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

1. Dec 12, 2014

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 $R$.

2. Dec 17, 2014