A proof of $\dim(R)+1 \leq \dim(R[T]) \leq 2 \dim(R)+1$ with the Coquand-Lombardi characterization of Krull dimension

Question

Question. Can we use the Coquand-Lombardi characterization$^1$ of Krull dimension to prove the well-known inequalities$^2$ $$\dim(R)+1 \leq \dim(R[T]) \leq 2 \cdot \dim(R)+1,$$ where $R$ is any commutative ring? The proof should in particular not use any prime ideals.

Background. The Coquand-Lombardi characterization of the Krull dimension of a commutative ring $R \neq 0$ can be phrased as $$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}}) + 1\right),$$ where $R_{\{x\}}$ denotes the the localization of $R$ at the multiplicative subset $x^{\mathbb{N}} (1+xR) \subseteq R$ (the lower boundary). More concretely, 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 $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ Notice that Coquand-Lombardi make the convention $\dim(0)=-1$, but that is not good since $\dim(R)+1 \leq \dim(R[T])$ would be wrong, we need to define $\dim(0) := -\infty$ (which is also compatible with the prime ideal definition).

In my previous question on this topic Neil Strickland made a very interesting comment:

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.

Indeed, you can prove basic things like fields have dimension $0$, PIDs have dimension $1$ (unless they are fields, of course), $\dim(R \times S) = \max(\dim(R),\dim(S))$, and $\dim(R) \geq \dim(S)$ if there is a surjective or localization homomorphism $R \to S$. Coquand-Lombardi demonstrated with their definition $\dim(K[X_1,\dotsc,X_n])=n$ when $K$ is a field. But it seems that my previous question was too optimistic. I think it will be more feasible to prove the inequalities stated above: the absence of a "Noetherian" condition will certainly make the claim more "formal".

The inequality $\dim(R)+1 \leq \dim(R[T])$ should be manageable, but I can't finish it. (A proof attempt appeared in one of the deleted answers over at SE/358423, but I think that was wrong.) The other inequality $\dim(R[T]) \leq 2 \cdot \dim(R)+1$ will be harder. With the classical definition of Krull dimension, this mainly relies on the Seidenberg's observation that no chain of three prime ideals in $R[T]$ can contract to the same prime in $R$ (which follows quickly from $\dim(K[T])=1$ for fields $K$). I guess we need to carry out a similar "it does not fit" trick here.

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$^1$ 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). Link to pdf

$^2$ A. Seidenberg, A note on the dimension theory of rings. Pacific J. of Mathematics, Volume 3 (1953), 505-512. Link to pdf

Answer 1

If the Krull dimension of $R$ is $k+1\ge 1$, it is not $\le k$. Hence there exists an $(x_0,\cdots,x_k)\in R^{k+1}$ such that $y:=x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)\ne 0$ for all $a_0,\cdots,a_k \in R$ and $m_0,\cdots,m_k \in \mathbb{N}$. Letting the $a_i$ range over $R[T]$ instead, the expression $y$ cannot be equal to the zero polynomial, as its constant term $y(0)\ne 0$. This yields $T^n \cdot (y+f\cdot T)\ne 0$ in $R[T]$ for all $f\in R[T]$ and $n\in \mathbb{N}$ (ánd all $a_0,\cdots,a_k \in R[T]$ and $m_0,\cdots,m_k \in \mathbb{N}$ as earlier). Indeed, its lowest degree term is $T^n\cdot y(0)$. The existence of a tuple $(T, x_0, \cdots, x_k)\in R[T]^{k+2}$ having this property means that $\mathrm{dim}(R[T])$ is not $\le k+1$.

The argument breaks down when $k<0$. For $k=-1$, i.e. $\mathrm{dim}(R)=0$, the inequality $\dim(R)+1 \leq$ $\dim(R[T])$ still holds. In fact, $\dim(R[T])=1$ then. (This easily follows by means of Th. 38 in Kaplansky's Commutative Rings.) For $k=-2$, the inequality fails when one agrees to the convention that the trivial ring should have dimension $-1$. (This convention is crucial for the general validity of the inductive formula $\dim(R)=$ $\sup_{x \in R} \left(\dim(R_{\{x\}}) + 1\right)$, which would fail for zero-dimensional $R$ if we'd put $\dim(0)=-\infty$.)

The second inequality, which does hold for the trivial ring too, is likely to be harder to tackle without resorting to prime ideals.

Answer 2

(As requested by OP, I convert my comment into an answer. However, I am not technically equipped on constructive mathematics, thus further edits are welcome, and I make it a community wiki.)

It turns out that these inequalities hold constructively in a very strong sense: neither any form of axiom of choice, nor the law of excluded middle, is necessary for this. See Coquand–Lombardi §XIII.7, in particular 7.13 and 7.15. By the way, this chapter also addresses Neil Strickland's question in some sense, namely, how far could one develop the dimension theory constructively?

The constructive nature implies that this statement holds internally in any topos $\mathcal T$. More precisely, [for experts to complete]