Let $R$ be a commutative ring. Then $\dim(R[T]) \geq \dim(R)+1$. Is there a classification of those commutative rings with the property $\dim(R[T])=\dim(R)+1$? Every Noetherian commutative ring has this property, but also every $0$-dimensional commutative ring (Noetherian or not) has this property (see here). A $1$-dimensional counterexample can be found here. In case it helps for the classification, we can also add closure properties and make the property stronger (which might simplify the problem). For example, is the class of commutative rings $R$ such that (every quotient of) (every localization of) (every finitely generated commutative algebra over) $R$ satisfies this dimension equality more easy to classify? The closure under finitely generated commutative algebras includes polynomial rings in particular, so that we also demand $\dim(R[T_1,\dotsc,T_n])=\dim(R)+n$ for all $n \in \mathbb{N}$.

Martin Brandenburg
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  • To generalize the 0-dimensional case, it also holds if the ring has weak global dimension $1$ after modding out nilpotents. – Badam Baplan Apr 26 '21 at 15:07
  • @BadamBaplan This is interesting, thank you. Would you mind writing this as an answer, including a proof? Right now, I have no clue how to prove this. – Martin Brandenburg Apr 26 '21 at 15:31
  • sure, i will write something up tomorrow when I have time or at least point you to sufficient references. – Badam Baplan Apr 27 '21 at 00:54
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    A keyword for you to look up could be "Jaffard ring", although that property is stronger than what you ask about, namely that Krull dimension is on its best behavior for all multivariate polynomial extensions (not just univariate) – Badam Baplan Apr 27 '21 at 01:02
  • Great! https://en.wikipedia.org/wiki/Jaffard_ring seems to be an answer then. There are several papers on this. I will accept it when you post it. – Martin Brandenburg Apr 27 '21 at 08:30
  • Martin - while I was collecting references for an answer I found this old math overflow post https://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras which probably covers anything I would write. Maybe flat dimension isn't discussed anywhere explicitly, but you can treat the problem locally (note flat dim $1$ = locally valuation domain) and certainly the case of valuation domains is in the literature... so I have nothing to add there either. I'm not sure what the policy is on (near-)duplicates across overflow and stack exchange? – Badam Baplan Apr 28 '21 at 16:38
  • Well a link to this discussion would be an answer as well :). Sometimes I make them CW (when I really have nothing else to say). Thank you a lot for your research! – Martin Brandenburg Apr 28 '21 at 16:58

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