Background: A referee has suggested a shorter proof of one of my results, but I'm having trouble justifying one of their assertions. The setting is that $A$ is a commutative ring, and the referee's suggestion is to reduce to the case that $A$ is local and finite (as a set) in the following way:

  • First assume that $A$ is finitely generated as a $\mathbb{Z}$-algebra.
  • Then form the set of ideals $\mathcal{J} = \{I \subseteq A\text{ ideal}: A/I\text{ is finite and local}\}$.
  • Claim that $\bigcap_{I\in\mathcal{J}} I = 0$.
  • Deduce that the canonical homomorphism $A \to \prod_{I\in\mathcal{J}} A/I$ is injective.
  • The result for $A$ follows from the result for each finite, local $A/I$.

I'm fine with each of these steps except for the middle one, that the intersection of all the ideals must be $0$. I can tell that $\mathcal{J}$ contains every maximal ideal, even every power of every maximal ideal, so $\bigcap \mathcal J$ must be contained in the Jacobson radical of $A$. Conversely, each ideal $I\in \mathcal{J}$ must be contained in a unique prime $\mathfrak{m}_I$, which must then be maximal since $A/I$ is finite, hence Artinian and dimension $0$. For every example I cook up, $\bigcap \mathcal J$ is zero, even when the Jacobson radical is nonzero, but I can't see why it must happen in general (or if it doesn't).

Precise question:

Let $A$ be a commutative, unital ring that is finitely generated as a $\mathbb{Z}$-algebra. Let $\mathcal{J} = \{I \subseteq A\text{ ideal}: A/I\text{ is finite as a set and local}\}$. Must it be the case that $\bigcap_{I\in\mathcal{J}} I = 0$?

  • 3
    $\begingroup$ More generally, every noetherian (commutative) ring is residually local artinian (the intersection of $I$ such that $A/I$ is local artinian is reduced to zero). Combined with the fact that f.g. commutative rings that are fields are finite. The proof consists of taking $x\neq 0$ and a maximal ideal $I$ not containing $x$, and show that $A/I$ is artinian. Namely a noetherian ring whose intersection of nonzero ideals is nonzero is artinian. This is easy using associated ideals. $\endgroup$
    – YCor
    Sep 13, 2022 at 20:29
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    $\begingroup$ Maybe helpful mathoverflow.net/questions/57515/… $\endgroup$ Sep 13, 2022 at 22:36
  • $\begingroup$ If $X\to Y$ is a morphism of finite type schemes (over $k=\mathbb{Z}$ or a field $k$) such that $X(A)\to Y(A)$ is a bijection for all "finite" $k$-algebras $A$ (either finite rings for $\mathbb{Z}$ or finite dimensional as $k$ vector spaces), then $X\to Y$ is an isomorphism. $\endgroup$
    – Kapil
    Sep 15, 2022 at 7:03

2 Answers 2


A proof of the proposed result that is similar (if not identical) to Peter Kropholler's and YCor's proofs can be derived from two well-known results, namely Lemma 2 and Theorem 3 below, together with the Artin-Rees lemma (alternatively [Theorem 18.4.v, 3]; see proof of Claim 5 and subsequent note).

We shall establish:

Claim 1. Let $R$ be a commutative unital ring. Let $\mathcal{I}$ be intersection of all ideals $I$ of $R$ such that $R/I$ is a local ring of finite cardinal. If $R$ is a finitely generated $\mathbb{Z}$-algebra, then $\mathcal{I}$ is $\{0\}$.

The main results we need are:

Lemma 2. [Lemma 4.8, 1]. A field which is finitely generated as a ring is finite.

Theorem 3. [Theorem 4.19 (Nullstellensatz, General form), 2]. Let $R$ be a Jacobson ring and let $S$ be a finitely generated $R$-algebra. Then $S$ is a Jacobson ring.

As an intermediate step, we shall prove:

Claim 4. Let $R$ be a finitely generated $\mathbb{Z}$-algebra. If $R$ is local, then $R$ is a finite ring.

Proof. Since $R$ is Noetherian, its unique maximal ideal $\mathfrak{m}$ is finitely generated. As $R$ is Jacobson by Theorem 2, the ideal $\mathfrak{m}$ is also the nilradical of $R$. Consequently, there is $n \ge 1$ such that $\mathfrak{m}^n = 0$, which shows in particular that $R$ is Artinian ($R$ is zero-dimensional and Noetherian). To conclude, it only remains to show that the residual field $R/\mathfrak{m}$ of $R$ is finite, which is given by Lemma 1.

The following result mentioned by YCor is instrumental. It can be proved by means of the Artin-Rees lemma as indicated by Peter Kropholler, or alternatively by using a result of Matlis'theory of injective modules of Noetherian rings [Theorem 18.4.v, 3].

Claim 5. Let $R$ be a commutative unital Noetherian ring. Then $R$ is residually local and Artinian, i.e., for every non-zero $x \in R$ there is an ideal $I$ of $R$ such that $x \notin I$ and $R/I$ is local and Artinian. (In other words, the intersection of all ideals $I$ such that $R/I$ is local and Artinian, results in the null ideal.)

Proof. Let $x \in R \setminus \{0\}$ and let $I$ be an ideal of $R$ maximal among the ideals of $R$ not containing $x$. Such an $I$ exists by Zorn's lemma. We shall prove that $\overline{R} = R/I$ is local. Let $\overline{x} = x + I$. By construction, we know that $\overline{x}$ is contained in every non-zero ideal of $\overline{R}$. It also follows from our assumptions on $x$ and $I$ that $\overline{R}\overline{x}$ is a simple $\overline{R}$-module, so that the annihilator $M$ of $\overline{x}$ is a maximal ideal of $\overline{R}$. We claim that there is $n \ge 1$ such that $M^n = \{0\}$. If the claim holds true, then any prime ideal of $\overline{R}$ contains a power of $M$ and hence is equal to $M$, which shows that $\overline{R}$ is local and Artinian. Reasoning by way of contradiction, we assume that $M^n \neq \{0\}$ for every $n \ge 1$. As $\overline{R}$ is Noetherian, we can apply the Artin-Rees lemma [Theorem 8.5, 3]. This lemma yields a positive integer $c$ such that $M^n \cap \overline{R} \overline{x} = M^{n - c}(M^c \cap \overline{R} \overline{x})$ for every $n > c$. Taking $n = c + 1$, we obtain that $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, which is the desired contradiction. Observe indeed that $M^c$ and $M^{c + 1}$ are non-zero by assumption, so that both ideals contain $\overline{x}$.

Note. In order to prove of Claim 5, we can use a result from the Matlis's theory of injetive modules over Noetherian rings instead of the Artin-Rees lemma. It goes as follows.

By construction, the ideal $\overline{R}\overline{x} \simeq \overline{R}/M$ is an essential $\overline{R}$-submodule of $\overline{R}$. Therefore the injective hull $E(\overline{R}\overline{x}) \simeq E(\overline{R} / M)$ contains $\overline{R}$. In particular, $1 \in E(\overline{R}\overline{x})$, so that $M^n \cdot 1 = \{0\}$ for some $n \ge 1$ by [Theorem 18.4.v, 3]. Thus $M^n = \{0\}$, which implies that $\overline{R}$ is local and Artinian, as desired.

Now we are in position to prove Claim 1.

Proof of Claim 1. Combine Claims 4 and 5.

[1] R. Swan, "Excision in algebraic K-theory", 1971.
[2] D. Eisenbud, "Commutative Algebra with a View Towards Algebraic Geometry", 1995.
[3] H. Matsumura, "Commutative Ring Theory", 1989.

  • $\begingroup$ I really appreciate you expanding on how the ideas connect together! At the time of writing you're mentioning there's an error in your proof of Claim 1: is it in deducing that $\overline x^2 = 0$? I don't see why that would follow, since the Jacobson radical of a ring can contain non-nilpotents in general. $\endgroup$ Sep 15, 2022 at 13:39
  • $\begingroup$ @OwenBiesel Dear Owen, the fact that $\overline{x}^2 = 0$ is correct (but it doesn't seem to be useful eventually). Hopefully, the changes that I just made in my tentative proof of Claim 1 will make this clearer. Note indeed that we have $\overline{x} \in M$, by construction, i.e. simply by means of our assumptions on $I$ and $x$. The (removed) error was in the reasoning right after this point. I'll come back to this incomplete proof in a day or two. $\endgroup$
    – Luc Guyot
    Sep 15, 2022 at 20:41
  • $\begingroup$ @OwenBiesel I finally got it. (I am willing to improve the presentation if anything remains obscure). $\endgroup$
    – Luc Guyot
    Sep 15, 2022 at 22:05
  • $\begingroup$ Thank you, this is very helpful! I follow you right up until the last line of the proof of Claim 5: letting $n = c+1$, don't we get $M^{c+1} \cap \overline{R}\overline{x} = M(M^c \cap \overline{R}\overline{x})$? Why does that mean $\overline{R}\overline{x} = M\overline{R}\overline{x}$? $\endgroup$ Sep 16, 2022 at 13:48
  • $\begingroup$ @OwenBiesel Thanks for your feedback. This is what we get indeed. As we have assumed that $M^{c + 1} \neq \{0\}$, we know that $M^{c + 1}$ contains $\overline{x}$. The same holds for $M^c$. Let's see if I can improve this last line. $\endgroup$
    – Luc Guyot
    Sep 16, 2022 at 14:47

It suffices to prove that when $A$ has a unique minimal ideal and is an essential extension of it then $A$ is finite. This will follow from the Artin-Rees property along with the Nullstellensatz.

Edit: in more detail and focussing on the 'middle step': let $x$ be a non-zero element of $A$. Let $I$ be an ideal maximal subject to not containing $x$ - chosen using Zorn's lemma. Then $A/I$ has a unique minimal ideal $J/I$ and this is an irreducible $A$-module so finite by a version of Hilbert's Nullstellensatz. Now use Artin-Rees to conclude that $A/I$ is finite.

  • $\begingroup$ Could you add a little more information? Which version of the Nullstellensatz are you using (there are so many), and how does the Artin-Rees lemma apply here? $\endgroup$ Sep 15, 2022 at 17:15
  • $\begingroup$ A version of the Nullstellensatz that meets the requirements says "Every field that is finitely generated as a ring is finite". This is also used in Luc Guyot's argument and he gives a reference. This can be applied to conclude that $J/I$ is finite because there is a maximal ideal $M$ of $A$ such that $A/M$ is isomorphic to $J/I$. The Artin Rees Lemma says that the $M$-adic topology on $A/I$ induced the $M$-adic topology on $J/I$ and hence there is a natural number $n$ such that $M^n$ annihilates $A/I$. This shows that $A/I$ is both finite and local. $\endgroup$ Sep 16, 2022 at 15:52

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