# Is every Noetherian Commutative Ring a quotient of a Noetherian Domain?

This was an interesting question posed to me by a friend who is very interested in commutative algebra. It also has some nice geometric motivation.

The question is in two parts. The first, as stated in the title, asks whether every Noetherian commutative ring a quotient of a Noetherian Domain? Geometrically, this question asks if every Noetherian affine scheme can be embedded as a closed subscheme of an integral scheme.

The second part of the question is that if the first part is answered in the affirmative, then is every (regular) Noetherian ring a quotient of a regular Noetherian domain? Geometrically, this asks whether every affine Noetherian scheme can be embedded as a closed subscheme of a smooth integral scheme.

We managed to make some progress on the first part of the question. We looked at finite products of Noetherian domains and showed that if the Noetherian domains A and B contain a common Noetherian subring C such that A and B are essentially of finite type over C, then $A \times B$ is a quotient of a Noetherian domain. But we weren't able to remove the essentially finite condition and the simplest example which we were unable to work out was $\mathbb{Q} \times \mathbb{C}$.

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I am a little confused about your second part. I usually understand regular to include finite Krull dimension, in which case Nagata's example mathoverflow.net/questions/21067/… violates at least the second statement of your second part. But it sounds like you have a broader definition of regular in mind? – David Speyer Apr 1 '14 at 15:45
I think the notion of regular I'm using just involves the hypothesis that the localization at each prime is a regular local ring, which is well-defined even in the absence of finite Krull-Dimension because each localization is finite Krull-Dimension. So Nagata's example is something I would call regular in the problem but I'm not sure what statement of the problem it violates. Could you elaborate a little on that please? – Siddharth Venkatesh Apr 1 '14 at 19:07
Oh, I see. Regular for you means that all local rings are regular local rings (of some finite Krull dimension). That makes a lot of sense. I'm used to defining "regular of dimension $d$" to mean local rings at maximal primes are regular of dimension $d$, and then "regular" means "regular of some dimension". – David Speyer Apr 2 '14 at 0:04
This might be stupid, but does anyone here can show that $\mathbb Q \times \mathbb F_p$ is the quotient of a Noetherian domain? What are examples of Noetherian domains which are not $\mathbb Q$-algebras but admit $\mathbb Q$ as a quotient? – Joël Apr 2 '14 at 0:16
@Joël The ring $\mathbb{Q}\times\mathbb{F}_p$ is the image of the Noetherian domain $\mathbb{Z}_{(p)}[x]$ under the map sending $x$ to $(1/p,0)$. – Julian Rosen Apr 2 '14 at 1:18

No for cardinality reasons.

Let $F$ a finite field and $G$ a field with cardinality strictly greater than the continuum. Then $F\times G$ is not the homomorphic image of a noetherian integral domain by lemma 2.1 in http://spot.colorado.edu/~kearnes/Papers/residue_final.pdf

Lemma 2.1. Let $R$ be a Noetherian integral domain that is not a finite field and let $I$ be a proper ideal of $R$. If $|R| = \rho$ and $|R/I| = \kappa$, then $\kappa + \aleph_0 \leq \rho \leq \kappa^{\aleph_0}$.

[note: I am not an expert and have not checked.]

[Edit by Joël: for convenience, I add the proof of $\rho \leq \kappa^{\aleph_0}$ taken from the cited article. Since $I$ is finitely generated, $I^n/I^{n+1}$ is a finite $R/I$-module, hence has cardinality at most $\kappa$ (resp. is finite if $\kappa$ is finite). Since $R/I^{n+1}$ is a successive extension of $I^k/I^{k+1}$ for $k=0,1,\dots,n$, the cardinality of $R/I^{n+1}$ is also at most $\kappa$ (resp. finite if $\kappa$ is finite). By Krull's lemma, $\cap_n I^n = 0$, so $R$ injects into $\prod_n R/I^n$ which has cardinality at most $\kappa^{\aleph_0}$, QED.]

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Thanks for the reference. – Siddharth Venkatesh Apr 2 '14 at 15:00
Interesting, this seems to contradict Will's answer (with his notation, take $R_1=\mathbb{F}_p$, $R=R_2=$ a very big field of characteristic $p$). Who is wrong? – abx Apr 2 '14 at 15:25
@abx I think Will is wrong. The proof of Lemma 2.1 is very short and simple, and could be given here for the sake of completeness (if user46855 agrees, I can edit his answer to add that proof). Great answer by the way. – Joël Apr 2 '14 at 18:12
Probably a stupid question, but why this result implies that $F \times G$ is not the quotient of a noetherian domain? – Ricky Apr 4 '14 at 7:58
@Ricky: Assume that $F \times G$ is a quotient of some noetherian domain $R$. Since $F$ is finite and $F$ is a quotient of $R$, the result tells you that the cardinality of $R$ is at most the continuum. But then $R$ cannot have a quotient that is as big as $G$. – Jérôme Poineau Apr 4 '14 at 8:22

I follow Olivier's suggestion and turn my comment on how to realize $\mathbb Q\times \mathbb C$ into an answer.

Fix $r>0$ and consider the ring $\mathbb Q_{r^+}[\![t]\!]$ of power series with coefficients in $\mathbb Q$ that converge on some neighborhood of the closed disc $D(0,r)$ of center 0 and radius $r$ in $\mathbb C$. It is noetherian (see Harbarter, Convergent Arithmetic Power Series). Actually, it is even a PID.

Now consider an evaluation map $f \mapsto f(z)$ with $z \in D(0,r)$. Its image is $\mathbb Q$ if $z=0$, $\mathbb R$ if $z$ is real and $\mathbb C$ otherwise. So we can get $\mathbb Q \times \mathbb C$ as a quotient.

Using the same kind of trick and replacing $\mathbb Q$ by $\mathbb Z$ or localizations of $\mathbb Z$, one should be able to construct a noetherian domain whose quotient is a given finite product of finite fields, finite extensions of $\mathbb Q_p$ or $\mathbb Q$, $\mathbb R$ or $\mathbb C$.

Of course the method has its limits and I have no idea how to realize $\mathbb Q \times \bar{\mathbb Q}$ for instance.

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Very nice answer. – Joël Apr 3 '14 at 23:20
That's a cool answer. Why would this not work with $Q$ replaced with $\overline{Q}$? Does the Noetherian property not hold for it? – Siddharth Venkatesh Apr 4 '14 at 2:06
@SiddharthVenkatesh: You are right, this should work with $\bar{\mathbb Q}$ instead of $\mathbb Q$ (although the ring we get is not explicitly in Harbater's paper). Actually, I have changed the wrong factor and should have written $\mathbb Q \times \bar{\mathbb Q}$ instead of $\bar{\mathbb Q} \times \mathbb C$. I will edit. – Jérôme Poineau Apr 4 '14 at 7:43