TITLE: Algebraic characterization of commutative rings of Krull dimension 1,2, or 3
QUESTION [6 upvotes]: A commutative ring $R$ (with $1$) is $0$-dimensional if and only if $R/\sqrt 0$ is von Neumann regular. Besides this result, there is a wealth of information about the algebraic structure of zero-dimensional rings.
I could not find any information about the algebraic structure of rings of higher Krull dimension. If the problem for general commutative rings is too hard, how about some special cases, such as when $R$ is Noetherian and/or an integral domain?

REPLY [3 votes]: A remark about the one-dimensional case. By the characterization of Krull dimension given by T. Coquand and H. Lombardi, dim($R$)$\,\leq 1$ iff $\forall_{x,y\in R}\exists_{a,b\in R,m,n\in\mathbb{N}}\,x^{m}(y^{n}(1+by)+ax)=0$. Here we may assume that neither $x$ nor $y$ is nilpotent or a unit.
If $R$ is reduced, we can require $m=1$.
If $R$ is a domain, the condition simplifies to $\exists_{b\in R,n\in\mathbb{N}}\,y^{n}(1+by)\in Rx$ for all non-zero non-units $x$ and $y$ in $R$.
If, in addition, $R$ is local with maximal ideal $\mathfrak m$, one gets dim($R$)$\,\leq 1\iff\forall_{0\neq x\in \mathfrak m}\,\mathfrak m=\sqrt {Rx}$. But this is of course quite trivial.