Let $f (x) \in \mathbb{Z}[x]$ be irreducible. Prove that $f(x)$ is primitive. Let $f (x) \in \mathbb{Z}[x]$ be irreducible. Prove that $f(x)$ is primitive.

My thought:-
Let $f(x)$ is not a primitive.Since $f(x)$ is not a primitive we can assume $\deg(f(x))>1$.Then there exist $k\in \mathbb{Z}$ with $|k|>1$ such that $f(x)=kg(x)$ . The units in $\mathbb{Z}[x]$ are $1$ and $-1$.Hence both $k$ and $g(x)$ are nonunit.   
Does my procedure is correct?
 A: Yes, it is correct. The key idea is this: $ $ the only units in $\,\Bbb Z[x]\,$ are those in $\,\Bbb Z,\,$ and this implies that, in $\,\Bbb Z[x],\,$ the associates of constants are constant.  Thus a nonconstant polynomial divisible by a constant $\,n\,$ cannot be associate to $\,n,\,$ hence the polynomial is reducible if $\,n\,$ is not a unit.
This hypothesis about units is all that is needed for the proof to succeed. Indeed, more generally 
Theorem $\ $ Suppose that $\,Z \subset R\,$ are rings such that every unit in $R\,$ is a unit in $Z.\,$ Then an element $\,f\in R,\ \color{#c00}{f\not\in Z}\, $ is reducible if it is divisible by a nonunit $\,n\in Z.$
Proof $\,\ $ Since $\,n\mid f\,$ we have  $\,f = n g\,$ for some $\, g\in R.\,$ Notice that $\,n\,$ remains a nonunit in $R,\,$ else, by hypothesis, $\,n$ unit in $R\,\Rightarrow\,n\,$ unit in $Z,\,$ contra hypothesis.  Further $\,g\,$ is a nonunit in $R\,$ else, by hypothesis, $\,g\,$ unit in $R\,\Rightarrow\,g\,$ unit in $Z,\,$ so $\,n,g\in Z\Rightarrow \color{#c00}{f = ng\in Z}.\  $ QED
