# If $R$ is UFD , then does $R \cong R[X,Y]$ imply $R \cong R[X]$?

$R$ is isomorphic to $R[X,Y]$, but not to $R[X]$ shows that it is possible to have commutative ring $R$ with unity such that $R \cong R[X,Y]$ but $R \ncong R[X]$.

My questions are: Is it possible to have an example of a commutative ring with unity $R$ such that $R \cong R[X,Y]$ but $R \ncong R[X]$, where

(i) $R$ is UFD ?

(ii) $R$ is a Valuation ring ?

UPDATE : Since $R[X]$ always has infinitely many maximal ideals for any commutative ring with unity $R$, so $R \cong R[X,Y]$ implies $R$ can't be local, hence question (ii) is meaningless. Question (i) still remains ...

• I don't' think any ring of the form $R[X,Y]$ is ever a valuation ring, because neither $X$ nor $Y$ could have higher valuation than the other, so certainly not in case (ii). – Will Sawin May 11 '18 at 15:57
• @WillSawin: I see your point ... indeed if $R \cong R[X,Y]$, then $R$ can't even be local I think, because $R[X]$ has infinitely many maximal ideals for any $R$ ... – user111524 May 11 '18 at 16:08