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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 $\chi_A$.

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Assume it is reducible. Then it stays so after adjoining a new variable $X$. I think over the field $K(X)$ you will find an irreducible polynomial $f$ of degree $n$, a candidate would be $f(T)=T^n-X$. –  doug May 18 '12 at 18:43
In case it's not obvious, this polynomial is always irreducible, by Eisenstein's criterion modulo $X$. –  Will Sawin May 18 '12 at 23:31

1 Answer 1

up vote 8 down vote accepted

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 $K$.

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My answer is essentially the same as Xogn Ambandl's comment. Moreover, I didn't mention a specific field extension with a degree n irreducible polynomial. –  Benjamin Steinberg May 18 '12 at 18:51
Hum, I didn't saw the possibility of evaluating in extension fields. Thanks for the neat argument. –  js21 May 18 '12 at 19:34

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