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

Question

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 $\mathbb{Q} \times \mathbb{C}$.

Answer 1

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.]

Answer 2

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.