# Irreducibility in finite field implies irreducibility in $\mathbb{Z}$?

Let $p$ be a prime, $f$ be a polynomial of $\mathbb{Z}[x]$. Suppose that $f$ is irreducible in $\mathbb{F}_p[x]$.

My question is : Is $f$ irreducible in $\mathbb{Z}[x]$ ?

This question is originally came from the example 4.29 from the notes Field and Galois theory of J.S.Milne, in that he gave an implication of " a polynomial is irreducibile mod 3(still reducible mod 2) then it is irreducible in $\mathbb{Z}$.

Update I have more question after reading the comments and answer below. So, if $f$ is monic, what is the condition for irreducible mod $p$ for some prime $p$ implying irreducibility of $f$ over $\mathbb{Z}[x]$ ? Must it be "irreducible modulo $p$ for all the prime $p$ (not for some) ?

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Not true, as the counterexamples below show. But if the leading coefficient of $f$ is not divisible by $p$, then yes (because then the degrees of any factors are preserved upon reduction modulo $p$). – Ted Jan 6 '13 at 4:37

To elaborate on Ted's comment, if the leading coefficient is not zero in the finite field, then this actually is true. The easiest way to see why is to consider the contrapositive:

$~\bf($irreducible in $\Bbb F_p[x]\implies$ irreducible in $\Bbb Z[x]\bf)\iff ($reducible in $\Bbb Z[x]\implies$ reducible in $\Bbb F_p[x]\bf)$

If $f(x)=a(x)b(x)$ is true in $\Bbb Z[x]$ then it is true in $\Bbb F_q[x]$; the only obstacle to the latter implication then is if one of the $a(x)$ or $b(x)$ is a scalar after reducing modulo $p$, which can't happen if the degree of $f$ is preserved by the modulo reduction.

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So, if a polynomial $f$ is irreducible mod $p$ for some prime $p$, and reducible mod $q$ for some other prime $q$, we can not say about the irreducibility of $f$ over $\mathbb{Z}[x]$, right ? I still have troubles with Milne's example. – knot Jan 6 '13 at 8:35

Certainly not. For example $x+px^2$ is irreducible mod $p$, as it is equivalent to $x$.

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Look at $3x^2-x-2$ with $p=3$.

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