When are immanants irreducible? For a partition $\lambda$ let $\chi_\lambda$ be the corresponding irreducible representation of the symmetric group $S_n$.
Let $\mathrm{Imm}_\lambda(x) = \sum\limits_{\pi \in S_n} \chi_\lambda(\pi) x_{1 \pi(1)} \dotsm x_{n \pi(n)}$ be the immanant corresponding to $\lambda$.

Question 1: Is $\mathrm{Imm}_\lambda$ irreducible over $\mathbb{Z}$? If not, in how many irreducible factors does it split depending on $\lambda$?

It seems for all partitions of $n$ for $n$ at most 5 those polynomials are irreducible.

Question 2: Let $R=\mathbb{Q}(x_{i,j})/I$ where $I$ is the ideal generated by the $\mathrm{Imm}_\lambda$. What is the Krull dimension of $R$?
The sequence starts with $2,6,12$ for $n=2,3,4$.

 A: They are irreducible. Actually the polynomials of the form
$$
P:=\sum_{\pi} F(\pi)x_{1\pi(1)}\ldots x_{n\pi(n)}
$$
are rarely reducible. Assume that $P=QR$. Consider both sides as the polynomials in the $i$-th column. Since $P$ has degree 1, it follows that one of $Q,R$ has degree 1 and another does not depend on the $i$-th column. That is, all columns are partitioned to $Q$-columns (which are represented in $Q$) and $R$-columns. Analogously the rows. Consider the element $x_{ij}$ where $i$ is the $Q$-row and $j$ an $R$-column (such $i$, $j$ exist since $Q$, $R$ are not-constant). We see that neither $Q$ nor $R$ depends on $x_{ij}$. Therefore we should have $F(\pi)=0$ whenever $\pi(i)=j$. But the irreducible characters can not satisfy this very restrictive property (since they depend only on the conjugacy class of $\pi$, we would get that $\chi({\rm id})=0$ when $i=j$ and $\chi(\pi)=0$ for all other permutations when $i\ne j$. The second option corresponds to a reducible (regular) character, the first is absurd.)
