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After a confusing session of hopping through Wikipedia articles, I started trying to summarize for myself some of the inclusions and relations among the many types of integral domains. Right now I'm just gathering and organizing facts from various sources, rather than going through all the proofs in detail. (To be honest, my knowledge of commutative algebra is embarrassingly poor, and I haven't even understood all of the definitions yet. But one has to start somewhere...)

Anyway, I have a few questions about things that I haven't been able to figure out on my own, and I'll start with this one:

Are there Krull domains that are neither Noetherian nor UFD?

(I would guess yes, but I haven't seen any examples. The standard example of a non-Noetherian Krull domain seems to be $\mathbf{Z}[X_1,X_2,\dots]$ with countably many variables, but that's a UFD unless I'm mistaken. And $\mathbf{Z}[\sqrt{-5}]$ is a non-UFD which is a Dedekind domain, hence a Krull domain and Noetherian.)

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HINT $\: $ UFDs are precisely the Krull domains with trivial class group. So it suffices to exhibit a non-Noetherian Krull domain with nontrivial class group. One way to do this (and much more) is to use Claborn's techniques for constructing Krull domains with arbitrary abelian class groups - see e.g. Section 14 of Fossum's book The Divisor Class Group of a Krull Domain, $1973$.

I don't have Fossum's book at hand, but according to Example $49$ of Hutchins: Examples of Commutative Rings, he uses $\rm\:T =\: R\big[\{x_i,y_i,u_i,v_i\}_{\:i\:\in\: I}\big]/(\{x_i\:u_i-y_i\:v_i\}_{\:i\:\in\: I})\:,\:$ for $\rm\:R\:$ Krull, and for $\rm\:I \ne \emptyset\:.\:$ $\rm\:T\:$ is a Krull domain, but not a UFD (it has divisor class group $\rm\:Cl(T)\cong\! Cl(R)\oplus \mathbb Z^I)\:.\:$ $\rm\:T\:$ is non-Noetherian if $\rm\:R\:$ is, or if $\rm\:I\:$ is infinite. Hence the sought example.

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    @George Yes, of course. Thanks. Hans: Yes, typo corrected.2011-10-07
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What about $\mathbf{Z}[\sqrt{-5}][X_1,X_2,\ldots]$? This is not a UFD and non-noetherian.

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    Actually, I believe that every polynomial extension of a Krull domain is Krull, and since $\mathbb{Z}[\sqrt{-5}]$ is a quadratic ring of algebraic integers (and hence Dedekind, and therefore Krull), it should follow that $\mathbb{Z}[\sqrt{-5}][X_1,X_2,\cdots]$ is Krull.2011-10-18