TITLE: Polynomial reducible modulo every integer
QUESTION [6 upvotes]: Hi,
let $f\in\mathbb{Z}[X]$ be a monic polynomial.  Assume that the reduction of $f$ modulo $m $ is reducible for all integers $m\geq 2$.
Q1:  Is $f$ reducible in $\mathbb{Z}[X]$ ?
I've thought about this question without making substantial progress. If $p$ is a prime number , $p\nmid disc(f)$, then $f$ mod $p$ is  separable, so if there is a nontrivial factorisation, we may find a factorisation into product of coprime monic polynomials. By Hensel's lemma, $f$ is reducible in $\mathbb{Z}_p[X]$.
If $p\mid disc(f)$, bad things could happen. For example, $X^4+1$ is reducible mod $2$, but irreducible mod $4$.
So I'm stuck here, and may be this approach will be not the right one but I have several related questions :
Q2: Assume that $p\mid disc(f)$. Does the fact that $f$ is reducible modulo $p^r$ for all $r\geq 1$ implies that $f$ is reducible in $\mathbb{Z}_p[X]$ ?
Q3: Let $f\in\mathbb{Z}[X]$ be a monic polynomial. Assume that $f$ is reducible in $\mathbb{Z}_p[X]$ for all prime integers $p$. Does $f$ is reducible in $\mathbb{Z}[X]$ ?
Thanks!
Greg

REPLY [17 votes]: The polynomial $x^4-72x^2+4$ is irreducible over $\mathbb{Z}$, but reducible modulo every integer. 
See E. Driver, P. A. Leonard and K. S. Williams Irreducible Quartic Polynomials with Factorizations modulo $p$
The American Mathematical Monthly
Vol. 112, No. 10 (Dec., 2005), pp. 876-890

REPLY [5 votes]: The answer to Q2 is yes, while the answer to Q1 and Q3 (which are equivalent by Q2 and CRT) is no.
Q2 is essentially formal.  The set of non trivial factorizations into monics of $f$ mod $p^r$ form an inverse system, and the inverse limit is nonempty (why?)  So take your compatible system of factorizations mod $p^r$ and take inverse limits coefficient by coefficient and you get a factorization over $\mathbb{Z}_p[x]$.
As for Q1/Q3, $f\in\mathbb{Z}[X]$ irreducible monic defines a number field $K=\mathbb{Q}[x]/(f)$ and $f$ is irreducible over $\mathbb{Z}_p$ if and only if there is only one prime of $K$ lying over $p$.  So to find a counterexample, you need to find a number field with more than one prime above every rational prime $p$.  This can easily be arranged, for example with $K$ a suitable biquadratic extension.