Questions tagged [noetherian]

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

554 questions
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Converse of : If $M$ is Noetherian then $M_m$ is Noetherian for every $m\in\operatorname{Max}(R)$.

Let $R$ be a Noetherian ring and $M$ be an $R$-module. We know that if $M$ is Noetherian then $M_m$ is a Noetherian $R_m$-module for every $m\in\operatorname{Max}(R)$. Now, I want to know if the ...
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Intersection of ideals equal to their product on noetherian ring

Let $R$ be a noetherian ring, $I\le R$ and $x\in R$ such that $\forall\frak{p}\in$Ass$(R/I)$, $x\not\in\frak{p}$. Show that $(x)\cap I=(x)I$. Obviously, we have the inclusion $(x)I\subseteq (x)\cap I$...
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Depth of a local ring

Let $(R,\mathfrak m)$ be a Noetherian local ring and $\mathfrak m$ is an associated prime of some $(x)\subset \mathfrak m$. I need to show that $\operatorname{depth}(\mathfrak m,R)\leq 1$. I need to ...
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Flaw in the proof that $M$ is noetherian given exact sequence?

Let $$0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0$$ be an exact sequence of $R$-modules. Then $M$ is noetherian if and only if $M'$ and $M''$ are. In my attempt of proving this I didn'...
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Is a finitely generated subring of a Noetherian ring also Noetherian?

Is a finitely generated subring of a Noetherian ring $R$ also Noetherian? Remark: In fact I'm interested in the case $R=\mathbb C[x_1,...,x_n]$.
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Grillet's “Abstract Algebra”, p. 148, ex. 3: Noetherian subring of a ring

Let $R$ be a Noetherian subring of a commutative ring $S$. Suppose that $S = (R\cup\{b_1,...,b_m\})$ for some $b_1,...,b_m \in S_n$. Then $S$ is Noetherian. I'm not sure how to approach this exercise....
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A nice way to phrase this theorem about ideals in Noetherian rings?

I came across this theorem: Theorem. Let $I$ be an ideal of a Noetherian ring $R$. Then there is $r \in \mathbb{N}$ with $\operatorname{rad}(I)^r \subseteq I$. Is there a concise way to state ...
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Let A be a Noetherian Ring. Prove that Sum anx^n is nilpotent iff an is nilpotent.

Let $A$ be a Noetherian Ring. Prove that $\sum_{n=0}^\infty a_n x^n$ is nilpotent iff $a_n$ is nilpotent for all $n\in\mathbb Z^+$.
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If $\mathfrak{p}$ is a prime such that $M_\mathfrak{p} \neq 0$, then $\mathfrak{p}$ contains an associated prime of $M$

I am studying from Serge Lang's Algebra (3rd edition), and in Chapter X Noetherian Rings and Modules, $\S2$ Associated Primes, we have the following proposition: Proposition 2.10. Let $A$ be ...
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Let R be noetherian. If ideal $I$ is maximal with respect to the property that $R/I$ is not of finite length, prove $I$ is prime.

I'm quite lost as to how to go about proving this. My instinct is to try find some sort of contradiction, but I couldn't formulate any concrete arguments. Any ideas to prove this?
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A Noetherian domain $R$ is a UFD iff $I_{fg}=\{h\in R \mid hg\in(f)\}$ is principal for all $f,g\in R$. [duplicate]

Let $R$ be Noetherian domain. For $f,g \in R$, define the ideal $I_{fg}=\{h\in R \mid hg\in(f)\}$. Prove $R$ is UFD iff $I_{fg}$ is principal for all $f,g\in R$. I'm stuck at this problem. The ...
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Consequence of epimorphism from Noetherian $R$-module

Let $R,S$ be a commutative rings with $1_R,1_S$ respectively. In the most commutative algebra one can find the following proposition. Proposition. Let $φ:R\twoheadrightarrow S$ be a ring ...
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$R_{\mathfrak{p}}$ is integrally closed in its total quotient ring, then $R$ is integrally closed in its total quotient ring.

It is known that if $R$ is a Noetherian ring, then an element $x\in K(R)$ in its total quotient ring belongs to $R$ iff the image of $x$ in $K(R)_{\mathfrak{p}}$ belongs to $R_{\mathfrak{p}}$ for ...
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The finite generation of $M$, to conclude $M$ is Noetherian (when $R$ is Noetherian).

We know the following proposition. Proposition. Let $R$ be a Noetherian/Artinian ring and $M$ an $R$-module. If $M$ is finitely generated, then the $R$-module $M$ is Noetherian module. I was ...
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Are unique prime ideal factorization domains locally noetherian?

In this question I asked: "Are unique prime ideal factorization domains noetherian?". In this answer Badam Baplan pointed out that locally noetherian domains are unique prime ideal factorization ...
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Are unique prime ideal factorization domains noetherian?

Let $A$ be a domain satisfying the following condition: If $\mathfrak p_1,\dots,\mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $\mathbb N^k$, ...
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Commutative Noetherian ring with distinct ideals having distinct index

Let $R$ be a Noetherian commutative, infinite ring with unity such that distinct ideals have distinct index i.e. if $I,J$ are ideals of $R$ and $I \ne J$ , then $R/I$ and $R/J$ are not bijective as ...
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$R$ noetherian, then every finitely generated $R$-module A has a resolution

I have seen in a definition of the book An Introduction to Homological Algebra (Weibel) the following: A ring $R$ is noetherian if every ideal is finitely generated. That is, every $R/I$ is ...
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Unique prime ideal factorization in noetherian domains?

[I changed the title and the body of the question. Below I explain why I did so, and paste the previous version.] Let (UPIF) (for "Unique Prime Ideal Factorization") be the following condition on a ...
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$\bigcap_{n\in\mathbb{N}} I^n = (0)$ if and only if no zero divisor of $R$ is of the form $1-z$ with $z\in I$.

Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that $$\bigcap_{n=1}^{\infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the ...
I wanted to prove the following equivalence. Consider $R$ a graded commutative Noetherian ring such that $R^{<0}=0$ and $M$ a graded, finitely generated $R$-module. Then $M^i=0$ for $i \gg 0$ if ...