Let $D$ be a commutative $k$-algebra which is a UFD, $k$ a field of characteristic zero.
Let $0 \neq a,b \in R$ be two irreducible elements of $D$, hence $\frac{D}{\langle a \rangle}$ is an integral domain and $\frac{D}{\langle b \rangle}$ is an integral domain. To exclude trivial cases, we also require that $\{a,b\}$ are linearly independent over $k$.
When $\bar{a}:=a+\langle b \rangle \in \frac{D}{\langle b \rangle}$ is irreducible?
Example: $D=k[x,y]$, $a=x$, $b=y$. Clearly, $\bar{x} \in \frac{k[x,y]}{\langle y \rangle}\cong k[x]$ is irreducible.
But I guess that generally, $\bar{a}$ may not necessarily be irreducible in $\frac{D}{\langle b \rangle}$.
Remarks:
(1) From our assumptions, it follows that $\gcd(a,b)=1$. Indeed, write $a=da', b=db'$, where $D \ni d=\gcd(a,b)$. But $\{a,b\}$ are irreducibles in $D$, hence $d \in k$ and we are done or $d \notin k$, hence $a',b' \in k$, so $\{a,b\}$ are linearly dependent over $k$, a contradiction to our assumption.
(2) If $\bar{a}$ is reducible, then there exist $u,v \in D-\langle b \rangle- k$ such that $\bar{a}=\bar{u}\bar{v}$, so $a-uv \in \langle b \rangle$. Therefore, there exists $w \in D$ such that $a-uv=wb$. This shows that $uv \in \langle a,b \rangle$, but we have no information about $\frac{D}{\langle a,b \rangle}$.
(3) Edit: According to the first comment, we should also require that $k$ is algebraically closed.
(4) This question is relevant. For $D=k[x,y]$, perhaps this question is relevant (but only if one of $\{a,b\}$ belongs to $k[x]$ or $k[y]$).
Any hint and comments are welcome!