Mathematics Stack Exchange
Mathematics Stack Exchange

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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How to prove that $k[x, xy, xy^2, \dotsc]$ is not noetherian?

Consider the subring $R=k[x,xy,xy^2,\ldots]$ of $k[x,y]$. I want to prove that $R$ is not noetherian. An ascending chain of ideals is the following: $$(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$$ It is intuitively clearly to me that this is an…
Mohan
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Is noetherianity a local property?

Let $R$ be a ring with finitely many maximal ideals such that $R_{\mathfrak m}$ ($\mathfrak m$ maximal ideal) is noetherian ring for all $\mathfrak m$. Is $R$ noetherian? I think $R$ has to be noetherian. Let $p_1 \subset p_2 \subset \cdots…
Sarjbak
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Converse to Hilbert basis theorem

Prove the converse to Hilbert basis theoren: If the polynomial ring $R[x]$ is Noetherian, then $R$ is noetherian.
mshj
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If a ring is Noetherian, then every subring is finitely generated?

Let $R$ be a commutative ring with $1$, and let $K$ be a field. We know that $R$ is Noetherian iff every ideal of $R$ is finitely generated as an ideal. Question 1: If $R$ is Noetherian, is every subring of $R$ finitely generated as a ring? Is there…
Leo
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How does one prove that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not Noetherian?

How does one prove that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not Noetherian? I let $(1, f_1, ..., f_n,...)$ be the $\mathbb{Z}$-basis of $\text{Int}(\mathbb{Z})$, the ring of rational polynomials sending $\mathbb{Z}$…
byesac
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Exercise $5.5.$M, Foundations of Algebraic Geometry by Ravi Vakil

In the following, $A$ is a Noetherian ring. The basic definition here is: A prime $p\subsetneqq A$ is said to be associated to $M$ if $p$ is the annihilator of an element $m$ of $M$. The exercise is: $5.5.$M. EXERCISE. (a) If $M$ is a finitely…
XiaYu
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Krull dimension of polynomial rings over noetherian rings

I want to prove the following theorem concerning Krull dimension: Theorem If $A$ is a noetherian ring then $$\dim(A[x_1,x_2, \dots , x_n]) = \dim(A) + n$$ where $\dim$ stands for the Krull dimension of the rings. Thus, $\dim(K[x_1,x_2, \dots ,…
Pipicito
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Ring of formal power series over a principal ideal domain is a unique factorisation domain

An exercise in my algebra course book asks to prove that if $R$ is a PID, then $R[[x]]$ is a UFD, where $R[[x]]$ is the ring of formal power series over $R$. After some failed attempts at proving the ACC I visited Wikipedia, which comments: If $R$…
Bart Michels
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Noetherian ring with finitely many height $n$ primes

If $R$ is a Noetherian commutative ring with unity having finitely many height one prime ideals, one could derive from the "Principal Ideal Theorem", due to Krull, that $R$ has finitely many prime ideals (all of height less than or equal to $1$). It…
karparvar
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Is an epimorphic endomorphism of a noetherian commutative ring necessarily an isomorphism?

Let $A$ be noetherian commutative ring with one, and let $f:A\to A$ be an epimorphic endomorphism of $A$. Is $f$ necessarily an isomorphism? ("An epimorphic endomorphism" means of course "an endomorphism which is an epimorphism". In this post…
Pierre-Yves Gaillard
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Is every commutative ring a limit of noetherian rings?

Let $\mathsf{Noeth}$ be the category of noetherian rings, viewed as a full subcategory of the category $\mathsf{CRing}$ of commutative rings with one. Let $A$ be in $\mathsf{CRing}$. Question 1. Is there a functor from a small category to…
Pierre-Yves Gaillard
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Intersection of powers of an ideal in a Noetherian ring

Given a Noetherian ring $R$ and a proper ideal $I$ of it. Is it true that $$\bigcap_{n\ge 1} I^n=0$$ as $n$ varies over all natural numbers? If not, is it true if $I$ is a maximal ideal? If not, is it true if $I$ is the maximal ideal of a local…
Dev Bappa
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Looking for a special class of ideals such that If every ascending chain of ideals from this class stabilizes, then $R$ is a Noetherian ring.

A commutative ring $R$ is called Noetherian if any one of the following holds: $1.$ Every ascending chain of ideals in $R$ stabilizes, that is, $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ implies the existence of $n\in\mathbb{N}$ such that…
Arpit Kansal
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If $M$ is a Noetherian $R$-module, then $R/\text{Ann}(M)$ is a Noetherian ring

Let $M$ be an $R$-module and $\text{Ann}(M)=\{r \in R: rm =0 , \forall m \in M\}.$ Suppose $M$ is Noetherian. Could anyone advise me on how to prove $R/\text{Ann}(M)$ is also Noetherian? Hints will suffice. Thank you.
Alexy Vincenzo
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The total ring of fractions of a reduced Noetherian ring is a direct product of fields

This is question 6.5 in Matsumura's "Commutative ring theory": How can I prove that the total ring of fractions of a reduced Noetherian ring is a direct product of fields?
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