# Example of a noetherian domain which is neither Dedekind domain nor UFD?

What is an example of a noetherian domain which is not Dedekind domain as well as not a UFD. I don’t have any clue with this. Also, we know that a noetherian domain is a factorisation domain! But is the converse true or are there any example of a factorisation domain which is not noetherian.

Thanks in advance for any kind of help!

• You need to practice your googling. Really. – Mariano Suárez-Álvarez Feb 16 '18 at 19:01
• @MarianoSuárez-Álvarez I tried "Noetherian but not Dedekind nor UFD" and the top result was this very question. Second result was this older, similar question: math.stackexchange.com/questions/1669093/… – Mr. Brooks Feb 17 '18 at 21:53

A simple way to get infinitely many examples: if $A$ is a Dedekind domain with nontrivial class group then $A[x_1,\ldots,x_n]$ is an example for all $n \geq 1$. For instance, $\mathbf Z[\sqrt{-5}][x_1,\ldots,x_n]$ fits your conditions. The explanation below may involve things you have not seen, but at least the examples are easily appreciated.

If $A$ is Noetherian then so is $A[x]$, and $\dim A[x] = \dim A + 1$ (Krull dimension). Since Dedekind domains are $1$-dimensional, if $A$ is a Dedekind domain then $A[x]$ is not Dedekind but it is still Noetherian, and the same is true of $A[x_1,\ldots,x_n]$ for all $n \geq 1$.

Do you know what a Krull domain is? Dedekind domains are the $1$-dimensional Krull domains, and if $A$ is a Krull domain then so is $A[x]$. Each Krull domain has a class group, UFDs are the Krull domains with trivial class group, and if $A$ is a Krull domain then $A[x]$ has the same class group as $A$; that generalizes the fact that if $A$ is a UFD then so is $A[x]$. Therefore if $A$ is a Krull domain that is not a UFD then so is $A[x]$, and the same is true of $A[x_1,\ldots,x_n]$ for all $n\geq 1$.

From the previous paragraphs, if $A$ is Dedekind then $A[x_1,\ldots, x_n]$ is Noetherian, has dimension $n+1$, and has the same class group as $A$, so when $A$ has a nontrivial class group then $A[x_1,\ldots,x_n]$ fits your conditions for all $n\geq 1$. This provides infinitely many examples from one example of a Dedekind domain that is not a UFD.

$k[x,y,z,w]/(xy-zw)$ works. It takes a bit to check.

• Can you specify which question you are referring to this example? – Riju Feb 16 '18 at 19:08
• Are you asking me if this is an example of a ring which is a factorization domain but not noetherian? – Mariano Suárez-Álvarez Feb 16 '18 at 19:09
• Yes I am again is this an example for the first question or the second question? – Riju Feb 16 '18 at 19:10
• Well, is this a noetherian ring? – Mariano Suárez-Álvarez Feb 16 '18 at 19:11
• It is noetherian ring because it is the quotient of a noetherian ring! – Riju Feb 16 '18 at 19:13

If $k$ is a field the ring $k[X_1,...,X_n,...]$ of polynomials in denumerably many indeterminates over $k$ is a non-noetherian factorization domain (even a UFD actually).

The DaRT query suggested a result:

$k[[x^2,x^3]]$

I am unpracticed with these conditions, but here's what I think: it's not a UFD because $x^6$ factors in two ways, and I think it's not Dedekind because $(x^6)$ factors in two ways.

$k[X_i\mid i\in\mathbb N]$ is a UFD that's not Noetherian. (I gave this earlier, then removed it, but I'd like to put it back.)

For a different example that isn't a UFD either, Anne Grams produced an example of a non-Noetherian atomic domain (DaRT query), which should be adequately cited there.

Update: I've updated the answer to reflect that a previously reported result ($\mathbb Q[x,y]_{(x,y)}$) was invalid. Owing to a typo in a particular entry, it was misclassified. It's been fixed for all affected rings. I owe a big debt of gratitude for uncovering this problem so that I could correct it. Thanks)

• Which is what don’t get it! So the first example is the one which is noetherian but not Dedekind nor a UFD. Is it? What about the second one? – Riju Feb 16 '18 at 18:45
• Sorry, didn't notice the separate request. If you read the query parameters in the link, the search indicates they are both Noetherian domains which aren't Dedekind or UFD. – rschwieb Feb 16 '18 at 18:46
• For once I'm not confident about these results, either. I do not understand why it responded that the first one is not a UFD. It's a localization of a UFD... This is, potentially, a mistake in the data, which I'll have to check up on. It seems plausible to me that the second isn't a UFD, but I don't know the justifcation :/ Krull dimension ensures the first isn't Dedekind, but apparently not so for the second one – rschwieb Feb 16 '18 at 18:47
• No, I also haven’t verified! I have to think! – Riju Feb 16 '18 at 18:48
• Yeah I think I posed the question wrong! I want a non noetherian domain which is a atomic domain but not a UFD. But in that case the example by grams is the one I am interested in! – Riju Feb 16 '18 at 20:44

I'm a bit late to the party, but here are a couple of examples that haven't been mentioned in the previous answers.

First example. The ring $$R=\mathbf{Z}[\sqrt{-3}] = \mathbf{Z}[X]/(X^2+3)$$ is a simple example of a Noetherian domain which is not integrally closed (since the element $$\frac12 + \frac12 \sqrt{-3}$$, which lies in $$R$$'s field of fractions $$\mathbf{Q}(\sqrt{-3})$$ but not in $$R$$, is a root of a monic polynomial in $$R[X]$$, namely $$X^2 - X + 1$$). Since Dedekind domains and UFDs are known to always be integrally closed, $$R$$ is therefore not a Dedekind domain and not a UFD.

Some background material for the second example:

• If $$A$$ is an integral domain with field of fractions $$K$$, let $$\operatorname{Int}(A)$$ denote the subring of $$K[X]$$ consisting of the polynomials which map $$A$$ into $$A$$. (So-called “integer-valued polynomials”, the name coming from the special case $$A=\mathbf{Z}$$.)
• A bounded factorization domain (BFD) is a factorization domain such that for each nonzero nonunit $$x$$ there is an $$N$$ such that all factorizations of $$x$$ have length at most $$N$$. Any UFD, such as $$\mathbf{Z}$$, is clearly a BFD.
• $$\operatorname{Int}(A)$$ is a BFD if and only if $$A$$ is a BFD. (This is Prop. VI.3.2 in the book Integer-Valued Polynomials by Cahen & Chabert.)

Second example. The ring $$\operatorname{Int}(\mathbf{Z})$$ is a BFD (since $$\mathbf{Z}$$ is a BFD), which is not Noetherian (Cahen & Chabert, Prop. V.2.7). In fact, it's a Prüfer domain (hence integrally closed) which is not a Dedekind domain (which implies that it's not a UFD either).

Similarly, the ring $$\operatorname{Int}(\operatorname{Int}(\mathbf{Z}))$$ is also a BFD (since $$\operatorname{Int}(\mathbf{Z})$$ is a BFD), which is not Noetherian. It's integrally closed but not a Prüfer domain; see this question.