Timeline for If $R$ is UFD , then does $R \cong R[X,Y]$ imply $R \cong R[X]$?
Current License: CC BY-SA 4.0
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| May 11, 2018 at 16:10 | history | edited | user111524 | CC BY-SA 4.0 |
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| May 11, 2018 at 16:08 | comment | added | user111524 | @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 $R$ ... | |
| May 11, 2018 at 15:57 | comment | added | Will Sawin | 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). | |
| May 11, 2018 at 15:43 | history | asked | user111524 | CC BY-SA 4.0 |