A question comes from irreducible polynomial?

Question

Let $D$ be a UFD with quotient field $F$. $P(X) \in D[X]$. We know if $P(X)$ is irreducible over $D$ then $P(X)$ is irreducible over $F$. But if $P(X)$ is irreducible over F, why is $P(X)$ probably not irreducible over $D$ (need $P(X)$ is primitive for the inverse proposition to be true). My problem is: $P(X)$ is irreducible over $F$, if $P$ is not irreducible over $D$, then we let $P(X)=f(X)g(X)$, $g$ and $f \in D[X]$, $g$ and $f$ are nonunit but $f$ and $g$ are also in $F[X]$, contradiction. Doesn't that mean the inverse proposition is true?

Answer 1

The key is that being a nonunit in $D[x]$ is not the same as being a nonunit in $F[x]$! For example, if $D = \mathbb Z$, then $2x + 2$ is irreducible over $\mathbb Q$ but reducible over $\mathbb Z$. Indeed, $2x + 2 = 2(x + 2)$, neither of which is a unit in $\mathbb Z[x]$. You can use the same idea for any UFD $D$ that is not a field.