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

Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization below. If you post something that doesn't mention the characterization, then it's not an answer and is offtopic.

Background. 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.

T. Coquand and H. Lombardi have found a surprisingly elementary characterization of the Krull dimension that does not use prime ideals at all.

T. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages)

You can read the article here.

For $$x \in R$$ let $$R_{\{x\}}$$ be the localization of $$R$$ at the multiplicative subset $$x^{\mathbb{N}} (1+xR) \subseteq R$$. Then we have

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

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 \eqref{1} 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 $$R$$.

• Note: I've asked this on stackexchange math.stackexchange.com/questions/358423 and was encouraged to ask this here. Jun 21, 2014 at 8:11
• I would like to know more generally what happens if you take the Coquand-Lombardi characterization as a definition and try to develop the basics of dimension theory from there. Jun 26, 2014 at 14:48
• @NeilStrickland: The following example can be found in Hutchins, Examples of Commutative Rings, Example 27: Let $k$ be a field and let $R = k(y)[[x]] \times_{k(y)} k$ the ring of those power series in $x$ with coefficients in $k(y)$ resp. $k$ for the constant term. Then $\dim(R)=1$ and $\dim(R[T])=3$. I've also read that actually every number between $\dim(R)+1$ and $2 \dim(R)+1$ may appear as $\dim(R[T])$. Aug 4, 2014 at 14:13
• @Neil: What Martin read (every number between the bound appers) you can also read, namely in A. Seidenberg, On the dimension theory of rings (II), Pacific J. Math. 4 (1954), 603-614. Aug 25, 2014 at 21:21
• The elementary characterization is also very useful in a topos-internal context. For example, a scheme $X$ is of dimension $\leq n$ if and only if, from the internal perspective of the little Zariski topos $\mathrm{Sh}(X)$, the (then plain old) ring $\mathcal{O}_X$ is of Krull dimension $\leq n$. See Proposition 3.13 of these sketchy notes. Feb 6, 2015 at 15:51