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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?

EDIT: Does the theorem assume that we're considering irreducibility mod $q$? Does this change anything?

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What do you (or what does the book) mean by "for some $p$ irreducible mod $q$"? –  Eric Auld Aug 4 '13 at 0:19
It might help if we could see the source of the proof. –  Gerry Myerson Aug 4 '13 at 0:29
@EricAuld The "irreducible mod $q$" describes $f$ considered mod $p$. The proof is contained in section II of Hall's paper "An Isomorphism Between Linear Recurring Sequences and Algebraic Rings". –  modnar Aug 4 '13 at 1:59
This is just a comment . I hope it could help. Consider prime p, let the constant is p, and the other coefficients are multiple of p, then by Eisenstein criterion, f is irreducible, so F is zero on such kinds of coefficients. And we can switch the prime p arbitrary, see if those informations imply theresult. –  user88908 Aug 4 '13 at 5:32
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