Let's recall a linear algebra fact: Let $A$ be an $n\times n$ matrix over a field $K$ and $\chi_A(t)$ be its characteristic polynomial. Then if $\chi_A(t)$ is reducible, $A$ would have a proper invariant subspace.

My intention is to find an analogy in the case of $m$ matrices. That is, given $m$ $n\times n$ matrices $A_1, A_2,\cdots , A_m$, can we find a polynomial $p$, with several variable maybe, so that if $p$ is reducible, then $A_1, A_2,\cdots , A_m$ have common invariant subspace.

A reasonable way to do this is to consider the polynominal $D(x_1,\cdots, x_m)=det(x_1a_1+\cdots+ x_mA_m)$. So clearly if D is irreducible, then $A_1, A_2,\cdots , A_m$ would have NO common invariant subspaces. However, it is mentioned by David E Speyer in here that the converse of this statement is in general not true.

My questions are:

(1) Where can I find the example to the case: $D(x_1,\cdots, x_m)$ is reducible while $A_1, A_2,\cdots , A_m$ have no common invariant subspace? If I choose $n$ large enough, is there such example also?

(2) Do we have other kind of polynomials suit this problem? For example, one can consider the polynomial $det(X_1\otimes A_1+\cdots +X_m\otimes A_m)$ whose variables are entries of $X_1,\cdots, X_m$ (These kind of function are semi-invariants of certain Kronecker quiver representation).