Let $f \in Z[X]$ primitive. Show that $(fQ[X]) \cap Z[X] = fZ[X]$.
Thanks!
I think I solved it.
Here is my solution:
Clearly $fZ[x] \subset (fQ[X]) \cap Z[X]$, so we need to prove the opposite.
Let $a \in (fQ[X]) \cap Z[X]$, then $a = fg$ for some non-constant polynomial $g \in Q[X]$.
$g \in Q[X]$, there is $c \in Q$ such that $cg \in Z[X]$ and primitive.
$h = cg, g = c^{-1}h, a = c^{-1}fh$
Can apply Gauss's Lemma: $f$ primitive, $h$ primitive implies $fh$ primitive.
It's easy to show that $c^{-1} \in Z$, so $g \in Z[X]$, $a = fg \in fZ[X]$