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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

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2 Answers 2

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

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The first example goes back to Hilbert. See also Guralnick (Robert), Schacher (Murray) and Sonn (Jack), Irreducible polynomials which are locally reducible everywhere, Proc. Amer. Math. Soc. 133 (2005), no. 11, 3171–3177. –  Chandan Singh Dalawat Oct 3 '11 at 14:04
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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.

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The field $K=\mathbf{Q}(\sqrt{13},\sqrt{17})$ would do, for example. I learnt this example in Cassels-Froehlich. The Galois group is $(\mathbf{Z}/2)^2$ of course, but the point is that only $13$ and $17$ ramify, and $13$ splits in $\mathbf{Q}(\sqrt{17})$ and $17$ splits in $\mathbf{Q}(\sqrt{13})$. –  Kevin Buzzard Oct 2 '11 at 20:35
    
Expanding on Kevin's answer, if $f(x)$ is monic irreducible in ${\mathbf Z}[x]$ then there's a nonobvious consequence of $f(x) \bmod p$ being irreducible for even one prime $p$. Write $G$ for the Galois group of $f(x)$ over ${\mathbf Q}$ and $H$ for the subgroup with fixed field ${\mathbf Q}(\alpha)$ for a root $\alpha$ of $f(x)$, so the coset space $G/H$ has size $\deg f$. If $f(x) \bmod p$ is irreducible for some prime $p$ then there's an element of $G$ with order $\deg f$ whose powers represent $G/H$. In particular, if ${\mathbf Q}(\alpha)$ is Galois over ${\mathbf Q}$ then $G$ is cyclic. –  KConrad Oct 3 '11 at 13:02
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The upshot of the previous comment is that if $K/{\mathbf Q}$ is any non-cyclic Galois extension and we take for $f(x)$ the minimal polynomial in ${\mathbf Z}[x]$ of any algebraic integer generating $K$ over ${\mathbf Q}$ then $f(x)$ is irreducible over ${\mathbf Q}$ but it's guaranteed that $f(x) \bmod p$ is reducible for every prime $p$ (if it were reducible for even one $p$ then ${\rm Gal}(K/{\mathbf Q})$ would be cyclic). The simplest such $K$ are biquadratic fields since their Galois groups over ${\mathbf Q}$ are products of two groups of order 2. –  KConrad Oct 3 '11 at 13:07
    
In Ottem's answer, for example, $x^4 - 72x^2 + 4$ has root $\sqrt{17}+\sqrt{19}$ and ${\mathbf Q}(\sqrt{17}+\sqrt{19}) = {\mathbf Q}(\sqrt{17},\sqrt{19})$ is biquadratic over $\mathbf Q$. –  KConrad Oct 3 '11 at 13:08
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KConrad, the questioner asked for $f$ reducible mod all integers, not just all primes. If the biquadratic extension had a prime $p$ with say "$e=f=2$" then $f$ would be reducible mod $p$, but irreducible modulo a sufficiently large power of $p$. –  user18237 Oct 3 '11 at 14:43
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