Let $K=\mathbb{Q}(a_1,a_2,\cdots)$ be an extension of $\mathbb{Q}$ where the $a_n$ are eigenvalues of an infinite family of operators $T_n$ who commute and whose characteristic polynomials have integer coefficients. Why is $K$ finite ? It is taken for granted in a draft I'm reading (which, besides, contains lots of typos and falsehoods).
The characteristic polynomial part means that the $a_i$ are algebraic numbers so $K$ would certainly be finite if there were a finite number of them, but this is not the case a priori, so what am I missing? I am not sure how to interpret the fact that the $T_n$ commute, this doesn't sound too relevant to me... If the $T_n$ are diagonalizable then we would get a common basis of eigenvectors but this still doesn't help. I assume the commuting property adds some kind of algebraic dependence between the eigenvalues but this is not obvious to me.
Any help or hints are welcome