Questions tagged [noetherian]

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

554 questions
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Example of a Noetherian domain which is not equidimensional

A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $\mathfrak p$ is defined to be Krull ...
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Proof Checking: Elements of Noetherian Domains may be factored into irreducibles

Edit: I have determined this initial proof to be incorrect. I have answered with what I believe to be a correct proof. Definition A Noetherian domain is a domain whose ideals are all finitely ...
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Artin-Rees property in commutative Noetherian rings with unit

I am trying to prove that if $R$ is a commutative Northerian ring with $1$, for all ideals $I$, $E$, then $E \cap I^n \subseteq EI$ for some $n\in \mathbb{N}$. I have tried to go in this direction, ...
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Proving Ascending Ideals Stabilize in PID

I am trying to understand the proof to the following theorem: Let $R$ be a principal ideal domain and let $I_{1}, I_{2}, ...$ be ideals in $R$ such that $I_{1} \subset I_{2} \subset \cdots$. Then ...
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An ideal of a Noetherian ring with one associated prime

Suppose that for a Noetherian ring $R$, we have an ideal $I$ such that $\operatorname{Ass}(R/I)$ has only one element, $P$. Then $\forall r \in R, p \in P, \; rp \in I \Rightarrow Rp = (p) \subset I$ ...
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Two definitions of an associated prime of an $R$-module

I have come across two definitions of an associated prime for an $R$-module $M$, one of which specifies that $R$ is Noetherian, however, I can't see the reason they would coincide. First one: A ...
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Every subset of ideals of $R$ has maximal element $\implies$ Every ideal of $R$ is finitely generated, if $R$ is Noetherian.

We know that a commutative ring $R$ is Noetherian, if for the increasing sequence of ideals $I_1\subseteq I_2 \subseteq...,\ \exists m\in \mathbb{Z}^+,\forall k\in \mathbb{Z}_{\geq m}:\ I_m=I_k$. ...
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Is this ring Noetherian? Artinian?

If R is a Noetherian ring, is $R[X]/((X-1)^2X)$ Noetherian? Artinian? So first I have to understand if it is Noetherian or Artinian, and then prove it or find chains of ideals that don't stabilize. So ...
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Confusion over finite generation and noetherian modules

There are a few relevant posts Error in proof that submodules of f.g. modules are f.g. Specific proof that any finitely generated $R$-module over a Noetherian ring is Noetherian. Finitely generated ...
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Is there a non-artinian noetherian ring whose non-units are zero-divisors?

Is there a non-artinian noetherian ring whose non-units are zero-divisors? Equivalent formulation: Is there a noetherian ring of positive dimension whose non-units are zero-divisors? [In this post, ...
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commutative Noetherian ring whose every maximal ideal is projective

Let $R$ be a commutative Noetherian ring. If every maximal ideal of $R$ is projective as an $R$-module, then is $R$ hereditary?
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$A$ Noetherian and $f:A \rightarrow A$ suryective then $f$ inyective [duplicate]

I think this must have been questioned before, but after searching, I couldn't find it. I thought of considering a set of $\{x_1, ..., x_n\}$ such that $\lt x_1, ... x_n\gt = A$. The hypotesis shows ...
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There are finitely many minimal prime ideals in every $B\subseteq \operatorname{Spec} A$ for $A$ noetherian?

It is a well known fact that for a noetherian ring $A$ there are finitely many minimal primes. Now I'm wondering if this is true for every subset in the primes of $A$. My question, specifically, would ...
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Why noetherian rings [duplicate]

While in undergraduate years, abstract stuctures are very confusing since they arise without many motivations most of the times, and find their living ground later on. I would like to understand why ...
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Hilbert's basis theorem original formulation.

Hilbert's basis theorem (1888) is usually stated as: "If R is a Noetherian ring, then R[X] is a Noetherian ring." This could not be the original formulation of the theorem since Noetherian rings ...
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Example of a non-zero module that has no associated primes

I'm currently reading up on Associated Primes and localization. I came across the following theorem. Let $M$ be an $R$ module. If $M = 0$ then $Ass(M)$ is empty. The converse is true if $R$ is a ...
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Is $X\times_ST$ a Noetherian scheme?

Let $X$ be a Noetherian scheme. Given two morphisms of schemes $\pi:T\to S$ and $f:X\to S$, if $\pi$ is flat, is $X\times_ST$ a Noetherian scheme?
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Application of forcing to first-order properties of rings

I'm not very well acquainted with forcing, just with the basic ideas; but I thought of the following proof, and since I'm definitely not comfortable with forcing I don't know if it's right, so my ...
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Is the monoid ring of a Noetherian monoid Noetherian?

Let $M$ be a monoid. Suppose that $M$ is left Noetherian, i.e. that every increasing chain of left ideals in $M$ stabilizes. Then is the monoid ring $\mathbb{C}[M]$ necessarily a left Noetherian ring? ...
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Show ring is Noetherian

Let $R$ be a commutative ring with identity. Suppose $\text{nilrad}(R)$ is a finitely generated ideal of $R$ and $R/\text{nilrad}(R)$ is isomorphic to finite product of fields. I want to show $R$ is ...
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Any submodule of any finitely generated $R$-module is finitely generated if $R$ is Noetherian

I am trying to prove that if $R$ is a Noetherian ring then any submodule of any finitely-generated $R$-module is also finitely-generated. What I have tried: I know that any finitely generated $R$-...
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On Noetherian ness as modules over base ring, of ideals, of an algebra

Let $R$ be a commutative ring with unity and $A$ be a commutative unital $R$-algebra. (i) Let $I$ be an ideal of $A$ and $a\in A$ be such that $I+Aa$ and $(I:a)=\{ x\in A : ax \in I\}$ are ...
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On the structure of $R_0$ in graded PID $R=\bigoplus_{n \ge 0} R_n$

Let $R=\bigoplus_{n \ge 0} R_n$ be a graded integral domain. If $R$ is a PID, then is $R_0$ a field ? Since $R$ is Noetherian, I know that $R_0$ is Noetherian and $R$ is a finitely generated $R_0$-...
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Commutative, infinite, Noetherian ring with no non-trivial nilpotent and no non-trivial idempotent

Let $R$ be a commutative, infinite reduced (no non-zero nilpotent) Noetherian ring with no non-trivial idempotent (so that the prime spectrum is connected under Zariski topology). Then is it true that ...
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Ring Homomorphisms and Noetherian…?

I am confused by the following corollary that appears in this book (just search "Corollary 7.8.11" for the appropriate page): Corollary 7.8.11 Let $f : A \to B$ be a ring homomorphism and suppose ...
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Jacobson local ring that is not Artinian

I am studying commutative algebra at the moment, so all rings are assumed commutative (and unital). Does there exist a Jacobson local ring $\newcommand{\mfm}{\mathfrak{m}}(A, \mfm)$ that is not an ...
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Proving a ring $A$, generated by Noetherian subring and nilpotent element, is Noetherian again.

I am studying some algebra during my spare time. In particular I am learning about Noetherian rings. A friend sent me the following excersise, and I am not able to solve it. Suppose that a ring $A$ ...
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Descending chain condition of maximal ideals for a domain

Let $R$ be an integral domain. $I\not= \{ 0 \}$ a maximal ideal. Is there an example where $I \supseteq I^2 \supseteq \ldots$ satisfies $I^n=I^{n+1} = \cdots$ for all $n \in \mathbb{N}$. I ...
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Example of non-Noetherian ring

Can you give an example of a non-noetherian ring $R[x_1,x_2]$ with an ideal which is not finitely generated?
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Tensor product of noetherian modules is also noetherian?

Here are my thoughts. I want to show that every submodule of this tensor product is finitely generated (I think that is easier than trying to show it satisfies the ascending chain condition) Also, ...
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I am studying the proof of this theorem (of T.Y.Lam). Namely: Let R be a ring for which rad R is nilpotent, and $R/radR$ is semisimple. Then for any R-module ${}_{R}M$, the following statements are ...
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Any module over a regular local ring has finite free resolution

In some notes I'm following, there is a statement (Clearly) over a regular local ring each finitely generated module has a finite free resolution Is the finitely generated assumption neccessary? ...
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$R$ is Noetherian semilocal. $J:=\operatorname{Jac}(R)$. If $R/\operatorname{nil}(R)$ is $J$-adically complete, then $R$ is $J$-adically complete

Let $R$ be a Noetherian ring with finitely many maximal ideals. Let $I$ be its nilradical, i.e. the intersection of all prime ideals, and $J$ be the Jacobson radical i.e. the intersection, hence ...
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$I$ be a finitely generated ideal of $R$ such that $R/I$ is a Noetherian and $R$ is $I$-adically complete. Then $R$ is Noetherian

Let $R$ be a commutative ring with unity and $I$ be a finitely generated ideal of $R$ such that $R/I$ is a Noetherian ring and $R$ is complete with the $I$-adic topology. Then how to show that $R$ is ...
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$R$ be Noetherian ring. Let $I \subseteq J$ be proper ideals of $R$. If $R$ is $J$-adically complete, then is $R$ complete $I$-adically?

Let $R$ be Noetherian ring. Let $I \subseteq J$ be proper ideals of $R$. If $R$ is $J$-adically complete, then is $R$ complete $I$-adically ? I was proceeding as follows: Let $\{x_n\}$ be an $I$-adic ...
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If $B$ is a Noetherian ring and $A$ is a subring of $B$, is $B$ Noetherian as an $A$ module?

Let $A\subset B$ and both $A$,$B$ are ring. Suppose given that $B$ is a noetherian ring then is it true that when we consider $B$ as $A$-module then it is again a noetherian $A$-module? My effort : I ...
Say you have a complete Noetherian local ring $(A,\mathfrak m_A,k_A)$ and a complete Noetherian local $A$-algebra $B$ with residue field $A$. Furthermore, suppose you can find a surjective local ...
Let $A$ be a Noetherian domain and $K$ its field of quotients. I would like to prove that, given $x \in K$, we have $x \in A$ iff $x\in A_p$ for every $p$ prime ideal associated to an $y\in A$, $y\ne0$...