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Questions tagged [noetherian]

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

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Noetherian/Artinian modules

Just trying to get my head around Noetherian and Artinian modules, I've come across this question, which I don't really know how to approach: Let $R=F[x,y]/(x^3)$ where $F$ is a field. Is R ...
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Does radical ideals $\sqrt{I_1}\subseteq\sqrt{I_2}\cdots$ becoming stationary imply ideals $I_1\subseteq I_2\cdots$ becoming stationary?

I wanted to solve the following problem: Let $R$ be a commutative ring with $1$ and $M$ a Noetherian $R$-module. Let $I=\operatorname{Ann}(M)$. Show that $R/I$ is a Noetherian ring. Now to show ...
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60 views

$M_p$ is artinian for all primes p, then M is artinian

Let $M$ be a finite $R$-module that is noetherian and such that $M_{\mathfrak{p}}$ is artinian for each $\mathfrak{p}\in \text{Spec}(R)$. Then $M$ is an artinian $R$-module. I have tried using a ...
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“De-localization” of a Noetherian module?

Let $R$ be a ring, $S\subset R$ a multiplicative subset, and let $M$ be a Noetherian $S^{-1}R$-module, then does there exist some Noetherian $R$-module such that $S^{-1}N\cong M$? What about if we ...
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Why $\ker \varphi|_N$ is finitely generated?

Let $R$ Noetherian and $M$ finitely generated over $R$, prove $M$ is Noetherian. Proof We can suppose WLOG that $M\cong R^{\oplus s}$. The proof goes by induction. For $s=1$ it's obvious. Suppose $...
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When does an integral group ring have finite global dimension?

Let $G$ be a finite group and $R=\mathbb{Z}[G]$ the integral group ring. If $G$ is such that $R$ is Noetherian (so $G$ polycyclic-by-finite) when does $R$ have finite global dimension? Another way of ...
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Summand Noetherian is Noetherian?

I want to prove if $S$ is a Noetherian ring, $R\subset S$ a subring, and there is an $R$-module homomorphism $\pi:S\to R$ such that $\pi$ is surjective and every element of $R$ is fixed by $\pi$ ($R$ ...
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315 views

If the localization of a ring at each prime ideal is Noetherian, does this imply that ring is Noetherian? [duplicate]

If the localization $R_p$ of a ring $R$ at each prime ideal $p$ in A is Noetherian, does this imply that $A$ is Noetherian? What we call such rings which is not Noetherian but localization at each ...
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209 views

What is an example of a Noetherian Semi-local ring with an infinite number of prime ideals?

This question comes from the naive belief that $|Spec(R)|\subset P(|Specm(R)|)$, which I now know is only true if $R$ is a Jacobson ring, which lead me to believe that Semilocal rings are ...
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Inverse image of a Noetherian module under a module morphism

During my exercise session of Commutative algebra we had the following question: Suppose $f: M \to N$ is a morphism of $R$-modules (where $R$ is a commutative ring with 1). Suppose $N$ is Noetherian, ...
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Intersection of all symbolic powers of a prime ideal

This is Exercise 8.37 in R.Y.Sharp's Steps in Commutative Algebra: Let $P$ be a prime ideal of the commutative Noetherian ring $R$. Prove that $$\bigcap_{n=1}^{\infty} P^{(n)}=\{ r\in R \mid \...
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Noetherian, Artinian on graded ring and localization

There's an exercise in Tom Marley's Graded Rings and Modules making me confused, stating $R$ is a nonnegatively graded local ring with $R_0$ being local. Let $M$ be the unique homogeneous maximal ...
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If every proper subring is noetherian then the ring is noetherian?

If every proper subring is noetherian then the ring is noetherian ? For example every proper subring of $\mathbb{Z} $ is $m\mathbb{Z}$ and we know that $m\mathbb{Z} $ is noetherian.
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The dimension modulo a principal ideal in a Noetherian local ring

Let $(A,\mathfrak m)$ be a Noetherian local ring and $f$ an element in the maximal ideal $\mathfrak m$. Is it true that $\dim A/(f) \geq \dim A - 1$? I don't really see a way to connect the two ...
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Every non-zero noetherian module has a simple quotient module.

Let $R$ be an arbitrary ring and $M$ a non-zero noetherian R-module. Show it has a simple quotient module. My idea: Let $S$ be the set of non-zero submodules of $M$. Then $M$ noetherian implies that ...
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228 views

Is any Finitely generated module over an Artinian ring , Noetherian?

In order to prove a statement, I need to prove the following claim : If $R $ is an (not necessarily commutative) Artinian ring and $M $ is a finitely generated $R $-module, then $M$ is Noetherian. ...
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Having a hard time wrapping my head around “submodules of quotient modules”

I'm working with an $R$-module $M$ which is both noetherian and artinian. The problem at hand is showing that there is a finite chain of submodules $$M = M_0 \supseteq M_1 \supseteq M_2 \supseteq ... \...
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302 views

A graded ring is Noetherian equivalence

If $R$ is a graded ring then $R$ is Noetherian if and only if $R_0$ is Noetherian and $R$ is finitely generated as an $R_0$-algebra. If $R$ is Noetherian then $R_0 = R/R_+$ is Noetherian and $R_+ $ ...
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Zero divisors of a reduced Noetherian ring

My goal is to show that a reduced Noetherian ring of dimension one is Cohen-Macaulay. My question is on why the reduced condition is imposed. Let $ \mathfrak{p} \in \operatorname{Spec}(R) $. If $ \...
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Check whether $\mathbb{Q}[X_{1},X_{2},…]/(f_{1},f_{2},…)$ is a Bézout domain, Noetherian, Factorial, or PID [closed]

Check whether $\mathbb{Q}[X_{1},X_{2},...]/(f_{1},f_{2},...)$ is a Bézout domain, Noetherian, Factorial, or PID with $f_{i} =X_{i}-X_{i+1}^2$. I don't really have a clue how to start. It would be ...
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When is $RS^{-1}$ a local ring?

Suppose we have a noncommutative ring $R$ and multiplicatively closed set that is both right Ore, and right reversible, i.e. it is a right denominator set. Now, we can localize $R$ at $S$ to form $RS^{...
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Over an artinian ring every nonzero module has a simple submodule?

I want to prove this : Over an artinian ring every nonzero module has a simple submodule. But the same statement for Noetherian rings is not true. Is there any hint how to show that? Thank you ...
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Does Cantor–Schröder–Bernstein theorem hold in the category of Noetherian modules?

If $M_1$ , $M_2$ are Noetherian modules such that $M_1$ is isomorphic to a submodule of $M_2$ and $M_2$ is isomorphic to a submodule of $M_1$ , then is $M_1 \cong M_2$ ?
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a Noetherian , semi-Artinian module is Artinian. [closed]

A semi-Artinian module is a module which every nonzero quotient of it contains a simple submodule. How can I prove that if a module is both semi-Artinian and Noetherian, it is Artinian? Thanks for ...
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Different version of the Hilbert basis theorem [duplicate]

How to prove the following version of the Hilbert basis theorem: $R$ is Noetherian if and only if $R[|x|]$ is Noetherian. Of course, in view of the isomorphism: $$\frac{R[|x|]}{(x)~R[|x|]} \simeq ...
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A noetherian $R$-module $M$ with submodules $L,N$ s.t $L\cap N={0}$ and $M/N\cong L$

With these circumstances, can we conclude that $M=L+N$ ? All the counter-examples that I know are not noetherian(e.g infinite vector spaces). And I know that if $M$ has finite length the answer is ...
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$M$ Noetherian $\implies $ $M'$ and $M''$ are noetherian.

Let $0\to M'\overset{\alpha }{\to} M\overset{\beta}{\to} M''\to 0$ an exact sequence. I want to show that if $M$ Noetherian, then $M'$ and $M''$ are also Noetherian. Let $\{M_i'\}$ a chain of $M'$. ...
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Is the direct product of quotient rings left Noetherian?

I have to figure out how to show the following ring is not left Noetherian: $\prod _{1<n\in \mathbb{N}}\mathbb{Z}_{n}$ It is the direct product of all quotient rings. I have been given the hint ...
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1answer
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Prove R/J is noetherian

I have a noetherian ring $R$ and $J\triangleleft R$. Then $R/J$ is noetherian? I suppose $I\triangleleft R/J$ and $\tilde{I}=\{r\in R:r+J \in I\}$. $\tilde{I}\triangleleft R$, $R$ is noetherian so ...
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In noetherian domains, do we have $x \in P^{n-1} \setminus P^n,xy \in P^n \implies y \in P$?

Let $R$ be a noetherian integral domain, let $P$ be a non zero prime ideal of $R$ and let $x,y \in R$. Assume that for some $n \geq 1$, we have $x \in P^{n-1} \setminus P^n$ and $xy \in P^n$. Does it ...
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1answer
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Equivalence to noetherian space uses Axiom of Choice?

One calls a topological space noetherian if every decreasing chain of closed subsets becomes stationary. In a lecture it was remarked that this is equivalent to every family $M$ of closed sets having ...
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A unique decomposition of an element to irreducible elements, not all of them primes

Let $D$ be a Noetherian integral domain which is not a UFD. Being Noetherian guarantees that every non-zero non-invertible element $d \in D$ can be written as a product of irreducible elements of $D$, ...
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$M$ be a finitely generated module over commutative unital ring $R$ , $N,P$ submodules , $P\subseteq N \subseteq M$ and $M\cong P$ , is $M\cong N$?

Let $R$ be a commutative ring with unity , $M$ be a finitely generated module over $R$ , let $N,P$ be submodules of $M$ such that $P\subseteq N \subseteq M$ and $M\cong P$ , then is it true that $M\...
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finitely many minimal prime ideals in $\mathbb{Z}[x]$?

In my commutative algebra course we are asked to (in the end) construct all prime ideals in $S=\mathbb{Z}[X]$. We are already given the minimal non-zero prime ideals of the form $(f)$ with $f$ ...
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Why $P_i+N$ is stationnary when $P_i$ are submodules of $M$?

Let $N\subset M$ be $R$-submodule. We suppose that $N$ and $M/N$ are Noetherian and we want to show that $M$ is also Noetherian. The proof start by: Let $P_i$ be a chain of submodule of $M$. Then $...
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If $R_3$ is a finitely generated $R_2$-module and $R_2$ is a finitely generated free $R_1$-module, can we say that $R_3$ is a free $R_1$-module?

Let $R_1 \subseteq R_2 \subseteq R_3$ be Noetherian rings. If $R_3$ is a finitely generated $R_2$-module and $R_2$ is a finitely generated free $R_1$-module, then $R_3$ is a finitely generated $R_1$-...
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Structure theorem for commutative Noetherian rings

We have Structure theorem for commutative Artin rings which is as follows : An Artinian ring $A$ is uniquely up to isomorphism a finite direct product of Artin local rings. I would like to know if ...
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Height of Product of Ideals [closed]

Let $R$ be any Noetherian ring and $I$ and $J$ be ideals in $R$. Is it possible to find an equation with $\mathrm{ht}(I)$ and $\mathrm{ht}(J)$ which give us $\mathrm{ht}(IJ)$?
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Is every finitely-generated $\mathbb{Z}$-module Noetherian?

I was working on proving that $\mathbb{Z}[\sqrt{-5}]$ was a noetherian ring without resorting to Hilbert Basis Theorem, and I was wondering if this generalized easily to when the ring $R$ is a $\...
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If the source of a morphism is a noetherian scheme, the morphism is quasi-compact.

I was wondering it while studying Hartshorne. In II.4.8 proof of the corollary (This part is about valuative criterion of properness.), it is written that a morphism from noetherian scheme is ...
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Is $\oplus_{\mathbb{N}}\mathbb{Z}_2$ Artinian but not Noetherian?

Just as in the title: I've seen the statement that Artinian rings are Noetherian several times (eg Commutative artinian ring is noetherian) but if we take $R=\oplus_\mathbb{N}\mathbb{Z}_2$, it seems ...
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For $k$ a field, $k[x]$ is Noetherian but not Artinian?

I have a question regarding some Commutative Algebra facts. I am using the book A Course in Commutative Algebra by Gregor Kemper. On page 24 he states that if $k$ is a field, the polynomial ring $k[x]...
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Local ring with finitely generated maximal ideal and $\cap_{n>0}\mathfrak{m}^n=\{0\}$ is Noetherian?

I have proven in the past that if $A$ is a local ring with maximal ideal $\mathfrak{m}$ so that $\mathfrak{m}$ is principal, and $\bigcap_{n>0}\mathfrak{m}^n = \{0\}$, then $A$ is necessarily ...
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810 views

a submodule of a noetherian module is also noetherian? [duplicate]

I was studing some algebra by my own and I was wondering if this propiety is true: Let $M$ be a module over a ring $A$ and $S\subseteq M$ a submodule, if $M$ is Noetherian, then $S$ is Noetherian. I ...
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62 views

Same as some localization at $\mathbb{Q}[X]$?

Consider $R=\mathbb{Z}[\{\frac{X}{p_n}\}_{n\ge1}]$ where $p_n$ is $n$'th prime. Suppose $f$ is an irreducible polynomial in $\mathbb{Q}[X]$. Then I need to show $R_{R\cap f\mathbb{Q}[X]}$ is ...
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1answer
207 views

About Fitting decomposition theorem of module

The theorem is that if we have a finite length module M (Noetherian and Artinian), and a map f is endomorphism. Then, we can decompose M = Ker($f^n$) $\oplus$ Im($f^n$). I understand the proof is ...
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1answer
156 views

An ascending chain of subgroups

Let $G$ be the restricted wreath product of two infinite cyclic groups, say $\langle g\rangle$ and $\langle x\rangle$. Say $B$ the base group of $G$ (suppose $x\in B$). Is it possible to find a ...
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1answer
355 views

A reduced Noetherian ring injects into a product of fields

I am trying to prove the following: If a Noetherian ring $R$ is reduced, then there is an injection from $R$ into a product of fields. What I know is that every associated prime of $R$ is minimal ...
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1answer
126 views

Power of an ideal in a Noetherian ring

Question is : Show that if $R$ is a Noetherian ring, and $I$ is an ideal such that there exists a unique prime ideal $\mathfrak{p}$ containing $I$, then $I$ contains some power of $\mathfrak{p}$. ...
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301 views

Unique minimal prime over an ideal

Krull's principal ideal theorem: Let $A$ be a Noetherian ring. Let $\mathfrak p$ be a prime ideal in $A$ of height $r$. Then there exist elements $a_1,...,a_r \in \mathfrak p$ such that $\mathfrak p$ ...