If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.

Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension (see here), which in particular does not use prime ideals at all. For $x \in R$ let $R_{\{x\}}$ be the localization of $R$ at $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have

$$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \qquad (\ast)$$

It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ You can use this to define the Krull dimension.

A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.

Question. Can we use the characterization $(\ast)$ of the Krull dimension by Coquand-Lombardi above to prove $\dim(R[T])=\dim(R)+1$ for Noetherian commutative rings $R$?

Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$.

Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is satisfied by Noetherian rings and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$.

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    $\begingroup$ At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows? $\endgroup$
    – user26857
    Apr 12, 2013 at 8:11
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    $\begingroup$ Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know. $\endgroup$ Apr 12, 2013 at 9:11
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    $\begingroup$ Shall I ask this on mathoverflow? $\endgroup$ Aug 26, 2013 at 12:43
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    $\begingroup$ MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1 $\endgroup$ Jun 21, 2014 at 19:55
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    $\begingroup$ Since the question is on MO now, should it be closed here? $\endgroup$ Jun 27, 2014 at 18:59

1 Answer 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

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