Is a local ring with a certain finite quotient necessarily noetherian?

Question

Let $R$ be a commutative unital ring, and let $p$ be a prime number.

Question 1: If $R$ is a local ring (with unique maximal ideal $m$) of characteristic $0$, $p \in m$, and $R/p$ is finite, does it follow that $R$ is noetherian?

Question 2: In case the answer to 1 is "no", does it hold if we additionally assume $R$ to be a domain?

Motivation: This came out of discussions to the question $\mathcal{O}_{\mathbb{C}_p}/ p \mathcal{O}_{\mathbb{C}_p}$ infinite. I have come to suspect that the answer is no to both, but I have not suceeded in cooking up counterexamples. I'm afraid it's been too long since I studied all the local ring regularity properties, and just hope there is no easy proof with Nakayama or something that I'm missing.

Answer 1

Let $S\subset\mathbb{Q}[x]$ be the subring consisting of polynomials with integer constant term. Note that $(p)$ is a maximal ideal of $S$ with $S/(p)\cong\mathbb{F}_p$. Let $R$ be the localization $S_{(p)}$. It is clear that $R$ is a local domain of characteristic $0$ with maximal ideal generated by $p$ and $R/(p)\cong \mathbb{F}_p$ is finite.

I claim, however, that $R$ is not Noetherian. Indeed, consider the chain of ideals $(x)\subset(x/p)\subset(x/p^2)\subset\dots$. These inclusions are all strict since $1/p\not\in R$ so $x/p^{n+1}$ is not a multiple of $x/p^n$ for any $n$.

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Here is a very general principle along these lines. Let $R$ be any non-Artinian ring. Then I claim $R$ has an elementary extension (over the first-order language of rings) which is not Noetherian. Indeed, add a sequence of constant symbols $x_0,x_1,x_2,\dots$ to the language and axioms saying $x_n$ is not in the ideal generated by $x_0,\dots,x_{n-1}$ for each $n$. These axioms are finitely satisfiable in $R$, since $R$ is not Artinian so there are arbitrarily long finite chains of ideals in $R$. So by compactness, there is an elementary extension of $R$ that satisfies all of them, and thus is not Noetherian.

In particular, applying this with (say) $R=\mathbb{Z}_{(p)}$ this gives an elementary extension of $\mathbb{Z}_{(p)}$ that is not Noetherian. The properties of being local, characteristic $0$, $p$ being in the maximal ideal, a domain, and $R/(p)$ having $p$ elements are all first-order, and thus are still true of this elementary extension.

Answer 2

(Sorry this might be incorrect, see below.)

(Set $\mathbb{N} := \mathbb{Z}_{\ge 0}$.) Let $\mathbb{Z}_{p}[[\{x_{n}\}_{n \in \mathbb{N}}]]$ be the $\langle \{x_{n}\}_{n \in \mathbb{N}} \rangle$-adic completion of $\mathbb{Z}_{p}[\{x_{n}\}_{n \in \mathbb{N}}]$. (As power series, we have the following description: Let $S$ be the set of functions $e : \mathbb{N} \to \mathbb{N}$ with finite support (i.e. $e(n) = 0$ for all but finitely many $n$); then an element of $\mathbb{Z}_{p}[[\{x_{n}\}_{n \in \mathbb{N}}]]$ can be identified with a function $F : S \to \mathbb{Z}_{p}$ such that for each "degree" $d$, there are only finitely many $e \in S$ such that $|e| = d$ and $F(e) \ne 0$.)

Set $R := \mathbb{Z}_{p}[[\{x_{n}\}_{n \in \mathbb{N}}]]/\langle \{x_{n} = px_{n+1}\}_{n \in \mathbb{N}} \rangle$. Then $R$ is local because it is a quotient of a local ring, and <error>$R/pR \simeq \mathbb{F}_{p}$ (since in every power series whose constant term is $0$, we can replace $x_{n} \mapsto px_{n+1}$ then divide by $p$)</error> (we can't do infinitely many replacements, for example in $x_{1} + x_{2}^{2} + x_{3}^{3} + \dotsb$). We have an injective ring map $\varphi : R \to \mathbb{Q}_{p}[[x]]$ sending $x_{n} \mapsto \frac{1}{p^{n}}x$ (in terms of the above description, we have $F \mapsto \sum_{d \in \mathbb{N}}(\sum_{|e| = d}F(e))x^{d}$), thus $R$ is an integral domain. We show that $\langle x_{n} \rangle \subsetneq \langle x_{n+1} \rangle$ for all $n$, so that $\langle x_{0} \rangle \subsetneq \langle x_{1} \rangle \subsetneq \langle x_{2} \rangle \subsetneq \dotsb$ is a strictly ascending chain of ideals of $R$. Note that each $x_{n}$ is a nonzerodivisor in $R$ since its image in $\mathbb{Q}_{p}[[x]]$ is a nonzerodivisor. Suppose $x_{n+1} = x_{n} g$ for some $g \in R$; then $x_{n+1} = x_{n+1} pg$ so $1 = pg$, i.e. $R$ contains $\frac{1}{p}$. But this contradicts the existence of a ring map $R \to \mathbb{Z}_{p}$ (for example, the one induced by the map $\mathbb{Z}_{p}[[\{x_{n}\}_{n \in \mathbb{N}}]] \to \mathbb{Z}_{p}$ sending $x_{n} \mapsto 0$ for all $n \in \mathbb{N}$).