A proof for $\dim(R[T])=\dim(R)+1$ without prime ideals?

Solution 1:

This may not be a direct answer to your question, but it is relevant to the problem and too long to share in the comments. You can use valuation overrings and valuative dimension instead of prime ideals and Krull dimension, respectively. Polynomial rings are well-behaved with respect to valuative dimension, and Jaffard rings are the natural context in which to study the condition you seek. All Noetherian rings and all Prufer rings are Jaffard, for example, and the class of Jaffard rings is much larger than that of the Noetherian rings. See, for example, https://reader.elsevier.com/reader/sd/pii/0022404988900278?token=D264232EFA26A1777308AB862E97D943396F8674B591A76662C5771929AA087440F2C1735F1714DCDE4D6A0C9CEAD688&originRegion=us-east-1&originCreation=20211113033250

See also:

https://en.wikipedia.org/wiki/Jaffard_ring

https://planetmath.org/jaffardring

https://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras/115481

https://www.jstor.org/stable/2373549

https://www.semanticscholar.org/paper/On-Jaffard-domains-David-Bouvier/5cbae52f1fcd0f67c27cb27fc43d589c07fefe31