A non-Noetherian local ring with finitely generated maximal ideal

Some time ago I found the following interesting lemma on the stacksproject:

Theorem: (tag/05GH) Let $I$ be a finitely generated ideal in a ring $R$ . Then the $I$ -adic completion $\widehat{R}$ is Noetherian if $R/I$ is.

Corollary: Let $R$ be a complete local ring with a finitely generated maximal ideal $\mathfrak{m}$ . Then $R$ is Noetherian.

This corollary turns out to be actually quite useful in some examples.

The natural question is whether this corollary holds without the completeness assumption. A positive answer in this direction is given by the following theorem:

Theorem: (Cohen) Let $R$ be a ring such that any prime ideal $\mathfrak p \subset R$ is finitely generated. Then $R$ is a Noetherian ring.
Proof: See Matsumura’s Commutative Ring Theory, Theorem 3.4.

However, it turns out that the answer is negative if one only assumes that all maximal ideals are finitely generated.

Proposition: There is a non-Noetherian local ring $R$ with a finitely generated maximal ideal $\mathfrak{m}$ .

I don’t know if there is an easy way to construct such a ring (see the update at the end of the post for some other examples). I’ll explain a construction that uses the notion of ultraproducts.

Before we start the construction, we choose an arbitrary non-trivial ultrafilter (any ultrafilter that contains a filter of cofinite sets) on the set of natural numbers $\mathbf{N}$ . We note that such an ultrafilter exists by Zorn’s lemma (ultrafilters are exactly maximal filters). In what follows, we will say just say “ultraproduct” instead of “ultraproduct with respect to the chosen nontrivial ultrafilter.”

Now we pick any local principal ideal ring $A$ with a non-nilpotent maximal ideal $\mathfrak{m}$ (for example, the ring $\mathbf{Z} _ {p}$ works). We define $R:=A_\ast$ to be the ultraproduct of the countable number of copies of $A$ (relative to the chosen ultrafilter).

We remind the reader that $A_\ast$ is defined as the quotient of the countable product $\prod_{i=1}^{\infty} A$ by the relation $(a_i)=(b_i)$ if the set of indices where $a_i=b_i$ lies in the chosen ultrafilter (we say that $a_i=b_i$ almost everywhere).

We have the natural surjective map

$\prod_{i \in \mathbf{N}} A \to A_\ast$

Lemma 1: Let $A$ be a field. Then the ultraproduct $A_ \ast$ is a field.

Proof: Consider any nonzero element $[(a)] \in A_ \ast$ and any of its representatives $a \in \prod_{i \in \mathbf{N}} A$ . Since $[(a)]$ is nonzero, the set of indices where $a_i = 0$ does not lie in the ultrafilter. This implies that the set of indices, where $a_i \neq 0$ lies in the ultrafilter. Therefore, the class $[(a)]$ is equal to the class $[(a ')]$ defined as $(a')_i = a_i$ if $a_i \neq 0$ and $a'_i=1$ otherwise . Then $a' \in (\prod_{\ i \in \mathbf{N}} A)^\times$ and the class $[(a') ^ {- 1}]$ is the inverse of $[(a')] = [(a)]$ . So the element $[(a)]$ is invertible.

Lemma 2: Let $A$ be a local ring with the maximal ideal $\mathfrak{m}$ . Then the ultraproduct $A_\ast$ is a local ring with the maximal ideal $\mathfrak{m}_\ast$ (ultraproduct of the ideals $\mathfrak{m}$ with respect to the ultrafilter chosen above) and the residue field $(A/\mathfrak{m}) _ \ast$ .

Proof: There is the natural map $\mathfrak{m}_\ast \to A_\sat$ . This map is obviously injective. The image is an ideal in $A_\ast$ since it consists exactly of the classes that have a representative $(a)$ , where all coordinates $a_i$ lie in the maximal ideal $\mathfrak{m}$ .

Step 1: The ideal $\mathfrak{m}_\ast$ is maximal. We note that to check that this ideal is maximal, it suffices to check that $A_\ast / \mathfrak{m}_\ast = (A/\mathfrak{m})_\ast$ because Lemma 2 guarantees that the right side of the equality is a field.

There is a natural map $A_\ast \to (A / \mathfrak{m})_\ast$ and $(\mathfrak{m})_\ast$ lies in the kernel of this map. This induces a natural map $A_\ast / \mathfrak{m}_\ast \to (A/\mathfrak{m})_\ast$ that we need to check is an isomorphism. Since surjectivity is clear, it remains to check injectivity.

Indeed, suppose the class $[(a)]$ is zero in $(A /\mathfrak{m})_\ast$ . This means that for almost all indices $i$ , we have $a_i \in \mathfrak{m}$ . Let us now define the element $(a')$ as $a'_i= a_i$ if $a_i \in \mathfrak{m}$ , and $a'_i = 0$ otherwise. So $[(a)] = [(a')]$ but now the class $[(a')]$ clearly lies in $\mathfrak{m}_*$ . Thus, $[(a)] \in \mathfrak{m}_\ast$ .

Step 2: The ring $A_\ast$ is local. It is enough to show that any element $[(a)]$ that does not lie in the maximal ideal $\mathfrak{m}_\ast$ is invertible.

Choose any element $[(a)] \notin \mathfrak{m}_\ast$ . Then the set of indices $i$ such that $a_i \in \mathfrak{m}$ does not lie in our ultrafilter. Therefore, the set of indices $i$ such that $a_i \notin \mathfrak{m}_\ast$ lies in the ultrafilter. Let’s do the standard trick with replacing the sequence $(a)$ with the sequence $(a')$ defined by the rule $a'_i = a_i$ , if $a_i \notin \mathfrak{m}_\ast$ , and $a'_i = 1$ otherwise. Then $[(a)] = [(a ')]$ and $[(a')]$ is clearly invertible with the inverse $[(a'^{- 1})]$ . This finishes the proof of Lemma 2.

Now let’s return to our case when $A$ is a local principal ideal domain with maximal ideal $\mathfrak{m} = (x)$ . Then Lemma 2 tells us that $R=A_ \ast$ is a local ring with maximal ideal $\mathfrak{m}_\ast$ . It follows from the definition that $\mathfrak{m}_\ast$ is principal and is generated by the element $[(x, x, x, x, x, x, x, …)]$ . Therefore, $R$ is a local ring with a principal maximal ideal. It remains to prove that $R$ is not Noetherian.

Recall that Krull’s Intersection Theorem says that if $(A, \mathfrak{m})$ is a Noetherian ring, the intersection $\cap_{n=1}^{\infty} \mathfrak{m}^n=(0)$ .We will show that this is not the case for $(R=A_\ast, \mathfrak{m}_\ast)$ .

Consider the element $t_1:= [(x, x^2, x^3, x^4, …)]$ in $A_\ast$ . Obviously, $t_1$ lies in the maximal ideal $\mathfrak{m}_\ast$ . Now we note that $t_1 = t_2:= [(x^2, x^2, x^3, x^4, …)]$ since the ultrafilter contains all cofinite subsets of $\mathbf{N}$ and so changing any finite number of elements does not change the class in the ultraproduct. But now $t_1 = t_2$ lies in $(\mathfrak{m}_\ast)^2$ . Similarly, the element $t_1 = t_3: = [(x^3, x^3, x^3, x^4, x^5, …)]$ lies in $(\mathfrak{m}_*)^3$ . Keep going to get that $t_1 = t_N := [(x^N, … \text{N times}, x^N, x^{N + 1}, x^{N + 2}, …)] \in (\mathfrak{m}_\ast)^{N}$ . Thus,

$t_1 \in \cap_ {i = 1} ^{\infty} (\mathfrak{m}_*)^i.$

The last thing left to check is that the constructed element $t_1$ is not zero. This is exactly the place where we need to use that $\mathfrak{m}$ is not nilpotent. Indeed, let us assume that $t_1 = 0$ . This means that the set of indices $i$ such that $x^i = 0$ lies in our ultrafilter. An empty set does not lie in an ultrafilter from the definition of an ultrafilter. Therefore, there is some $M$ such that $x^M = 0$ . But then this contradicts the fact that $\mathfrak{m} = (x)$ is not nilpotent. So $t_1 \neq 0$ .

Thus, we have verified that $A_\ast$ is a local ring with finitely generated (even principal!) maximal ideal, but is not Noetherian!

Update: Gleb Terentiuk explained in the comments that there is an easier construction. Also, look here for some other examples.

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Gleb Terentiuk

2 years ago

For another example let us consider the ring $A$ of germs of $C^{\infty}$ functions on the real line near $0$ . Strictly speaking, elements of $A$ are equivalence classes but let me ignore it for simplicity and work with small enough neighborhoods. In this case $\mathfrak{m}$ consists of such $f \in A$ that $f(0)=0$ . Note that if $f \in \mathfrak{m}$ then by the Hadamard’s lemma one can write $f(x) = x g(x)$ where $g(x)$ is a smooth function. Thus, the maximal ideal is generated by one element $x$ . Let us show that $\bigcap_{i \ge 0} \mathfrak{m}^i \ne (0).$

To do this, let us check that if $f^{(i)}(0) = 0$ for $i \le n$ , then $f(x) \in \mathfrak{m}^{n+1}$ . Let us proceed by induction, use the Hadamard’s lemma to write $f(x) = x g (x)$ where $g (x)$ is a smooth function. Note $f^{(k)} (x) = k g^{(k-1)} (x) + x g^{(k)} (x)$ . Therefore, $g^{(i)}(0) = 0$ for $i \le n-1$ . Thus, $g(x) \in \mathfrak{m}^{n}$ by induction hypotesis and therefore $f(x) \in \mathfrak{m}^{n+1}$ .

In particular, if $f(x)$ is a function with $f^{(k)} (0) = 0$ for any $k$ , then $f(x) \in \bigcap_{i \ge 0} \mathfrak{m}^i$ . Therefore, $\bigcap_{i \ge 0} \mathfrak{m}^i$ is non-zero due to existence of non-analytic smooth functions. In particular, the ring $A$ is non-noetherian and its maximal ideal is generated by one element.

Author

Bogdan Zavyalov

Reply to Gleb Terentiuk

Thanks! Updated the post.

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