A short proof for dim(R[T])=dim(R)+1?

[Ganesh Ujwal]

If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$. For noetherian $R$, 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. It states:

For $l \in \mathbb{N}$ we have $\dim(R) \leq l$ if and only if for all $x_0,\dotsc,x_l \in R$ there are $a_0,\dotsc,a_l \in R$ and $m_0,\dotsc,m_l \in \mathbb{N}$ such that $x_0^{m_0} (\dotsc ( x_l^{m_l} (1+a_l x_l)+\dotsc)+a_0 x_0)=0$.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 $K$-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.

Can we use this elementary characterization of the Krull dimension to give a new short proof of $\dim(R[T])=\dim(R)+1$ for noetherian commutative rings $R$?

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

 

[jvicens]

Thank you for sharing this interesting characterization of the Krull dimension! It is always exciting to see new and potentially simpler proofs of well-known theorems.

To answer your question, I believe it is possible to use this characterization to give a new proof of $\dim(R[T])=\dim(R)+1$ for noetherian commutative rings $R$. Here is an outline of how it could be done:

1. First, we can use the characterization to show that for any noetherian ring $R$, we have $\dim(R) \leq n$ if and only if for any $x_1,\dotsc,x_n \in R$, there exist $a_1,\dotsc,a_n \in R$ and $m_1,\dotsc,m_n \in \mathbb{N}$ such that $x_1^{m_1} (\dotsc (x_n^{m_n} (1+a_n x_n)+\dotsc)+a_1 x_1)=0$.

2. Next, we can use induction on $n$ to show that for any noetherian ring $R$, we have $\dim(R) \leq n$ if and only if for any $x_1,\dotsc,x_n \in R$, there exists $a \in R$ and $m \in \mathbb{N}$ such that $x_1^m (1+ax_2)=0$.

3. Now, let $R$ be a noetherian commutative ring and let $x_1,\dotsc,x_{n+1} \in R[T]$ be any elements. By induction, we can assume that $\dim(R) \leq n$. Then, using the characterization, we can find $a \in R$ and $m \in \mathbb{N}$ such that $x_1^m (1+ax_2)=0$ in $R[T]$. This implies that $x_1^m (1+a(Tx_2))=0$ in $R[T,T^{-1}]$. Since $R[T,T^{-1}]$ is also a noetherian ring, we can use the characterization again to find $b \in R[T,T^{-1}]$ and ##k \in \mathbb{N}