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).




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Precise question:



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> 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 $\bigcap\_{I\in\mathcal{J}} I = 0$?
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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?
A proof of the proposed result that is similar (if not identical) to [Peter Kropholler's](https://mathoverflow.net/a/430392/84349) and YCor's proofs can be derived from two *well-known* results, namely **[Lemma 2](https://mathoverflow.net/questions/57515/a-finitely-generated-mathbbz-algebra-that-is-a-field-has-to-be-finite)** and **Theorem 3** below, together with the **[Artin-Rees lemma](https://en.wikipedia.org/wiki/Artin%E2%80%93Rees_lemma)** (alternatively [Theorem 18.4.v, 3]; see proof of Claim 5 and subsequent note).


We shall establish:



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> **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](https://en.wikipedia.org/wiki/Local_ring) of finite cardinal. If $R$ is a finitely generated $\mathbb{Z}$-[algebra](https://en.wikipedia.org/wiki/Algebra_over_a_field#Generalization:_algebra_over_a_ring), then $\mathcal{I}$ is $\{0\}$.
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The main results we need are:



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> **[Lemma 2](https://mathoverflow.net/questions/57515/a-finitely-generated-mathbbz-algebra-that-is-a-field-has-to-be-finite).** [Lemma 4.8, 1]. A field which is finitely generated as a ring is finite.
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> **Theorem 3.** [Theorem 4.19 (Nullstellensatz, General form), 2].
> Let $R$ be a [Jacobson ring](https://en.wikipedia.org/wiki/Jacobson_ring) and let $S$ be a finitely generated $R$-[algebra](https://en.wikipedia.org/wiki/Algebra_over_a_field#Generalization:_algebra_over_a_ring).
> Then $S$ is a Jacobson ring.
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As an intermediate step, we shall prove:



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> **Claim 4.** Let $R$ be a finitely generated $\mathbb{Z}$-algebra. If $R$ is local, then $R$ is a finite ring.
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> *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](https://en.wikipedia.org/wiki/Nilradical_of_a_ring) 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.
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The following result mentioned by YCor is instrumental.
It can be proved by means of the [Artin-Rees lemma](https://en.wikipedia.org/wiki/Artin%E2%80%93Rees_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].



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> **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.)
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> *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 $\overline{x}$.
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**Note.** In order to prove of Claim 5, we can use a result from the Matlis's theory of [injetive modules](https://en.wikipedia.org/wiki/Injective_module) over Noetherian rings instead of the Artin-Rees lemma. It goes as follows.



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> 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](https://en.wikipedia.org/wiki/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.
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Now we are in position to prove Claim 1.



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> *Proof of Claim 1.* Combine Claims 4 and 5.
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[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.


