[Math] $f$ irreducible over $\mathbb{Z}_{p}$ implies $f$ is irreducible over $\mathbb{Q}$

Question

Let $f \in \mathbb{Z}[x]$ be a non-constant polynomial and let $p$ be a prime number which is not a divisor of the leading coefficient of $f$. I need to prove that if $f$ is irreducible over $\mathbb{Z}_{p}$, then $f$ is irreducible over $\mathbb{Q}$.

I decided to approach this proof by proving the contrapositive: i.e., that $f$ is reducible over $\mathbb{Q}$ $\implies $ $f$ is reducible over $\mathbb{Z}_{p}$.

Now, it is certainly the case that $f$ is irreducible over $\mathbb{Q}$ if and only if it is irreducible over $\mathbb{Z}$. Therefore, if $f$ reducible over $\mathbb{Q}$ implies that $f$ is reducible over $\mathbb{Z}$.

So, the proof therefore reduces to showing that $f$ reducible in $\mathbb{Z} \implies$ $f$ reducible in $\mathbb{Z}_{p}$.

To that effect, suppose $f$ is reducible in $\mathbb{Z}$. Then, $\exists g,h \in \mathbb{Z}[x] $ such that $f(x) = g(x)h(x)$. Now, the fact that $p$ is not a divisor of the leading coefficient of $f$ means that when we reduce $\,f \mod p$ (call this result $\overline{f}$), $f$ and $\overline{f}$ have the same degree.

From this point, I am not sure how to proceed. It seems to make sense to me intuitively that if we reduce $\, f \mod p$, we can reduce $\, g \mod p$ and $\, h \mod p\,$ as well so that we will have shown that $\overline{f}$ is factorable/reducible. But, I'm not sure how to show this in a rigorous way. Could anyone please help me on how I should proceed?

Thank you.

Answer 1

Suppose $f$ is reducible over $\mathbb{Z}$. Then $f$ has degree $n>1$ and $f(x) = g(x)h(x)$ with degree $g(x) = j>0$, degree $h(x) = k >0$.

Suppose also that $p \not| a_n$, the coefficient of $x^n$ in $f(x)$.

We have $a_n = b_j c_k$, where $b_j$ and $c_k$ are the leading coefficients of $g$ and $h$.

$\mathbb{Z}_p$ is an integral domain (in particular, it's a field), so $a_n \not\equiv 0$ mod $p$ implies that $b_j \not\equiv 0$ mod $p$ and $c_k \not\equiv 0$ mod $p$. Then deg $g$ mod $p$ > 0 and deg $h$ mod $p$ > 0. So $f(x)$ is reducible over $\mathbb{Z}_p$.

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By the way, maybe this is obvious, but be careful: the statement is

$\exists$ a $p$ such that $f$ irreducible over $\mathbb{Z}_p$ $\implies$ $f$ irreducible over $\mathbb{Q}$.

So the contrapositive is

$f$ reducible over $\mathbb{Q}$ $\implies$ $f$ reducible over $\mathbb{Z}_p$, $\forall$ $p$.

It didn't matter here, but sometimes leaving out the quantifiers can cause confusion on this particular test of irreducibility.