I'm trying to prove this result from Hungerford's Algebra (Lemma 6.13):
Let $D$ be a unique factorization domain with quotient field $F$ and $f$ a primitive polynomial of positive degree in $D[X]$. Then $f$ is irreducible in $D[X]$ if and only if $f$ is irreducible in $F[X]$.
The proof starts with "Suppose $f$ is irreducible in $D[X]$ and $f = gh$ with $g,h \in F[X]$ and $\deg(g) \geq 1$, $\deg(h) \geq 1$". But isn't a polynomial of degree $\geq 1$ always not a unit? We have in fact this result:
Let $D$ be an integral domain. The units in $D[X]$ are precisely the constant polynomials that are units in $D$.
Considering $g,h \in F[X]$ with $\deg(g) \geq 1$, $\deg(h) \geq 1$ doesn't imply $f$ is reducible? We have $f(X) = g(X) \cdot h(X)$ with $g,h \in F[X]$ not units. Where am I mistaken?