It is well known that if $R$ is a UFD with field of fractions $K$ and $f \in R[x]\setminus R$, then the following holds:
$f$ is irreducible in $R[x] \iff f$ is primitive and irreducible in $K[x]$
So far, I've always seen the term "primitive" and thus the statement with $R$ being a UFD. I'm wondering what happens in the general case of $R$ being an arbitrary integral domain. It is pretty easy to show the following ($f$ being a non-constant polynomial):
$f$ is irreducible in $R[x] \implies f$ primitive
$f$ is irreducible in $R[x] \impliedby f$ primitive and irreducible in $K[x]$
So the interesting point for me here is the following:
$f$ is irreducible in $R[x] \implies f$ irreducible in $K[x]$.
My question is: Under which condition is the last-mentioned statement true?
When I tried to find a counterexample over a non-UFD which is at least integrally closed, I've found $f=2x^2-2x+3 \in \mathbb Z[\sqrt{-5}][x]$, which should be irreducible over $\mathbb Z[\sqrt{-5}]$, but reducible over the field of fractions $\mathbb Q(\sqrt{-5})$ due to $f=\frac12(2x-1-\sqrt{-5})(2x-1+\sqrt{-5})$.
My next question is: Is there a monic polynomial being irreducible over $\mathbb Z[\sqrt{-5}]$, but reducible over $\mathbb Q(\sqrt{-5})$? All I know is that it must necessarily have degree $4$ (otherwise it would have a root in $\mathbb Q(\sqrt{-5})$ and thus in $\mathbb Z[\sqrt{-5}]$ because $\mathbb Z[\sqrt{-5}]$ is integrally closed).