# For a finite-type $\mathbb{Z}$-algebra $A$, is the intersection of all ideals $I$ such that $A/I$ is finite and local necessarily zero?

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 $$\bigcap_{I\in\mathcal{J}} I = 0$$?

• 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.
– YCor
Sep 13, 2022 at 20:29
• Maybe helpful mathoverflow.net/questions/57515/… Sep 13, 2022 at 22:36
• 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. Sep 15, 2022 at 7:03

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.

• 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. Sep 15, 2022 at 13:39
• @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. Sep 15, 2022 at 20:41
• @OwenBiesel I finally got it. (I am willing to improve the presentation if anything remains obscure). Sep 15, 2022 at 22:05
• 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 $\overline{R}\overline{x} = M\overline{R}\overline{x}$? Sep 16, 2022 at 13:48
• @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. 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.

• 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? Sep 15, 2022 at 17:15
• 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. Sep 16, 2022 at 15:52