Let $f(x)$ and $g(x)$ be polynomials with integer coefficients such that $f(x)>0$ and $g(x)>0$ for all real values of $x$. Suppose that for every integer $n$, if $f(x)=0$ has a solution mod $n$, then $g(x)=0$ has a solution mod $n$. What can generally be said about $f(x)$ and $g(x)$?

My guess is that $g(x)$ must have a factor $h(x) \in \mathbb{Z}[x]$ such that $f(x)=h\circ k(x)$ for some $k(x) \in \mathbb{Z}[x]$.

  • 4
    $\begingroup$ Without the positivity condition, non-isomorphic number fields with the same Dedekind zeta function should give counterexamples. I don't know if the positivity condition would rule these out but I suspect not. $\endgroup$ – Felipe Voloch Apr 28 '18 at 21:30
  • $\begingroup$ Yes, without the positivity conditions, one can take $f(x)=x$ and $g(x)$ to be any intersective polynomial. $\endgroup$ – Marco Apr 28 '18 at 21:53

The question becomes more tractable if we assume that $f$ and $g$ are monic irreducible polynomials, and we restrict to $n$'s coprime to the discriminants of $f$ and $g$. In this case, by Hensel's lemma and standard facts on discriminants, the modified question remains unchanged if we restrict to primes $n$ coprime to the discriminants of $f$ and $g$.

Let us consider the corresponding number fields $K=\mathbb{Q}[x]/(f(x))$ and $L=\mathbb{Q}[x]/(g(x))$ whose discrimants divide the discriminants of $f$ and $g$, respectively. Then, in the special case when $L$ is Galois (i.e. $L$ contains all the roots of $g$), the modified condition holds for $f$ and $g$ if and only if $K$ contains $L$. This follows from Prop. 15 in Ch. VIII-5 of Weil: Basic number theory.

Example. Let $p$ be a prime congruent to $1$ mod $4$. Let $f(x)=x^{p-1}+\dots+1$ be the $p$-th cyclotomic polynomial, and let $g(x)=x^2-p$. Consider an arbitrary prime $n\nmid 2p$. Then, $f(x)$ has a root modulo $n$ if and only if $n$ is congruent to $1$ mod $p$, and $g(x)$ has a root modulo $n$ if and only if $n$ is a quadratic residue mod $p$. Indeed, in this case, $K=\mathbb{Q}(e^{2\pi i/p})$ is the $p$-th cyclotomic field and $L=\mathbb{Q}(\sqrt{p})$ is its unique quadratic subfield, and we just recorded the splitting primes in $K$ and $L$.

  • $\begingroup$ It seems the positivity condition did not matter in this argument. Can one use the Inclusion theorem to derive the same conclusion for general irreducible f(x) and g(x)? I am in particular interested in the case $f(x)=x^2+1$. $\endgroup$ – Marco Apr 28 '18 at 23:13
  • 3
    $\begingroup$ @Marco: The inclusion theorem cannot be extended to general monic irreducible $f(x)$ and $g(x)$, for the reason explained by Felipe Voloch's comment (under your post). But $f(x)=x^2+1$ is a special case, for which I don't know the answer. The question is interesting! $\endgroup$ – GH from MO Apr 29 '18 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.