I am currently reading Juliusz Brzezinski's book "Galois Theory Through Exercises", in which he states Gauss's Lemma as a useful test for the irreducibility of certain polynomials.
However, the statement of the lemma in the book is as follows:
A nonconstant polynomial with integer coefficients is reducible in $\mathbb{Z}[X]$ if and only if it is reducible in $\mathbb{Q}[X]$. More exactly, if $f ∈ \mathbb{Z}[X]$ and $f = gh$, where $g, h ∈ \mathbb{Q}[X]$, then there are rational numbers $r, s$ such that $rg, sh ∈ \mathbb{Z}[X]$ and $rs = 1$, so $f = (rg)(sh)$.
In the proof of the lemma the book states that
Of course, if a polynomial $f ∈ \mathbb{Z}[X]$ is reducible in $\mathbb{Z}[X]$, then it is reducible in $\mathbb{Q}[X]$.
But I must be missing something here, as this implication does not seem obvious to me. In fact, I would have thought that the implication doesn't always hold since the polynomial $f(X)=2X+2=2(X+1)$ is reducible in $\mathbb{Z}[X]$ (as neither $2$ nor $(X+1)$ are units in $\mathbb{Z}[X]$) but $f(X)$ is irreducible in $\mathbb{Q}[X]$ because $2$ is a unit in $\mathbb{Q}[X]$.
I have seen Gauss's Lemma before, without incident, where it was stated with the condition that the polynomial must also be primitive, and I wonder if this is the source of my confusion.
Apologies if this is a stupid question, any help would be greatly appreciated!