has a name in algebraic geometry.