Questions tagged [noetherian]
For questions on Noetherian rings, Noetherian modules and related notions.
771
questions
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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 ...
0
votes
1answer
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.
5
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1answer
106 views
+50
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 ...
0
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1answer
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 ...
1
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0answers
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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0answers
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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1answer
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_{\...
0
votes
1answer
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 ...
0
votes
2answers
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 $I$ ...
1
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3answers
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 ...
2
votes
2answers
90 views
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 ...
0
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0answers
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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3answers
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 ...
7
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1answer
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 $X$ ...
3
votes
0answers
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 ...
2
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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 ...
0
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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 ...
0
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3answers
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?
1
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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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0answers
124 views
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 ...
0
votes
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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0answers
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 ...
0
votes
1answer
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 $R$-...
0
votes
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. ...
1
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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 $\iff$ $M$ ...
0
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0answers
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 ...
1
vote
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 \...
0
votes
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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1answer
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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0answers
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 $F\subset k[x_1,...,x_n]$, $Z(F)\subset \mathbb{A}^n$...
1
vote
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 ...
0
votes
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 ...
1
vote
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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0answers
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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0answers
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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1answer
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 ...
1
vote
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 ...
3
votes
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 ...
2
votes
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 ...
4
votes
1answer
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)...
1
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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?
0
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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 $...
2
votes
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 $k \not\in M$. ...
1
vote
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
...
2
votes
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 ...
0
votes
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 $A_P$-...
-1
votes
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_{...
0
votes
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 ...
