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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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    $\begingroup$ What do you (or what does the book) mean by "for some $p$ irreducible mod $q$"? $\endgroup$ – Eric Auld Aug 4 '13 at 0:19
  • $\begingroup$ It might help if we could see the source of the proof. $\endgroup$ – Gerry Myerson Aug 4 '13 at 0:29
  • $\begingroup$ @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". $\endgroup$ – modnar Aug 4 '13 at 1:59
  • $\begingroup$ 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. $\endgroup$ – user88908 Aug 4 '13 at 5:32
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Let $F$ be as in the theorem.

Let $(z_1,\ldots, z_k)\in\Bbb Z^k$ be arbitrary.

Let $p,q$ be any two distinct primes. It is well-known that there exist irreducible polynomials $\in\Bbb F_q[X]$ of arbitrary degree, i.e., there is $g(X)=x^k-b_1X^{k-1}-\ldots -b_k\in\Bbb Z[X]$ such that $g(X)\bmod q\in \Bbb F_q[X]$ is irreducible. By the Chinese Remainder Theorem, we find $a_1,\ldots, a_k\in\Bbb Z$ such that $$\tag1a_i\equiv z_i\pmod p$$ and $$\tag2a_i\equiv b_i\pmod q.$$ From the $(1)$, we conclude that $F(z_1,\ldots, z_k)\equiv F(a_1,\ldots, a_k)\pmod p$, whereas $(2)$ tells us that $X^k-a_1X^{k-1}-\ldots -a_k$ is irreducible in $\Bbb Z[X]$. Therefore, $F(a_1,\ldots, a_k)=0$ and so $$F(z_1,\ldots, z_k)\equiv 0\pmod p.$$ As $p$ was an arbitrary prime, it follows that $$ F(z_1,\ldots, z_k)=0.$$ As this holds for all $z_1,\ldots ,z_k\in\Bbb Z$, we ultimately conclude that $F$ is the zero polynomial.

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