Questions tagged [noetherian]

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

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23 views

$\mathbb{Z}+x \cdot \mathbb{Q}[x]$ is not noetherian

Consider the ring $R=\mathbb{Z}+x \cdot \mathbb{Q}[x]$. I need to check whether it is noetherian or not. A well known $\mathbb{Q}+x\cdot \mathbb{R}[x]$ is not noetherian, so I think my ring $R$ is ...
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19 views

Is QG a Galois extension of Q? [closed]

We know that if G be a finite group then ZG is a Noetherian domain (Z - integers). QG (Q - qoutient) be a field. Is QG a Galois extension of Q? Probably is a finite extension.
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Artinian, but not Noetherian module: a wonderful example.

I found this wonderful example here. I did not understand some details, could you help me understand? The questions will be asked within the example. Let $p\in\mathbb{N}$ be a prime number. Consider ...
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29 views

A question about a Noetherian ring starting from a subring

Proposition. Let $A$ be a subring of $B$, $A$ be a Noetherian, and $B$ finitely generated as an $A-$ module. Then $B$ is a Noetherian ring. Proof. Since $B$ is finitely generated as an $A-$ module, ...
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1answer
37 views

Proposition 9.2 (iv)=>(v) Atiyah: why principal?

So far I understood everything except for (iv)=>(v). I think I'm almost done with help of supplementary note. But I still cannot understand that the image of a in A/m^n implies a is principle. Can ...
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18 views

Exact functors in local cohomology

In local cohomology the ideal transform functors with respect to a pair of ideals are defined by $D_{I,J}(-)=\underset{\textbf{a} \in \tilde{w‎‎} ‎(I,J)‎} {\varinjlim}\,\,\text D_{\textbf{a}}(-)$. ...
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33 views

Prove that if $\mathrm{gr}(A)$ is Noetherian without zero-divisors, then so is $A$.

Let $A$ be a filtered commutative algebra and $\mathrm{gr}(A)$ the associated graded algebra. Prove that if $\mathrm{gr}(A)$ is Noetherian without zero-divisors, then so is $A$. Associated graded ...
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36 views

Direct sum of completions is faithfully flat

Let $R$ be a commutative Noetherian ring (with unity). For every maximal ideal $\mathfrak{p}$, denote by $R_{\mathfrak{p}}$ the completion of $R$ at $\mathfrak{p}$. I wish to show that $\bigoplus_{\...
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18 views

$S^{-1}M$ is a Noetherian $S^{-1}R$ module

Let $R$ be a ring, $S$ a multiplicative subset and $M$ a Noetherian module. Show that $S^{-1}M$ is a Noetherian $S^{-1}R$ module. My attempt: Let $K$ be a submodule of $S^{-1}M$. Then $\varphi^{-1}_S ...
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166 views

If $R$ is a Noetherian ring and $I$ a proper ideal, then the $R$-module $R/I$ has finite length iff the associated primes of $I$ are maximal [closed]

Let $R$ be a Noetherian ring and $I$ a proper ideal in $R$. We need to show that the $R$-module $R/I$ has finite length if and only if the set $Ass(I)$ of the associated primes of the ideal $I$ ...
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66 views

Is the ring $\mathbb{Z}_{(2)}$ Noetherian or Artinian? If it is Noetherian, what is its Krull dimension? [closed]

First of all, $\mathbb{Z}_{(2)} = \{ \frac{a}{b} \ : \ a,b \in \mathbb{Z}, 2 \nmid b\}$ is the local ring of $\mathbb{Z}$ at $(2)$. I was wondering, which are the prime ideals of that ring? If there ...
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Is the ring $\mathbb{C}[x,y]/(x^2,y^3)$ Artinian? If not, what is its Krull dimension?

So, I know that the ring $\mathbb{C}[x,y]/(x^2,y^3)$ is Noetherian, since the ring $\mathbb{C}[x,y]$ is Noetherian. In order to prove that the ring is not Artinian, I've tried finding a prime ideal ...
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22 views

Characterization of the Jacobson radical for non-Noetherian rings

For a unital ring $R$, I'd like to show the following equality: $$ \prod_{\mathfrak{m}\text{ maximal left ideal}}\mathfrak{m} = \{x\in R : \text{for all }y\in R, 1+yx\in R^\times\}$$ (i.e., prove ...
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124 views

Suppose that $R$ is not a noetherian ring. Then can we have an ideal in $R[x_1,…,x_n]$ which is not finitely generated?

I know that if $R$ is noetherian then the statement holds true by Hilbert basis theorem. However I am looking for a example where it doesn't hold true if $R$ is not noetherian. I was specifically ...
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129 views

Why do we need Noetherianness to conclude that a connected scheme is integral iff its local rings are?

I'm trying to proof/understand the following Statement: If $X$ is a Noetherian and connected scheme, then $X$ is integral if and only if $X$ has integral stalks. It can quickly by verified that if $X$ ...
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59 views

non noetherian rings

There are many questions about Examples of a non-Noetherian rings on this site, but this is a bit different. Let $f:R\to S$ be a homomorphism and R be Noetherian. If $f$ is surjective, then $S$ is ...
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1answer
37 views

$S_0$ Noetherian but not $S$

Question: I want to show that $S_0$ being Noetherian does not imply that $S$ is Noetherian. Thoughts: Here, a ring $S$ is graded if $S=S_0\oplus S_1\oplus\dots\oplus S_n\cdots$ of the additive abelian ...
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1answer
25 views

Zero dimensional quotient of a local Noetherian ring

Let $(R,m)$ be a local Noetherian integral domain. Let $I$ be an ideal in $R$, so $I \subseteq m$. Then $R/I$ is also a local Noetherian ring, see (and the image of any surjective ring homomorphism of ...
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70 views

$\mathbb Q$ is not a finitely generated $\mathbb Z$-module [closed]

Is it valid to say that $\mathbb Q$ is not a finitely generated $\mathbb Z$-module because $\mathbb Q$ is not finitely generated since $\mathbb Q$ is not Noetherian?
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1answer
34 views

On a Krull-intersection type problem for certain two generated ideals in local rings

Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $x,y\in \mathfrak m$ such that $y$ is not a zero-divisor on $R$. Then, is it true that $\cap_{n=1}^\infty (x,y^n) \subseteq (x)$ ? By Krull ...
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1answer
45 views

Powers of maximal ideal in a Noetherian local ring

$\newcommand{Spec}{\operatorname{Spec}}$ Let $R$ be a local Noetherian ring with maximal ideal $m$. Show that if $\Spec(R)\ne\{m\}$ then for every positive integer $n$, $m^n\ne m^{n+1}$. Also what ...
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Classification of commutative rings $R$ satisfying $\dim(R[T])=\dim(R)+1$

Let $R$ be a commutative ring. Then $\dim(R[T]) \geq \dim(R)+1$. Is there a classification of those commutative rings with the property $\dim(R[T])=\dim(R)+1$? Every Noetherian commutative ring has ...
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1answer
62 views

Intersection of ideals is zero and quotients are noetherian

Assume $R$ is a ring and $I_1,...,I_n\subseteq R$ ideals s.t $R/I_i$ is noetherian for every $i$, and $\bigcap_{i=1}^nI_i=\{0\}$, then $R$ is noetherian. My attempt: I defined the natural ...
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45 views

Powers of maximal ideal in local ring with a single prime ideal.

Let R be a non zero, local, Noetherian ring with $\mathfrak{m}$ the maximal ideal of R. If we assume that $\mathrm{Spec}R=\{\mathfrak{m}\}$, what can we say about powers of $\mathfrak{m}$ ? I have ...
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30 views

Equivalent condition for Noetherian ring

If a commutative ring $R$ is Noetherian, then every finitely generated $R$-module has a resolution by finitely generated free $R$-modules. It goes as follows: Start with a finitely generated $R$-...
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1answer
24 views

$A/\alpha$ is Noetherian as an $A$-module but how to get it is Noetherian as an $A/\alpha$-module?

I meet a problem in commutative algebra of Atiyah. Pprposition $6.6.$ Let $A$ be Noetherian (resp. Artinian), $\alpha$ an ideal of $A$. Then $A/\alpha$ is a Noetherian (resp. Artinian) ring. Proof. ...
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1answer
41 views

Proving: M finitely generated as an R-module and $R/\text{Ann} _R(M)$ noetherian implies M noetherian

Let $R$ be a commutative Ring, $M$ an $R$-module, we define $\text{Ann} _R(M) := \{ r \in R : rm = 0 \text{ }\forall m \in M\} $. As an exercise we have to show: "$M$ noetherian $\iff$ $M$ ...
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37 views

$\mathbb{R}[x,y]/(y^2+x)$ is not a noetherian $\mathbb{R}$-module

I've been given the task to find whether or not $R=\mathbb{R}[x,y]/(y^2+x)$ is noetherian as a $\mathbb{R}$-module. I've been thinking of using the usual submodule of powers of $x$ to prove that it ...
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1answer
73 views

Wrong use of Zorn's lemma in equivalence of noetherian properties.

I was wondering wether the proof of the follwing statement I found online was correct. Let $M$ be an $R$-Module ($R$ commutative and unital). (a) Every ascending sequence $N_0 \subset N_1 \subset \...
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2answers
43 views

How to show this ring is Noetherian?

I got the following exercise: Let $W$ be a finite-dimensional $\Bbb{R}$-vector space. Let $\Bbb{R}_W=\Bbb{R}\times W$. Define addition and multiplication by $(r,w)+(s,v)=(r+s,w+v)$, $(r,w)*(s,v)=(rs,...
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17 views

Is $ R =\{a_{2n} x^{2n} + … + a_2 x^2 + a_0\mid n \in \Bbb N\}$, subring of $\Bbb Z[x]$, noetherian?

Is $ R =\{a_{2n} x^{2n} + ... + a_2 x^2 + a_0\mid n \in \Bbb N\}$}, subring of $\Bbb Z[x]$, noetherian ?
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38 views

Pullback on $\mathbb{A}^{n-1}$ of $\pi$ : $Z(F) \rightarrow \mathbb{A}^{n-1}$

I am self studying some introductory algebraic geometry and the author of lecture notes make the following claim without explaining the reason. Let $F\subset k[x_1,...,x_n]$, $Z(F)\subset \mathbb{A}^n$...
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1answer
41 views

Example of Noetherian module over a nonnoetherian domain

I was looking for following two examples: A divisible module $M$ over a commutative domain $R$ such that $R$ is Noetherian but $M$ is not. $M$ is Noetherian but $R$ is not. I observed that if we ...
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1answer
67 views

If $R$ is a Noetherian ring, $I,J,Q$ are its ideals, $Q$ is primary and $IJ \subseteq Q$ , then $I \subseteq Q$ or $J^n \subseteq Q$ for some $n$

We consider a Noetherian commutative ring $R$ and $I,J,Q$ ideals of $R$. If $Q$ is a primary ideal and $IJ \subseteq Q$, then we need to show that either $I \subseteq Q$ or that there is a positive ...
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1answer
44 views

Proving that ${{X}_{m}}=\{P\in {{k}^{n}}\left| {{f}_{1}}(P)=…={{f}_{m-1}}(P)=0 \;\;\hbox{and} \;\; {{f}_{m}}(P)=1\} \right. $ is eventually empty.

We consider a field $k$, $n \in \mathbb{N}$ and ${{f}_{1}},{{f}_{2}},...\in k[{{x}_{1}},...,{{x}_{n}}]$. For every integer $m\ge 2$, we define ${{X}_{m}}=\{P\in {{k}^{n}}\left| {{f}_{1}}(P)=...={{f}_{...
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60 views

How to show that ${\rm Ass}(M_{\mathfrak p})=\{\mathfrak p\}$?

$R$ Noetherian, $M$ finitely generated module. If $P$ is minimal over $\operatorname{ann}M$, how is $M' = \ker (M \to M_P)$ a $P$-primary submodule? While using this question and answer to solve ...
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50 views

A result concerning local rings

The following result appears in the book 'Commutative Algebra with a View Toward Algebraic Geometry' by David Eisenbud: Theorem 18.16* Let $(R,P)$ be a regular local ring, and let $(A,Q)$ be a local ...
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97 views

Are the rings of the functions $\mathbb{R}\to\mathbb{R}$ and $\mathbb{Z}_n\to\mathbb{Z}_n$ Noetherian? [closed]

I want to check which of the following rings are noetherian: a) the ring of the functions $\mathbb{R}\to\mathbb{R}$. b) the ring of the functions $\mathbb{Z}_n\to\mathbb{Z}_n$, $n>1$. This is the ...
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2answers
69 views

If $R$ is a UFD, then $R[x]$ is Noetherian?

If $R$ is Noetherian, then $R[x]$ is Noetherian. However, if $R$ is a UFD, then $R$ might not be Noetherian. $\DeclareMathOperator{\Frac}{Frac}$ I want to show that if $R$ is a UFD, then $R[x]$ is a ...
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2answers
95 views

Semisimple Rings are Noetherian

I have seen the statement that if $R$ is a commutative ring and $R$ is a semisimple ring, then $R$ is Noetherian several times while reading through some algebra resources; however, I have never seen ...
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1answer
21 views

Sufficient condition for complete local ring to be the completion of a certain subring

The context is two statements from this paper: A quasi-local ring $(R, M \cap R)$ contained in a complete local ring $(T,M)$ is Noetherian and has completion $T$ provided the map $R \to T/M^2$ is ...
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82 views

Are finitely generated subrings of $\mathbb{H}$ Noetherian?

Let $R$ be a finitely generated subring of the ring of real quaternions $\mathbb{H}$, that is, $R$ is the subring generated by a finite subset of $\mathbb{H}$. I want to show (or find a counterexample)...
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0answers
13 views

Can a laminar family of sets be ordered by a Noetherian relation?

Every pair in a laminar family of sets is either disjoint or related by containment. Can the pairs which are related by containment be ordered by a Noetherian relation?
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1answer
38 views

References for trivial extension of a ring by a module

I'm reading a note having this definition: Let $R$ be a ring amd $M$ an $R$-module. The trivial extension of $R$ by $M$, written by $R\ltimes M$, is the $R$-algebra formed by endowing the direct sum $...
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2answers
90 views

Prove that an element in the quotient field is actually in the ring itself [duplicate]

Let $R$ be a noetherian integral domain and $K$ the quotient field of $R.$ Suppose $f \in K.$ Suppose for each maximal ideal $M$ of $R,$ we can find $h,k\in R$ so that $f = h/k$ and $k \not\in M$. ...
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2answers
98 views

Questions on ring fixed by automorphisms

Suppose $A$ is an integral domain and $F$ is its field of fractions. Let $G \leq Aut(F)$ be a group of automorphisms of $F$. Assume $g(a) \subseteq A$ for all $g \in G$ and let $A^G$ be the fixed ring ...
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1answer
61 views

Locally noetherian schemes

I'm reading Algebraic Geometry written by R. Hartshorne. There is a proposition in section 3, I have few problems with a part of its proof A scheme $X$ is locally noetherian iff for every open subset ...
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1answer
97 views

Projective finitely generated module over noetherian ring

Let $A$ be a noetherian ring and $M$ be a finitely generated $A$-module, I want to prove the following M is projective $\iff$ for all $P\subset A$ prime ideal the localisation $M_P$ is a free $A_P$-...
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1answer
57 views

Locally principal Ideal in a Noetherian semilocal domain [closed]

Let $R$ be a commutative Noetherian domain with only finitely many maximal ideals $\mathfrak m_1,...,\mathfrak m_n$. If $I$ is an ideal of $R$ such that $I_{\mathfrak m_i}$ is a principal ideal of $R_{...
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1answer
41 views

Reflexive modules and the canonical map $M\to \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$

Let $R$ be a commutative Noetherian ring. For any $R$-module $M,N$, there exists a canonical map $d_{M,N}: M \to \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$ sending an element $m \in M$ to the ...

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