I'm trying to understand a proof of the following theorem (from section II of Hall's paper *An Isomorphism Between Linear Recurring Sequences and Algebraic Rings*):

If $F(a_1, \ldots, a_k)$ is a polynomial in $a_1, \ldots, a_k$ with integer coefficients and $F = 0$ whenever $a_1, \ldots, a_k$ are integers such that $f(x) = x^k - a_1x^{k - 1} - \cdots - a_k$ is irreducible, then $F \equiv 0$.

The proof provided was that taking $f(x)$ modulo $p$ for some $p$ (prime?) irreducible mod $q$, we find that $F = 0$ (modulo $p$). Since $F \equiv 0$ (mod $p$) for any appropriate $a_1, \ldots, a_k$ for arbitrary $p$, we have that $F \equiv 0$.

I'm confused about the first part of the proof because irreducibility over integers doesn't imply irreducibility modulo $p$ for any prime $p$. Also, does it make sense to consider $q$ to be an arbitrary positive integer?

Does the theorem assume that we're considering irreducibility mod $q$? Does this change anything? Also, what does "e" on p. 200 (after Theorem 4.1) refers to? There was no mention of it earlier in the paper and I don't see why it should be the constant $e$.

Note: This is cross-posted from math.SE (and slightly edited): https://math.stackexchange.com/questions/459173/if-fa-1-ldots-a-k-0-whenever-a-1-ldots-a-k-are-integers-such-that-fx