Question. Is the $F$-polynomial of an indecomposable quiver representation irreducible?
Here the $F$-polynomial is the generating function of the Euler characteristics of quiver Grassmannians, that is,
$$F_M =\sum_{e_1,...e_n} \chi( \mathrm{Gr}_{e_1,...e_n} M) x_1^{e_1} ... x_n^{e_n}$$
It is known that the $F$-polynomial of the direct sum of $M_1$ and $M_2$ is the product of $F_{M_1}$ and $F_{M_2}$.
No.
Consider the Kronecker quiver: two vertices with two arrows between them. Consider the representation with dimension vector $(3,3)$ given by the matrices $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.$$
The corresponding $F$-polynomial is computed in section 5 of Caldero-Zelevinsky, up to a monomial change of variables. (I won't try to get the monomial change of variables right.) It is $$\begin{matrix} y^3 & & & \\ + 3 y^2 & + x y^2 & & \\ + 3y & + 4 xy & +x^2 y & \\ +1 & +3x & +3x^2 & +x^3 \end{matrix}$$
This factors as $$\begin{pmatrix} y & \\ +1 & + x \end{pmatrix} \begin{pmatrix} y^2 & & \\ +2y & & \\ +1 & + 2x & x^2 \end{pmatrix}.$$
Numerical experimentation reveals an interesting pattern: Let $F(d)$ be the $F$-polynomial of the indecomposable representation with dimension vector $(d,d)$. Then $F(n)$ appears to be prime if and only if $n+1$ is prime. Presumably, this should be an obvious consequence of the Chebyshev polynomial formulas in Sherman-Zelevinsky, but I don't see it right now.
I can think of various ways you could try to salvage this, but I'll leave that project to you.
