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

Question

Please read this first before answering. This is not the right place for you to advertise your favorite proof of the dimension formula. This question is only concerned with a proof using the Coquand-Lombardi characterization below. If you post something which doesn't mention it, it's not an answer, offtopic, and will probably be deleted by moderators.

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.

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 $R$.

Answer 1

Let me try.

I assume $\dim(R) > k$ if and only if there are $x_0\ldots x_k\in R$, such that for any $m_0\ldots m_k\in\mathbb N$ and any $a_0\ldots a_k\in R$ $x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)\ne0.$ Fix this sequence $x_0\ldots x_k\in R.$

Now extend this sequence with $T: x_0\ldots x_k, T$ from $R[T]$ and note that $x_0^{m_0} (\cdots ( x_k^{m_k} (T^N(1+a_{k+1}T)+a_k x_k)+\cdots)+a_0 x_0)\ne0,$ because coefficient for $T^N$ on the left hand side is not zero given the choice of $x_i$. Note there no multiplications of polynomials on the left hand side, except in the $T^N(1+a_{k+1}T)$ and remaining occurrences of polynomials are $a_i$. Thus $\dim(R[T]) > k + 1$

If this proof is incorrect, maybe proof of special case for $\dim R=0$ would help progress.