Let $A$ be a unique factorization domain, and let $K = S^{−1}A$, where $S = A \setminus \{0\}$, be the field of fractions of $A$. Show that if $f ∈ A[X]$ is primitive and its image in $K[X]$ is prime, then $f ∈ A[X]$ is also prime. Is this still true without the assumption that $f$ be primitive?
I think I need to prove by contradiction and use Gauss Lemma if $f,g \in A[X]$ is primitive then so is $fg$. But I am unable to figure it out. Can anyone give me some hints? Thank you in advance!
Suppose you have $f\mid gh$ in $A[X]$. Then $fk=gh$, for some $k\in A[X]$.
Write $g=ag_0$, $h=bh_0$, $k=ck_0$, with $a,b,c\in A$ and $g_0$, $h_0$, $k_0$ primitive, so that $$ cfk_0=abg_0h_0 $$ Up to associates, we can assume $c=ab$, because $fk_0$ and $g_0h_0$ are primitive.
Therefore we can restart with $f\mid gh$, $f\nmid g$, with $g$ and $h$ primitive.
Since $f$ is prime in $K[X]$, it must either divide $g$ or $h$ in $K[X]$.
Suppose, without loss of generality, $g=f\tilde{g}$, for some $\tilde{g}\in K[X]$. Collecting denominators we can write $\tilde{g}=\frac{a}{b}g_1$, where $a,b\in A$ and $g_1$ primitive in $A[X]$. Then $$ bg=afg_1 $$ and, since $fg_1$ is primitive, we can assume $a=b$ (up to associates), so $g=fg_1$ in $A[X]$.