# If $f(x) \in D$ integral domain and $f(x)$ is irreducible over $F$ field but reducible over $D$, factorization?

Suppose that $D$ is an integral domain and $F$ is a field containing $D$. If $f(x) \in D[x]$ and $f(x)$ is irreducible over $F$ but reducible over $D$, what can you say about the factorization of $f(x)$ over $D$?

This doesn't make sense to me, how can $D \subseteq F$ and be irreducible in $F$ but reducible in $D$? The same polynomials in $D$ are in $F$, so any polynomial product representation of $F$ is irreducible.

The trick is that there are elements of $D[x]$ which become units in $F[x]$, so there are factorizations which are trivial in $F[x]$ but nontrivial in $D[x]$. For instance, with $2x$ factors nontrivially in $\mathbb{Z}[x]$ as $2\cdot x$, but it is irreducible in $\mathbb{Q}[x]$ since $2$ is a unit there. In general, if $f(x)$ is irreducible over $F[x]$, then for any factorization in $D[x]$, one of the factors must become a unit in $F[x]$: that is, one of the factors must have degree $0$