Is it true that if a polynomial $p(x)$ is irreducible over $\mathbb{Q}$ then it is also irreducible over $\mathbb{Z}$?

Question

I'd expect that it'd be trivially true because $\mathbb{Z} \subset \mathbb{Q}$, so any factorisation in $\mathbb{Z}$ would also be a valid factorisation in $\mathbb{Q}$. However, I realise that I could be missing something.

Thanks in advance.

Answer 1

Ultimately this depends on your definition of irreducibility. Taking the definition used on both wikipedia and MathWorld, where a polynomial is irreducible if it cannot be factored into two non-constant polynomials you're pretty much correct, assuming that $p(x) \in \mathbb{Z}[x]$.

Suppose for contradiction that $p(x) = f(x)g(x)$ with $f(x),g(x) \in \mathbb{Z}[x]$. Then $f(x),g(x) \in \mathbb{Q}[x]$ also, and so we have a factorisation in $\mathbb{Q}[x]$, a contradiction.

Answer 2

This statement is false. For example, $p(x)=2x-2$ is obviously irreducible over $\mathbb{Q}$ but it is reducible over $\mathbb{Z}$ since $p(x)=2(x-1)$ and $2,x-1$ are not units in $\mathbb{Z}[x]$.

However the converse is true by Gauss' lemma.