a ring R, a subring S of R, and an element u ∈ S ⊂ R such that u is a unit in R; but u is NOT a unit in S and u is irreducible in S

Question

I'm looking for a ring R where the unit a is not a unit for the subring S, but is irreducible in S.

I'm unsure of how to approach this problem, I've tried several different types of rings but can't make any progress.

Answer 1

$R=\mathbb{Q}(x)$, $S=\mathbb{Q}[x]$ and $a=1+x^2$ works.

This $R$ is a field, so all non-zero elements are units. $a$ is irreducible in $S$ because otherwise it would have to have roots in $\mathbb{Q}$.

Answer 2

Let $R = \mathbb{Q}$ and $S = \mathbb{Z}$. $u = 3$ is a unit in $R$ since has inverse $u^{-1} = \frac{1}{3}$, but in $S$ has no inverse. Still $u$ is irreducible in $S$ or $3$ would be irreducible in $\mathbb{Z}$, but $3$ is a prime number.