If $F(a_1,\ldots,a_k)=0$ whenever $a_1,\ldots,a_k$ are integers such that $f(x)=x^k-a_1x^{k-1}-\cdots-a_k$ is irreducible, then $F\equiv0$

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

• He means take primes $p,q$, polynomials $f_p,f_q$ such that $f_p$ is arbitrary modulo $p$ and $f_q$ is irreducible modulo $q$, then find $f$ which is $f_p \pmod p, f_q \pmod q$ by CRT. Then $f$ is irreducible (a factorization of $f$ gives a factorization of $f_q$) so $F=0$ and thus $F$ vanishes identically $\pmod p$ and (as $p$ is arbitrary) $F$ vanishes identically. This should have been answered in MSE... – Felipe Voloch Aug 7 '13 at 3:14
• Thanks for the clarification! I thought so too, but there weren't any answers on MSE even a while after the question was posted. – modnar Aug 7 '13 at 3:42

Pick two distinct primes $p$ and $q$. Now choose any monic irreducible polynomial $f_q$ of degree $k$ modulo $q$ (i.e. in ${\bf F}_q$); these exist. Also choose $a_1, \dots, a_k$ arbitrary mod $p$ and let $f_p = x^k - a_1 x^{k-1} + \dots + (-1)^k a_k$. Then by CRT, there is a common lift $f(x) \in {\bf Z}[x]$ which agrees with $f_p$ mod $p$ and with $f_q$ mod $q$. Since $f_q$ is irred, so is $f$. So $F(a_1, \dots, a_k) = 0$ for every such choice, and so $F \equiv 0 \pmod p$ etc.
Aso for your second question, I'm not sure, but it looks like you pick $h$ to be one of the $h_i$ from equation (4.3) and then $e$ is the associated $e_i$.