I have to find a condition on a polynomial $f(x) \in \mathbb{Z}[x]$ such that $f(x)$ is reducible in $\mathbb{Z}[x]$ but irreducible over $\mathbb{Q}$.
My attempt: Let $f(x)$ be reducible over $\mathbb{Z}$ then for some $a(x), b(x) \in \mathbb{Z} [x]$, we have $$f(x)=a(x). b(x)$$
where $0\leq$deg($a(x)$), deg($b(x)$) $<n$ ,and neither $a(x)$ nor $b(x)$ unit in $\mathbb{Z}[x]$
and I know that units in $\mathbb{Z}[x]$ are $1, -1$.
To ensure that this polynomial is reducible over $\mathbb{Z}$ but not over $\mathbb{Q}$ at least one of $a(x)$ or $b(x)$ has to be a constant different from $1,-1$
which clearly means that $f(x)$ cannot be primitive, because if it were so then it wasn't possible to write it as a product of a non unit constant integer and a polynomial with the same degree.
So, $f(x)$ has to be a non primitive polynomial
But, at the back of the textbook I'm using the answer is given"$f(x)$ is primitive"
Can someone tell me what I did wrong to arrive at the opposite conclusion?
Thanks a lot!
P.S.: Alternate solutions/hints are welcome.
The polynomial $f(x)=2x-2$ is reducible in $\mathbb{Z}[x]$, because both $2$ and $x-1$ satisfy the definition.
On the other hand, $2x-2$ is irreducible in $\mathbb{Q}[x]$.
So it should be easy to find the condition: can there be other cases?
(Probably the book missed a “not”.)
Let's assume that $f(x)\in\mathbb{Z}[x]$ is irreducible when considered in $\mathbb{Q}[x]$, but not in $\mathbb{Z}[x]$. Then we can write a proper factorization $$ f(x)=a(x)b(x) $$ in $\mathbb{Z}[x]$, with neither $a(x)$ nor $b(x)$ being invertible. Since this is also a factorization in $\mathbb{Q}[x]$, one of the two factors (we can assume it is $a$) is invertible in $\mathbb{Q}[x]$, so it is a non zero constant in $\mathbb{Z}$.
Therefore the constant $a$ divides all coefficients of $f$ in $\mathbb{Z}$ and so $f$ is not primitive.
Conversely, a non primitive polynomial of positive degree is always reducible in $\mathbb{Z}[x]$.