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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?

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    $\begingroup$ Cross-posted: math.stackexchange.com/questions/1134804/… $\endgroup$ – user26857 Feb 6 '15 at 17:40
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    $\begingroup$ Yes, I asked the same question on mathstackexchange. I hope there is a better chance of getting an answer here. $\endgroup$ – Andrew Chiriac Feb 6 '15 at 17:44
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    $\begingroup$ A noetherian, integrally closed domain of Krull dimension 1 is called a Dedekind ring, and has many nice properties. If you drop the "integrally closed" assumption, you'll get all sorts of singularities, so there is no hope for a structure theorem. This becomes much worse of course in higher dimension. $\endgroup$ – abx Feb 6 '15 at 17:48
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    $\begingroup$ A question is whether, say, having Krull dimension $\le d$ can be characterized by a 1-st order sentence, or by a family thereof. The answer is, I think, negative for $d\ge 1$, because any ultrapower of $\mathbf{Z}$ has infinite Krull dimension; yet this make sense to ask if it is in a more restricted class (among noetherian rings? among finitely generated algebras over PIDs?). Note that the sentence in Matthé's answer is not 1st order in the language of rings, because of the quantifier $\exists n,m$. $\endgroup$ – YCor Jun 14 '18 at 14:43
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Ok, let me try to say something intelligent about the 1-dimensional case.

As has already been noted, the normal Noetherian 1-dimensional rings are finite products of Dedekind domains. So what about the non-normal case (say for reduced rings)?

Pass to the normalization (dimension = 1)

Suppose that $R$ is an excellent purely 1-dimensional reduced ring. Let $R^{\mathrm{N}}$ be the normalization. So we know that $R^{\mathrm{N}}$ is a finite product of Dedekind domains. How do we reconstruct $R$ from it?

It follows that $R$ is always the pullback of a diagram $\{ R^{\mathrm{N}} \to R^{\mathrm{N}}/I \leftarrow S \}$ where $R^{\mathrm{N}}/I$ and $S$ are zero-dimensional (which say we've already classified) and the map $S \to R^{\mathrm{N}}/I$ is injective. To see this, see my answer here and my answer here.

On the other hand, if $A$ is Noetherian normal and purely 1-dimensional, $A/I$ and $B$ are zero dimensional, and $B \to A/I$ is injective, then $\{ A \to A/I \leftarrow B \}$ gives you a 1-dimensional reduced Noetherian ring. So this is at some level a classification (in the second answer, a canonical choice of the $A/I$ and $B$ is given).

On the other hand, if the map $B \to A/I$ is not injective (but still a map between zero dimensional rings), then this should actually classify 1-dimensional rings that are generically reduced.

Two dimension and higher

Edit: I originally said this was hopeless, but maybe that's too pessimistic. There are approaches at some level. Here are some vague thoughts and directions to look up for references.

Ok, so probably it be reasonable to localize (and maybe also complete). Then what do you do? There are some perhaps even reasonable results. It might be of course you have to leave the world of algebraic structure and pass to geometric structure.

Resolution of singularities (if it is known to exist) says something, but doesn't tell you everything you might want (certainly the structure of the exceptional divisor, which is heavily studied, is less information than isomorphism after completion).

One can talk about the equisingularity theory as introduced by Zariski (which tries to break up the variety into strata where all the singularities are the same). This might be really relevant to you and there are plenty of recent papers on it in the commutative algebra. This is still a really hard area in my opinion.

I remember Hans Schoutens talking about a version of this problem in an AMS special session once, the most relevant version of the paper I can find is this one: click HERE. I don't quite see how one could use the sort of thing that Schoutens proves but maybe if it was further developed (or maybe you see ways to use it).

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There are some interesting thoughts about this in the paper "A short proof for the Krull dimension of a polynomial ring" by Coquand and Lombardi. I like to rephrase the key idea as follows. Put $P_d(A)=A[x_0,\dotsc,x_d]$, and say that an element $f\in P_d(A)$ is comonic if it is a single monomial (with coefficient $1$) plus a sum of lexicographically higher monomials (with arbitrary coefficients). Then $A$ has dimension at most $d$ iff for every $A$-algebra homomorphism $P_d(A)\to A$, the kernel contains a comonic polynomial. (This is equivalent to Coquand and Lombardi's Theorem 1, by an argument which is implicit in their Corollary 3.)

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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.

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