If $\overline{f}(x)$ is irreducible in $\mathbb{Z}_p[x]$ then $f(x)$ is irreducible in $\mathbb{Z}[x]$

Question

Let $f(x)=a_0+\dots +a_n x^n \in \mathbb{Z}[x]$. Let $p$ be a prime with $p \nmid a_n$.

We define $\overline{f}(x)=\overline{a_0}+\dots +\overline{a_n} x^n \in \mathbb{Z}_p[x]$

How can I show that if $\overline{f}(x)$ is irreducible in $\mathbb{Z}_p[x]$ then $f(x)$ is irreducible in $\mathbb{Z}[x]$ and therefore also in $\mathbb{Q}[x]$ ?

Could you give me some hints?

Answer 1

If you factorize $f=gh$ in $\Bbb Z[x]$, then $\bar f=\bar g\bar h$ in $\Bbb Z_p[x]$.

Answer 2

What you want to prove is false, for instance $ P = p T^2 + (p+1)T +1 = (pT+1)(T+1)$ is not irreducible in $\mathbf{Z}[T]$ but its reduction $Q = T+1$ mod $p$ is obviously irreducible in $(\mathbf{Z}/p \mathbf{Z}) [T]$...

You need other conditions on your polynomial $P$ from $\mathbf{Z}[T]$ than only its irreducibility to ensure that its image $Q$ in $(\mathbf{Z}/p \mathbf{Z}) [T]$ will be irreducible...

For instance if $Q$ and $P$ have the same degree, then $P$ will be irreductible in $\mathbf{Q}[T]$ if its image $Q$ in $(\mathbf{Z}/p \mathbf{Z}) [T]$ is irreducible. This completes your initial unsufficient hypothesis.

You have another sufficient condition, the so-called Einsenstein criterion : if $p$ divides all coefficient of $P$ but not the one from the higher degree power of $T$ and if $p^2$ does not divide $P(0)$, then again, you have irreducibility over $\mathbf{Q}[T]$.

Now, having irreducibility in $\mathbf{Q}[T]$, how do you get irreducibility over $\mathbf{Z}[T]$ ?

Here come the primitive polynomials. (Theory valid over any factorial ring.) You can see this for a start, but the result is the following : a polynomial with integer coefficients is irreducible over $\mathbf{Z}[T]$ if and only if it is irreducible over $\mathbf{Q}[T]$. And it is entirely based on the Gauss lemma.

So you have at least two conditions.