A polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$

Question

Is it true that a polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$? I read this in "Polynomials" by Victor Prasolov and they also give a proof of this, but I had previously read in a book from my country that if a polnomial $f\in \mathbb{Z}[X]$ is irreducible over $\mathbb{Z}$ then it is irreducible over $\mathbb{Q}$(this is Gauss' lemma), but they went on to say that if $f\in \mathbb{Z}[X]$ is a primitive polynomial which is irreducible over $\mathbb{Q}$ only then is it also irreducible over $\mathbb{Z}$.
So, my question is : in the second case, does the polynomial really need to be primitive?

Answer 1

I am going to give you a counterexample of the statement in the title.

It is false.

For instance, look at the polynomial $p(x)=2x+2\in\Bbb Z[x]\subset \Bbb Q[x]$. If you consider $p(x)$ as a polynomial with integer coefficient, then it is reducible as $2(x+1)$ because $2$ is not a unit in $\Bbb Z$ - it is not invertible in $\Bbb Z$. If you look at $p(x)$ as a polynomial in $\Bbb Q[x]$ instead, then it is irreducible. Indeed, even if you write $2(x+1)$, $2$ is a unit in $\Bbb Q$.

Answer 2

Prasolov uses a nonstandard definition of irreducibility over $\mathbb Z$ (page 48):

He goes on to discuss the content of a polynomial and then to state that "a polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$". In the proof he assumes that the polynomial has content $1$.