# Non-Noetherian one-dimensional domain with a finitely generated maximal ideal

Does there exist a non-Noetherian integral domain of Krull dimension one that has a finitely generated maximal ideal?

There is an example of a non-Noetherian valuation ring of Krull dimension one here but its maximal ideal appears to be infinitely generated.

In the first part I'll point you to a large class of counterexamples. In the second part I'll briefly show why you can't find counterexamples in $$1$$-dimensional valuation domains.

Part (1) -- Counterexamples of the form $$A + xB[x]$$

Let $$A \subseteq B$$ be an extension of rings. The ring $$R = A + xB[x]$$ is the subset of the ring of univariate polynomials over $$B$$ which have constant term in $$A$$.

Rings of the form $$A + xB[x]$$ are very helpful for producing counterexamples of the sort you desire because they afford us great control over both Krull dimension and chain conditions in largely independent ways, while leaving us with some clear choices for prime ideals.

Here are some useful facts$$^1$$ that will help us construct many counterexamples:

Let $$R = A + xB[x]$$
Fact 1: $$R$$ is noetherian iff $$A$$ is noetherian and $$B$$ is finitely generated as an $$A$$-module.
Fact 2: If $$B$$ contains the fraction field of $$A$$, then $$\dim(R) = \dim(A) + \dim(B[x])$$.
Fact 3 If $$A \subseteq B$$ is an algebraic extension, then $$xB[x]$$ is a height $$1$$ prime of $$R$$.

Fact $$2$$ points us to any extension of fields $$F \subseteq L$$, guaranteeing that $$R$$ is $$1$$-dimensional. If we choose $$L$$ to be algebraic over $$F$$, then Fact 3 guarantees us a height $$1$$ prime $$xB[x]$$ which is of course principally generated and, in light of $$R$$ being $$1$$-dimensional, is maximal. All that remains is to ensure that $$L$$ is not finitely generated as an $$F$$-module, which by Fact $$1$$ ensures that $$R$$ is not Noetherian.

From these considerations, lots of natural examples emerge. For example, take $$A$$ to a be field such that its algebraic closure $$B$$ is not a finite extension, and $$A + xB[x]$$ will be a non-noetherian domain of dimension $$1$$ with a principal maximal ideal generated by $$x$$. As a specific example, you could take $$A = \mathbb{Q}$$ and $$B = \bar{\mathbb{Q}}$$, the ring of algebraic numbers. Or to generalize, take $$D$$ to be any unique factorization domain which is not a field, $$A=K$$ to be its field of fractions, and $$B = \bar{K}$$ to be the algebraic closure of $$K$$ (its a nice exercise to show that this is not a finite degree field extension).

Part $$(2)$$ -- Any non-noetherian 1-dimensional valuation domains has a non-finitely generated maximal ideal

Let $$V$$ be a $$1$$-dimensional valuation ring with fraction field $$K$$. Let's sketch a proof that if its maximal ideal is finitely generated, then it's noetherian.

The only overrings of valuation rings are localizations at prime ideals, so a $$1$$-dimensional valuation ring has no overrings, and it follows that for every ideal $$I$$ of $$V$$, $$(I :_K I) = V$$, since $$(I :_K I) := \{k \in K \mid kI \subseteq I \}$$ has the structure of a ring between $$V$$ and $$K$$. Thus $$V$$ is completely integrally closed, and in particular it is archimedean, i.e. $$\bigcap_n a^nV = 0$$ for all non-unit $$a \in V$$. Now since the maximal ideal $$\mathfrak{m}$$ of $$V$$ is f.g., and $$V$$ is Bezout, $$\mathfrak{m}$$ is even principal. The argument concludes by observing that any local ring $$(R, \mathfrak{m})$$ in which $$\mathfrak{m}$$ is principal and $$\bigcap_n \mathfrak{m} = 0$$ must be a principal ideal ring. This follows because the assumptions guarantee that for any $$a \in R$$, there's a largest $$n$$ such that $$a \in \mathfrak{m}^n \setminus \mathfrak{m}^{n+1}$$, hence $$a = um^n$$ for some unit $$u$$, and it's quick to see that any ideal is principally generated.

$$^1$$ The 2nd and 3rd facts are demonstrated in the 1994 paper Krull and Valuative Dimensions of the $$A + B[x]$$ Rings, by Fontana, Izelgue, and Kabbaj, as Theorem 2.1 and Corollary 1.4, respectively. The first fact I don't have a reference for, but the proof is not hard.

• For proving Part (2) one can also invoke Cohen characterization of noetherian rings using prime ideals. – user26857 Aug 9 at 20:13
• @user26857 hmm yea that's a good point, and for this purpose a much better explanation than the one I gave. I tend to think of this as a special case of the fact that the localization of a completely integrally closed domain at a divisorial prime is a DVR, which I guess got me sidetracked. – Badam Baplan Aug 9 at 20:27