Let $K$ be a field, and $F\_K$ be the fraction field of the polynomial ring $R\_K$ in $n^2$ indeterminates $X\_{11},X\_{12},...,X\_{nn}$ over $K$.
Now set $A = (X\_{ij})\_{i,j} \in M\_n (F\_K)$, and let $\chi\_A$ be the characteristic polynomial of $A$. 


**Question** : Is it always true that $\chi\_A$ is irreducible over $F\_K$ ?


Some thoughts :


* Since $R\_K$ is a UFD, it is sufficient to check irreducibility in $R\_K[X]$.
* If $ |K| \geq n$, then $\chi\_A$ is separable (by evaluation of $A$ to a diagonal matrix with distinct diagonal entries).
* If $K[X]$ contains some irreducible $f$ of degree $n$, then this is true (just evaluate $A$ to the companion matrix of $f$ and observe that the evaluation morphism cannot increase degrees). For example, this yields the result when $K = \mathbb{Q}$ or when $K$ is a finite field.
* More generally, a non trivial factor of $\chi\_A$ is unlikely to exist, since it would give a generic factorization of the characteristic polynomial of a $n \times n$ matrix with coefficients in $K$.


Even in the case $K=\mathbb{C}$, I cannot prove (or disprove) the irreducibility of $\chi\_A$.



Characteristic polynomial of a generic n*n matric
If it were reducible, all $n\times n$ matrices over any field extension of $K$ would have reducible characteristic polynomials. But consider the companion matrix of an irreducible polynomial over some not algebraically closed field extension of $K$.


