Questions tagged [noetherian]

For questions on Noetherian rings, Noetherian modules and related notions.

A (commutative) ring is called Noetherian if every ascending chain of ideals becomes stationary. For non-commutative rings, the notions of left- and right-Noetherian exist, and they apply to left and right ideals respectively. A Noetherian non-commutative ring is both left- and right-Noetherian.

More generally, a module is Noetherian if each ascending chain of submodules becomes stationary.

A vector space is Noetherian if and only if it is of finite dimension, and "Noetherian" can in a vague way be considered as a generalization of "finite-dimensional."

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A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?

Background. If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$ , where $\dim$ denotes the Krull dimension . If $R$ is Noetherian, we have equality. Every proof of this fact I'm aware of uses quite a bit of commutative…
Martin Brandenburg
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Commutative non Noetherian rings in which all maximal ideals are finitely generated

In commutative rings we have the following Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated. From this Theorem I am looking for commutative rings $R$ in which every maximal ideal is finitely generated but $R$…
Tomi Albertini
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Why is the localization of a commutative Noetherian ring still Noetherian?

This is an unproven proposition I've come across in multiple places. Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian. Why is this? I thought about taking some chain of submodules $$…
Buble
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A non-noetherian ring with noetherian spectrum

Question 1: Does such a ring exist? Note: The definition of a noetherian topological space is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite number of steps ( Question 2: is this equivalent to saying…
fosco
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If $R$ is a Noetherian ring then $R[[x]]$ is also Noetherian

Let $R$ be a Noetherian ring. How can one prove that the ring of the formal power series $R[[x]]$ is again a Noetherian ring? It is well-known that the ring of polynomials $R[x]$ is Noetherian. I try imitating the standard proof of the fact by…
Pooya Ronagh
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Noetherian ring with infinite Krull dimension (Nagata's example).

I just started to read about the Krull dimension (definition and basic theory), at first when I thought about the Krull dimension of a noetherian ring my idea was that it must be finite, however this turned out to be wrong. I am looking for an…
user117449
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A non-noetherian ring with all localizations noetherian

If for a ring $A$ every localization $A_\mathfrak{p}$ by a prime $\mathfrak{p}\subseteq A$ is noetherian, is it true that $A$ is noetherian? I believe not but I can't find a good counterexample.
Daniele A
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When is a tensor product of two commutative rings noetherian?

In particular, I'm told if $k$ is commutative (ring), $R$ and $S$ are commutative $k$-algebras such that $R$ is noetherian, and $S$ is a finitely generated $k$-algebra, then the tensor product $R\otimes_k S$ of $R$ and $S$ over $k$ is a noetherian…
Heidi
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Why are Noetherian Rings important?

I know they are important in abstract algebra, but why do people study them? Why are they so important to study? Do they make certain things easier to understand?
user114442
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Subring of a finitely generated Noetherian ring need not be Noetherian?

A common example showing that a subring of a Noetherian ring is not necessarily Noetherian is to take a polynomial ring over a field $k$ in infinitely many indeterminates, $k[x_1,x_2,\dots]$. The quotient field is then obviously Noetherian, but the…
Emilia
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What is a clever proof of Hilbert's basis theorem?

So I am studying commutative algebra at the moment and I have come across the proof of the Hilbert Basis Theorem (the proof I have is the same as the one in Reid's Undergraduate Commutative Algebra ). I can't see how I would ever have thought of…
user183870
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Is the ring of holomorphic functions on $S^1$ Noetherian?

Let $S^1={\{ z \in \Bbb{C} : |z|=1 \}}$ be the unit circle. Let $R= \mathcal{H}(S^1)$ be the ring of holomorphic functions on $S^1$, i.e. the ring of functions $f: S^1 \longrightarrow \Bbb{C}$ which can be extended to an holomorphic function on an…
Crostul
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Primary ideals of Noetherian rings which are not irreducible

It is known that all prime ideals are irreducible (meaning that they cannot be written as an finite intersection of ideals properly containing them). While for Noetherian rings an irreducible ideal is always primary, the converse fails in general .…
Alex Becker
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What is an easy example of non-Noetherian domain?

Keep in mind, I'm strictly an amateur, though a very old one. I learned about imaginary numbers barely two years ago and ideals a year ago, and I'm still decidedly a novice in both topics. In the university library, I was looking at Modules over…
Mr. Brooks
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"categorical" proof of a seemingly symmetric statement about Noetherian/Artinian modules

There are two statements which to me seem rather symmetric: Let $A$ be a ring, $M$ an $A$ -module, and $f : M \to M$ . If $M$ is Noetherian and $f$ is surjective, then $f$ is injective. If $M$ is Artinian and $f$ is injective, then $f$ is…
MT_
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