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 $R$.

Martin Brandenburg
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    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? –  Apr 12 '13 at 08:11
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    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. – Martin Brandenburg Apr 12 '13 at 09:11
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    This characterization of $\dim R\le l$ does not look like first-order to me, at least in the language of rings: it uses quantifiers over $\mathbb N$ and exponentiation with the exponent as an argument, which is not a ring operation. – Alexey Jul 26 '13 at 12:35
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    Indeed. By the compactness theorem, it can be seen that ${\rm dim}R\le l$ is not a first order statement for any finite $l$. – George Lowther Jul 26 '13 at 16:35
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    Shall I ask this on mathoverflow? – Martin Brandenburg Aug 26 '13 at 12:43
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    MO copy of the question: http://mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1 – Martin Sleziak Jun 21 '14 at 19:55
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    Since the question is on MO now, should it be closed here? – Martin Brandenburg Jun 27 '14 at 18:59
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    bourbaki's proof of this theorem uses only general results, not more partiuclar results such as krull IT – syzygy Jul 02 '16 at 21:00
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    If you can explain in a few words, what is T and what does it mean +1 in your equation? – usiro Mar 19 '17 at 10:40
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    I guess T is a set of polynomial variables, like $T= \{x\}$ gets you $R[x]$ polynomials in $x$ with coefficients in the ring $R$, $T= \{x_1, x_2, ... , x_n\}$ gets you the polynomials in $n$ variables, and so on. I'm not sure if OP specificies T to be numerable or even finite. +1 means just that, +1, the dimension is an integer number, so it makes sense to add 1 to it. – Werner Germán Busch Mar 21 '17 at 01:48
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    For my own education, what do you mean when you write "first order property"? What is the definition of a first order property? – Malkoun Mar 22 '17 at 17:46
  • Might I ask why the dimension definition using prime ideals isn't elementary enough? The proof isn't short, but it isn't terrible to see why the dimension increases by 1. Is there a reason this other construction is the one you'd prefer to use? – Chickenmancer Apr 14 '17 at 00:30
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    The proof of the theorem is clear and we leave it as an exercise to the reader. – Emma Sep 08 '18 at 14:06
  • This might not be too helpful, but have you tried searching through Gathmann's old notes on algebraic geometry? We're going through his notes in my algebraic geometry this year and he gives a lot of alternative proofs to commutative algebra facts :D – Foobanana Dec 06 '18 at 16:47
  • @Foobanana I know them. There is no such proof. – Martin Brandenburg Dec 22 '19 at 14:24
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    @Malkoun https://en.wikipedia.org/wiki/First-order_logic – Martin Brandenburg Dec 22 '19 at 14:24
  • This question has been referred to at https://math.meta.stackexchange.com/questions/33453/answered-on-math-overflow-tag – Gerry Myerson Apr 15 '21 at 23:55
  • @moderators: Please make this question CW. – Martin Brandenburg Apr 27 '21 at 00:28
  • What is the difference between MO and MSE? – user71207 Jul 12 '21 at 01:55
  • @WernerGermánBusch Of course $T$ is just *one* variable. – Martin Brandenburg Jul 15 '21 at 13:52

1 Answers1


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:






Jesse Elliott
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