Questions tagged [noetherian]

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

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125 views

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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1answer
83 views

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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2answers
360 views

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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1answer
439 views

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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1answer
824 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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1answer
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
214 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
165 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
369 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
128 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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1answer
306 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 $\mathfrak p$ ...
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0answers
179 views

Proving that $A$ is a Noetherian ring when $M$ is a faithful Noetherian $A$-module

Let $M$ be a Noetherian $A$-module (where $A$ is a commutative ring). I want to show that $M$ is faithful implies $A$ is Noetherian. I am very aware that this question is fully answered in both of ...
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3answers
442 views

Is the function ring $C^{\infty}(M)$ noetherian? [closed]

Let $M$ be a smooth manifold and $C^{\infty}(M)$ be its function ring. Is this a noetherian ring?
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1answer
45 views

$I$ be an ideal of a Noetherian ring $R$ , then does there exists $ n \in \mathbb N$ such that $(\sqrt I)^n \subseteq I$ ?

Let $I$ be an ideal of a Noetherian ring $R$ , then is it true that $\exists n \in \mathbb N$ such that $(\sqrt I)^n \subseteq I$ ? I can clearly see this to hold if $\sqrt I$ is a principal ideal , ...
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1answer
139 views

If $A/I \cong A/J$ as rings and $I\subseteq J,$ then $I=J.$ [duplicate]

Let $A$ be a commutative Noetherian ring and let $I$ and $J$ be ideals of $A.$ Suppose that $I\subseteq J$ and that $A/I \cong A/J$ as rings. I want to prove that $I=J.$ Observations so far: 1) If ...
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37 views

Is the module $\left<1,x,\dots, x^{s-1}\right>$ noetherian?

In order to prove the Hilbert base theorem ($A$ noetherian $\Longrightarrow$ $A[x]$ noetherian) consider an ideal $I\subset A[x]$, and $(f_1,\dots,f_k)$ such that the leading coefficients of the $f_i$ ...
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1answer
90 views

Does there exist a Noetherian domain (which is not a field ) whose field of fractions is ( isomorphic with ) $\mathbb C$ ?

Does there exist a Noetherian domain which is not a field , whose field of fractions is ( isomorphic with ) $\mathbb C$ ?
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1answer
74 views

$C$ be a subring of $B$ which is again a subring of $A$ , let $A,B,C$ be Noetherian and $A \cong C$ , then is $A \cong B$?

Let $C$ be a subring of $B$ which is again a subring of a commutative ring $A$ , also suppose all of $A,B,C$ are Noetherian and $A \cong C$ , then is it true that $A \cong B$ ? If the claim is not ...
2
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1answer
40 views

Does an ideal in $k[x_1, \dots, x_n]$, $k$ a field, contain only finitely many irreducible elements?

Let $k$ be a field and consider a nonunital ideal $I$ in $k[x_1, \dots, x_n]$. Since $k$ is Noetherian, we may write $I = (f_1, \dots, f_k)$ where each $f_i$ is nonconstant irreducible. Is it true ...
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2answers
397 views

Show $\mathbb{Z}G$ is a noetherian ring for finite group $G$

Let $G$ be a finite group. Show that $\mathbb{Z}G$ is noetherian as a ring. It's noetherian as a $\mathbb{Z}$-module (this follows because $\mathbb{Z}$ is noetherian and $\mathbb{Z}G$ isomorphic to ...
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2answers
175 views

Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is $I=(0)$?

Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is it true that $I=(0)$ ? I know that surjective ring endomorphism on Noetherian ...
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2answers
145 views

Rings with given “dimension” (in a specific sense, related to generators of ideals)

Let $R$ be a noetherian ring. If $I \trianglelefteq R$ is an ideal of $R$, we define $\mu(I)$ to be the minimal$^{(1)}$ number of generators of $I$ (this is well-defined since $R$ is noetherian). Then ...
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0answers
211 views

Let $R$ be a commutative Noetherian ring with unity such that every maximal ideal is principal. Then is $R$ a PIR?

Let $R$ be a commutative Noetherian ring with unity such that every maximal ideal is principal. Then is it true that every ideal of $R$ is principal ? So I think I should go like this: Suppose not. ...
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1answer
602 views

If $R$ is a commutative ring satisfying ascending chain condition on finitely generated ideals, then is $R$ Noetherian?

If $R$ is a commutative ring unital ring satisfying ascending chain condition on finitely generated ideals then is $R$ Noetherian ? So we would have to prove a.c.c. for any ideals or by Cohen's ...
3
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1answer
149 views

$R$ be an infinite commutative ring such that $R/I$ has only finitely many ideals for every non-zero ideal $I$ , what can we say about $R$?

It is known that if $R$ is an infinite commutative ring such that for every non-zero ideal $I$ , $R/I$ is finite then $R$ is a Noetheian domain . It is also known that if $R$ is a PID then for every ...
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2answers
68 views

$(R,\mathcal m)$ be a Noetherian local ring and let $P$ be a prime ideal of $R$. If $P^2$ is a prime ideal of $R$, then $P=0$

Let $(R,\mathcal m)$ be a Noetherian local ring and let $P$ be a prime ideal of $R$. If $P^2$ is a prime ideal of $R$, then $P=0$. I was thinking to use Nakayama lemma as: $R_P$ is local with $PR_P$ ...
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1answer
64 views

$R$ be a Noetherian domain , $t\in R$ be a non-zero , non-unit element , then is it true that $\cap_{n \ge 1} t^nR=\{0\}$?

Let $R$ be a Noetherian domain, $t\in R$ be a non-zero, non-unit element, then is it true that $$\bigcap_{n \ge 1} t^nR=\{0\} \text{?} $$ It almost feels like the nilradical (which is zero for any ...
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1answer
342 views

Annihilator of maximal ideals

Let $R$ be a Noetherian ring. Suppose all the nonzero proper ideals of $R$ have nonzero annihilators. Show that if $M$ is a maximal ideal of $R$, then $\exists$ $x \in R $ such that $M$ = $\mathrm{ann}...
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1answer
91 views

Noetherian Rings Definition, countability?

In my book (Jantzen, Algebra, 2014), Noetherian rings are defined by three equivalent conditions. I wonder how the first two can be equivalent: Every ascending chain of ideals $(a_1) \subset (a_2) \...
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56 views

Is it Noetherian and can you find an ideal which is non-finitely generated in it?

I have a lot of confusion in polynomial ring. Please help me explain it. Firstly, as we know, if $R$ is a ring then $R$ is sub-ring of $R[X]$ and $R[X]$ is a sub-ring of $R[[X]]$. I just wanna ask ...
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1answer
187 views

$R$-module $M$ which is artinian, but not noetherian [duplicate]

As I' m currently dealing with artinian and noetherian modules, I' m asking myself whether there is an artinian module which is not noetherian. I think so, but even after I thought a lot about this, ...
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1answer
112 views

Field is an Artinian module

I am going through theorem 2.14 in Eisenbud's Commutative Algebra. Given a ring $R$ that is Noetherian, all of whose prime ideals are maximal, we want to prove that $R$ is Artinian. Assume that $R$ ...
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1answer
417 views

Every ideal is the sum of principal ideals implies ascending chain of ideals is finite?

I am looking at this problem at the moment: If R is a commutative ring with 1, in which every ideal is the sum of finitely many principal ideals $(I=r_1R+r_2R+...+r_nR)$ , show that this implies, ...
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0answers
342 views

Proof of Krull's intersection theorem with Taylor expansion

I took a commutative algebra course last semester (with Kaplansky's book), and I've learned about Krull's intersection theorem. In the course, we proved it without using Artin-Rees Lemma. I heard that ...
2
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1answer
386 views

Noetherian vector space is finite-dimensional

Given a field $k$, and a $k$-vector space $V$ which is noetherian as $k$-module, I want to show that $V$ is finite-dimensional. Is it correct that this follows because since $V$ is noetherian, every ...
2
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2answers
353 views

Example of non-noetherian ring whose spectrum is noetherian and infinite

A topological space is noetherian if it satisfies the descending chain condition for its closed subsets. Let be $R$ a commutative ring and let $\mathrm{Spec}(R)$ its spectrum with Zariski topology. I ...
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64 views

Integral closure and $\bigcap \mathfrak{a}^n$

Let $R$ be a domain such that $\bigcap_{n=1}^\infty \mathfrak{a}^n=0$ holds for all proper ideals $\mathfrak{a}$ of $R$ (this holds, for example, if $R$ is Noetherian). Let $K$ be the quotient field ...
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0answers
70 views

Quotient of free algebra - Noetherian/Artinian?

Let $k$ be a field. Is the following ring noetherian/artinian? $$k\left<x,y,z\right>/\left<y^2 - xyz^5\right>$$ I ran into this problem and have no idea how to deal with it. Please help!
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0answers
160 views

$\mathbb{Z}[\sqrt{10}]$ is noetherian

How can we prove that $\mathbb{Z}[\sqrt{10}]$ is noetherian except by using Hilbert basis theorem? How can we find a sequence of ideals that satisfy the ACC?
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3answers
5k views

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 ...
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2answers
316 views

Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian, are then $S$ and $R/I$ also Noetherian?

Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian are then $S$ and $R/I$ also Noetherian? I have done the following: $R$ is Noetherian iff each increasing sequence of ideal $...
2
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1answer
298 views

direct sum of noetherian modules is noetherian

We say that an $A$-module $M$ is Noetherian if all of its submodules are finitely generated. Having that definition in mind can anyone give me some hints to prove that if $M$ and $N$ are Noetherian ...
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1answer
92 views

$M$ is a $\Bbb Z/36 \Bbb Z$-module and $\bar 6 \cdot M = \{0\}$. Prove: $M$ is Noetherian $\iff$ it's Artinian

Suppose that $M$ is a $\Bbb Z/36 \Bbb Z$-module such that $\bar 6 \cdot M = \{0\}$. Prove that $M$ is Noetherian $\iff$ $M$ is Artinian. I managed to prove the forward direction: if $M$ is Noetherian,...
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1answer
191 views

Are Noetherian Rings Goldie Rings?

We know that a ring $R$ is said to be Right-Goldie if $R$, as a right $R$ module, satisfies: (i) $R$ has finite uniform dimension; (ii) every ascending chain of right annihilators of $R$ terminate. ...
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1answer
47 views

If every ideal I in R is contained in a finite series of ascending ideals, prove that only finitely many ideals of any kind contain I.

If a Noetherian ring is defined by the fact that all ideals are contained within a finite series of ascending ideals, how does this prove that the initial ideal is contained within finitely many ...
2
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1answer
190 views

If $I$ is finitely generated nilpotent and $R/I^{n-1}$ is noetherian then $R$ is noetherian

If $I$ is a finitely generated ideal of a commutative ring $R$ with $1$ such that $I^n = \{0\}$ and $R/I^{n-1}$ is noetherian, then $R$ is also noetherian. I don't know what I should do. If I can ...
0
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1answer
349 views

If $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$

Let $R$ be a commutative unitary ring and $I,J$ be two ideals of $R$. I need to prove that if $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$ It seems that a direct ...
2
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1answer
236 views

System of parameters in Noetherian local rings

I'm trying to understand the theorem for systems of parameters in Noetherian local rings, which says: Let $R$ be a Noetherian local ring with maximal ideal $m$. Then there exists an $m$-primary ideal ...
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1answer
15 views

The product $JR$ of a proper ideal $J$ and the nontrivial ideal $R$

There is a proposition on page 390 of Hungerford's ``algebra'' (1974) as follows: Proposition 4.6. Let $J$ be an ideal in a commutative ring $R$ with identity. Then $J$ is contained in every maximal ...
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1answer
257 views

$R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ Noetherian?

Let $R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ a Noetherian ring ?