Is it true that $\forall f \in \mathbb{Z}[x],\exists p$ prime number, s.t. $ f $ is reducible in $\mathbb{F}_p$? I've seen this problem before but I can't find it. And when I try to understand the picture of $\operatorname{Spec}(\mathbb{Z}[x])$. I want to have a certain answer about this problem
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2How about $f=1$? Maybe there is some additional hypothesis on $f$ that you want? – Eric Wofsey Jul 02 '19 at 03:08
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Obviously this is not true for polynomials of degree $\leq 1$. It is true for any polynomial $f$ of degree $>1$. To prove this, suppose first that there exists $n\in\mathbb{Z}$ and a prime $p$ which does not divide the leading coefficient of $f$ such that $p\mid f(n)$. Then when we reduce mod $p$, $x-n$ is a factor of $f$, and $f$ still has degree $>1$, so $f$ is reducible.
It remains to prove that such $n$ and $p$ exist. Let $p_1,\dots,p_m$ be all the primes dividing the leading coefficient of $f$. If no such $n$ and $p$ exist, then $f(n)$ has the form $\pm p_1^{d_1}\dots p_m^{d_m}$ for all $n\in\mathbb{Z}$. Note that the number of numbers of this form in an interval $[-N,N]$ is at most $$2\cdot \prod_i \log_{p_i}(N)<C\log(N)^m$$ for some constant $C$. On the other hand, if $f$ has degree $d$, then $|f(n)|$ grows like $n^d$ (within a constant factor) as $n$ gets large, and so the number of values of $f(n)$ in $[-N,N]$ must be at least $DN^{1/d}$ for some constant $D>0$. When $N$ is sufficiently large, $DN^{1/d}>C\log(N)^m$, and so there must be some value $f(n)$ in $[-N,N]$ which does not have the form $\pm p_1^{d_1}\dots p_m^{d_m}$.
Eric Wofsey
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1An alternative : $a_k= \prod_j p_j^{e_j}$ the leading coefficient, $f(l) \ne 0$, take $d$ large enough such that $f(l) \not \equiv 0 \bmod p_j^d$, let $g = \gcd(f(l), a_k^\infty)$, then for every $n$, $f(l + n a_k^d) \equiv f(l) \bmod p_j^d$ so $\frac{f(l + n a_k^d)}{g}$ is coprime with $a_k$ and any of its prime divisors works – reuns Jul 02 '19 at 03:28
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