I've been looking at the statements I found on these two Stack Exchange answers and I've been trying to prove them. The first claim is:
If $R$ is a PID and $0\neq r\in R$ is irreducible, then $R[X]/(r)\cong (R/(r))[X]$
The second claim follows from the first:
If $k$ is a field, and $p(X)$ is an irreducible in $k[X]$, then $K[X,Y]/(p(X)) \cong (K[X]/(p(X)))[Y]$
Attempt to prove claim 1: It looks like I can prove this using the First Isomorphism Theorem by finding a ring morphism $R[X] \hookrightarrow (R/(r))[X]$ with $(r)$ as the kernel. However the only one I can think of is the following:
Let $f:R\rightarrow R/(r)$ be the projection $f(x)=\bar{x}$. Then we have the morphism of rings $\phi:R[X]\rightarrow (R/(r))[X]$ with the kernel $(r)[X]$ described as $\phi(a_nx^n+\cdots+a_1x+a_0)=\phi(a_nx^n)+\cdots+\phi(a_1x)+\phi(a_0)$ $=f(a_n)x^n+\cdots+f(a_1)x+f(a_0)=\bar{a_n}x^n+\cdots+\bar{a_1}x+\bar{a_0}$
Any help to prove this is appreciated!
edit: Further comment that I am interested whether the isomorphism hold only when $R$ is a commutative, unitary ring with any ideal $I$. There is utility in pulling out the '$[X]$' and treating any quotient of a ring polynomial as a ring polynomial with coefficients that are quotients i.e. $R[X]/I\cong (R/I)[X]$. Note I'm aware $R[X]/I[X] \cong (R/I)[X]$ holds.