# Questions tagged [noetherian]

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

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### $\mathbb{Z}+x \cdot \mathbb{Q}[x]$ 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 ...
1answer
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### 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 ...
1answer
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### 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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### 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 ...
0answers
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### 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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### 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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2answers
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### 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 $I$ ...
3answers
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### 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 ...
2answers
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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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### 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 ...
3answers
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### 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 ...
1answer
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### 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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### 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 ...
1answer
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### $S_0$ 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 ...
1answer
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### 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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### $\mathbb Q$ 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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### 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 ...
1answer
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### 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 ...
1answer
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### 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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### 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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### 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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### $A/\alpha$ 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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### 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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### $\mathbb{R}[x,y]/(y^2+x)$ 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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