TITLE: Minimal cardinality of a field where a polynomial has a root
QUESTION [13 upvotes]: Let $P(n)$ be the set of all monic polynomials of degree $n$ with integer coefficients, such that all coefficients have absolute value at most $2^n$.
Given a positive integer $n$ let us define $A(n)$ as the minimal number $m$ such that for any $f \in P(n)$ there exists a field $k$ with at most $m$ elements such that for some $x\in k$ we have $\bar f(x)=0$, where $\bar f$ is the reduction of $f$ modulo the characteristic of $k$.
Q: What is the asymptotic behaviour of $A(n)$? In particular is it true that $\liminf_{n} \frac{A(n)}{n} = 0$ ?

REPLY [5 votes]: No to the final question, stealing an idea from Andreas Thom. For all primes $p$ and degrees $d$ with $p^d>m$, there is a degree $d$ monic polynomial over $\mathbb F_p$ that has no roots in any field of characteristic $p$ of order $\leq m$. Indeed, an irreducible polynomial of degree $d$ does the trick.
Applying this, for any degree $d$ with $2^d > m$, there is a degree $d$ monic polynomial over $\mathbb Z/ P_m \mathbb Z$ where $P_m= \prod_{p\textrm{ prime}, p \leq m} p $ which does not have any roots in any field of cardinality at most $m$.
We can now lift this to an integral polynomial, with coefficients $<P_m$. So any $n$ with $2^n \geq P_m$ and $2^n > m$ has $A(n)>m$. Obviously the first condition is the interesting one, so we have $$n \approx \log_2 P_m =  \sum_{p \textrm{ prime}, p \leq m} \frac{\log p}{\log 2} \approx \frac{m}{\log 2}$$ by the prime number theorem.
So $A(n) $ is greater than, approximately, $n \log 2$