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 be a finitely generated ideal in a ring
. Then the
-adic completion
is Noetherian if
is.
Corollary: Let be a complete local ring with a finitely generated maximal ideal
. Then
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 be a ring such that any prime ideal
is finitely generated. Then
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 with a finitely generated maximal ideal
.
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 . 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 with a non-nilpotent maximal ideal
(for example, the ring
works). We define
to be the ultraproduct of the countable number of copies of
(relative to the chosen ultrafilter).
We remind the reader that is defined as the quotient of the countable product
by the relation
if the set of indices where
lies in the chosen ultrafilter (we say that
almost everywhere).
We have the natural surjective map
Lemma 1: Let be a field. Then the ultraproduct
is a field.
Proof: Consider any nonzero element and any of its representatives
. Since
is nonzero, the set of indices where
does not lie in the ultrafilter. This implies that the set of indices, where
lies in the ultrafilter. Therefore, the class
is equal to the class
defined as
if
and
otherwise . Then
and the class
is the inverse of
. So the element
is invertible.
Lemma 2: Let be a local ring with the maximal ideal
. Then the ultraproduct
is a local ring with the maximal ideal
(ultraproduct of the ideals
with respect to the ultrafilter chosen above) and the residue field
.
Proof: There is the natural map . This map is obviously injective. The image is an ideal in
since it consists exactly of the classes that have a representative
, where all coordinates
lie in the maximal ideal
.
Step 1: The ideal is maximal. We note that to check that this ideal is maximal, it suffices to check that
because Lemma 2 guarantees that the right side of the equality is a field.
There is a natural map and
lies in the kernel of this map. This induces a natural map
that we need to check is an isomorphism. Since surjectivity is clear, it remains to check injectivity.
Indeed, suppose the class is zero in
. This means that for almost all indices
, we have
. Let us now define the element
as
if
, and
otherwise. So
but now the class
clearly lies in
. Thus,
.
Step 2: The ring is local. It is enough to show that any element
that does not lie in the maximal ideal
is invertible.
Choose any element . Then the set of indices
such that
does not lie in our ultrafilter. Therefore, the set of indices
such that
lies in the ultrafilter. Let’s do the standard trick with replacing the sequence
with the sequence
defined by the rule
, if
, and
otherwise. Then
and
is clearly invertible with the inverse
. This finishes the proof of Lemma 2.
Now let’s return to our case when is a local principal ideal domain with maximal ideal
. Then Lemma 2 tells us that
is a local ring with maximal ideal
. It follows from the definition that
is principal and is generated by the element
. Therefore,
is a local ring with a principal maximal ideal. It remains to prove that
is not Noetherian.
Recall that Krull’s Intersection Theorem says that if is a Noetherian ring, the intersection
.We will show that this is not the case for
.
Consider the element in
. Obviously,
lies in the maximal ideal
. Now we note that
since the ultrafilter contains all cofinite subsets of
and so changing any finite number of elements does not change the class in the ultraproduct. But now
lies in
. Similarly, the element
lies in
. Keep going to get that
. Thus,
The last thing left to check is that the constructed element is not zero. This is exactly the place where we need to use that
is not nilpotent. Indeed, let us assume that
. This means that the set of indices
such that
lies in our ultrafilter. An empty set does not lie in an ultrafilter from the definition of an ultrafilter. Therefore, there is some
such that
. But then this contradicts the fact that
is not nilpotent. So
.
Thus, we have verified that 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.
For another example let us consider the ring
of germs of
functions on the real line near
. Strictly speaking, elements of
are equivalence classes but let me ignore it for simplicity and work with small enough neighborhoods. In this case
consists of such
that
. Note that if
then by the Hadamard’s lemma one can write
where
is a smooth function. Thus, the maximal ideal is generated by one element
. Let us show that 
To do this, let us check that if
for
, then
. Let us proceed by induction, use the Hadamard’s lemma to write
where
is a smooth function. Note
. Therefore,
for
. Thus,
by induction hypotesis and therefore
.
In particular, if
is a function with
for any
, then
. Therefore,
is non-zero due to existence of non-analytic smooth functions. In particular, the ring
is non-noetherian and its maximal ideal is generated by one element.
Thanks! Updated the post.
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