For question on noetherian rings and modules and related notions.

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$R$ 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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35 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
39 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
23 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 $M$ = $ann(x)$ (...
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14 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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48 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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$R$-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
21 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
32 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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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 ...
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1answer
41 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 ...
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2answers
104 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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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
37 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
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$\mathbb{Z}[\sqrt{10}]$ 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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289 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
62 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 $...
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1answer
32 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
36 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
34 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
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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 ...
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1answer
28 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 ...
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1answer
28 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 ...
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1answer
32 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
10 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
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On those integers $n>1$ such that any commutative ring with identity having exactly $n$ ideals is a PIR

Convention : All rings are commutative with unity unless stated otherwise. By ideals we will mean to include $\{0\}$ and $R$ also. Let us call an integer $n>1$ a "principal number" if any ring $R$...
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1answer
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$R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ Noetherian? [closed]

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 ?
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2answers
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Artinian - Noetherian rings and modules suggest study guide

What text or any document that has gathered this part of Algebra theory. Thanks. Pd: I seek on variety's book of commutative algebra but the subject is partially dealt
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1answer
22 views

Question about proof of Krull principal ideal theorem

How can we explain the following step in the proof of Krull principal ideal theorem: $l\{ ((z):x^n)/(z) \}$ or $l\{ ((x^n):z)/(x^n) \}$ is finite? $l(M)$ - length of module.
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1answer
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About a short proof of Krull principal ideal theorem

How from this theorem I can get a proof of Krull principal ideal theorem? I understand that w.l.g. we can prove it for a Noetherian local ring. But why we can consider that $(x)$ is $M$-primary?
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1answer
45 views

If $A$ is a commutative, unitary ring and $I$ an ideal of $A$ such that $I$ and $A/I$ are Noetherian rings, then $A$ is Noetherian?

If $A$ is a commutative, unitary ring and $I$ an ideal of $A$ such that $I$ and $A/I$ are Noetherian rings, then $A$ is Noetherian ? I know just that if $A$ is Noetherian then $A/I$ is ...
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1answer
39 views

Noether's normalization lemma in practice (example)

I would like to know how to use the Noether's normalization lemma in practice. Noether's normalization lemma Let $k$ an infinite field, and $k[a_1,\dots ,a_n]$ be a finite $k$-algebra. There ...
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2answers
204 views

A Bézout UFD is a PID. [duplicate]

Let $R$ be an integral domain and a Noetherian U.F.D. with the following property: for each couple $a,b\in R$ that are not both $0$, and that have no common prime divisor, there are elements $u,v\in ...
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0answers
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Proof of commutative Artinian ring is Noetherian

I think that I have a proof, but it seems much simpler than all proofs that I can find on the internet. Hence I suppose that there must be a mistake in my proof. The commutative ring $R$ is ...
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1answer
55 views

Let $R$ be a ring and $I$ be a finitely generated nilpotent ideal. If $R/I$ is noetherian (resp. Artinian) then $R$ is so.

Let $R$ be a ring and $I$ be a finitely generated nilpotent ideal. If $R/I$ is noetherian (resp. Artinian) then $R$ is so. In between step is $I^j/I^{j+1}$ is noetherian (artinian) $\forall j$. I am ...
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1answer
37 views

Example of a non noetherian module $M$ s.t. $M/IM$ is noetherian

What is an example of a non noetherian module $M$ s.t. $M/IM$ is noetherian? What is an example of a non artinian module $M$ s.t. $M/IM$ is artinian? What I was thinking that $k[x_1,...,x_n,...]$ is ...
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1answer
28 views

Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring. [duplicate]

Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring. I was trying to show that $R/I$ is a submodule of $M^n$ if $M=\langle f_1,...,f_n\rangle$. Now $M$ is ...
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1answer
75 views

Noetherian Lemma Contradiction

There is a noetherian version of Higman's Lemma which says If $X$ is a noetherian poset, then $X^*$ is noetherian. Now I was thinking, given $X=\{1\}$ we get $X^*=\{1,11,111,\dots\}$ by the ...
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1answer
54 views

Let $A$ a ring such that $0=m_1m_2…m_k$ where $m_i$ are maximal ideals of $A$. [closed]

Let $A$ a ring such that $m_1m_2...m_k=0,$ where $m_i$ are maximal ideals of $A$. Then $A$ is Noetherian if and only if $A$ is Artinian.
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1answer
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Can every element of a Noetherian ring be expressed as a product of irreducible elements?

Let $X$ be a Noetherian ring. Question: Can every element of $X$ be expressed as product of irreducible elements? I'm not assuming that $X$ is a Noetherian integral domain, only $X$ is Notherian....
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1answer
71 views

Noetherian rings have only finitely many minimal prime ideals. [duplicate]

We say that $p$ is minimal prime if It does not contain any other prime. Assume that $A$ is Noetherian ring Question: $A$ has only finitely many minimal primes. any suggestions please
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43 views

A module over a noetherian ring and its dual

Let $R$ be a noetherian ring, $M$ an $R$-module and $M^*$ be its dual. Is it true that if $M$ is noetherian then $f(x) = 0$ for all $f \in M^*$ implies $x =0$? How about the other way round? Ok, so ...
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why is this subring noetherian?

Let $R_1$ be a noetherian commutative ring and $R_2 \subset R_1$ be a commutative subring. We know that in general $R_2$ need not be noetherian. But I'm asked to prove that if there's a $R_2$-module ...
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2answers
71 views

Noetherian submodules and isomorphisms

Suppose $M$ is an $A$-Module, and $N$ is a submodule of $M$. Let $f:N\to M$ be an $A$-module epimorphism. How could I show that if $N$ is noetherian, then $f$ is an isomorphism? Thanks
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102 views

the ideal contains some power of the unique prime ideal containing it

Let $R$ be a Noetherian ring, and $I$ an ideal such that there exists a unique prime ideal $\mathfrak{p}$ containing $I$. Show that $I$ contains some power of $\mathfrak{p}.$ In this question, I ...
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2answers
58 views

Subgroups of $\mathbb Z^5$

Is there a nice way to see that any subgroup of $\mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z$ can have at most 5 generators? I know that $\mathbb Z$ is Noetherian, so ...
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1answer
27 views

Chains of ideals contained in maximal ideal in non-Noetherian commutative ring

Given a maximal ideal $M$ in a non-Noetherian commutative ring $R$, I'm trying to determine whether or not there can exist infinite strictly ascending chains of ideals of $R$ contained in $M$. I know ...
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1answer
60 views

Polynomial ring in infinitely many variables over a noetherian ring is coherent

If $R$ is noetherian, show that the polynomial ring of infinite variables $R[x_1,x_2,...]$ is coherent, i.e. every finitely generated ideal is finitely presented. I don't really know how to get ...
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53 views

System of parameters in a Noetherian local ring

I'am studying "system of parameters" on Commutative Ring Theory by H. Matsumura. There is a theorem about height of an ideal and the number of generators of the ideal. In the proof of the part (ii), ...
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1answer
73 views

$R\subset A\subset R[X]$, $A$ is Noetherian. Is $R$ Noetherian?

Let $R\subset A\subset R[X]$ be commutative rings and suppose $A$ is Noetherian. Is $R$ Noetherian? I guess the answer is yes. Can we say from this relation that $A[X]=R[X]$? If yes, then by Hilbert ...