1. algebraic geometry
  2. algebraic topology
  3. analysis of PDEs
  4. category theory
  5. classical analysis and ODEs
  6. combinatorics
  7. commutative algebra
  8. complex variables
  9. differential geometry
  10. dynamical systems
  11. functional analysis
  12. general mathematics
  13. general topology
  14. geometric topology
  15. group theory
  16. information theory
  17. K-theory and homology
  18. logic
  19. mathematical physics
  20. metric geometry
  21. number theory
  22. numerical analysis
  23. operator algebras
  24. optimization and control
  25. probability
  26. quantum algebra
  27. representation theory
  28. rings and algebras
  29. spectral theory
  30. statistics theory
  31. symplectic geometry

Non-noetherian local ring with finitely generated maximal ideal

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

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

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: Matsumura “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 UPD 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 nontrivial 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_* to be the ultraproduct of the countable number of copies of A (relative to the chosen ultrafilter).

We remind the reader that A_* 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

    \[\prod_{i \in \mathbf{N}} A \to A_*\]

Lemma 1: Let A be a field. Then the ultraproduct A_ * is a field.

Proof: Consider any nonzero element [(a)] \in A_ * 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_* is a local ring with the maximal ideal \mathfrak{m}_* (ultraproduct of the ideals \mathfrak{m} with respect to the ultrafilter chosen above) and the residue field (A/\mathfrak{m}) _ *.

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

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

There is the natural map A_* \to (A / \mathfrak{m})_* and (\mathfrak{m})_* lies in the kernel of this map. This induces the natural map A_* / \mathfrak{m}_* \to (A/\mathfrak{m})_* that we need to check that it is an isomorphism. Surjectivity is clear. So only the injectivity needs to be checked.

Indeed, let the class [(a)] go to zero in (A /\mathfrak{m})_*. 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}_*.

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

Choose any element [(a)] \notin \mathfrak{m}_*. 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}_* 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}_*, 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 the maximum ideal \mathfrak{m} = (x). Then Lemma 2 tells us that R=A_ * is a local ring with the maximal ideal \mathfrak{m}_*. It follows from the definition that \mathfrak{m}_* is principal and is generated by the element [(x, x, x, x, x, x, x, …)]. Therefore, R is a local ring with a principal maximum ideal. It remains to prove that it cannot be Noetherian.

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

Consider the element t_1:= [(x, x^2, x^3, x^4, …)] in A_*. Obviously, t_1 lies in the maximum ideal \mathfrak{m}_*. 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}_*)^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}_*)^{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 the non-nilpotency condition on the ideal \mathfrak{m}. 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 non-nilpotency of the ideal \mathfrak{m} = (x). So t_1 \neq 0.

Thus, we have verified that A_* is a local ring with a principal maximal ideal, but is not itself Noetherian!

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


Notify of
Most Voted
Newest Oldest
Inline Feedbacks
View all comments
Gleb Terentiuk
1 month 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.

1 month ago

Test of LaTeX in comments. Inline equation a=b. Display equation

    \[c = d.\]