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Questions tagged [noetherian]

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

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Every ideal contains a power of it's radical in a Noetherian ring.

Let $I$ be an ideal in a Noetherian ring $R$. Prove that there exists a positive integer $N$ such that $\text{rad}(I)^N ⊂ I$. [Hint:Let $\text{rad}(I)=⟨g_1,\ldots,g_k⟩$,and suppose $g_i^{n_i} \in I$....
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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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Primary decomposition of modules - uniqueness proof

Let $M$ be $A$-module, $A$ commutative ring, and $N$ submodule and let $$N=Q_1\cap\dots\cap Q_r=Q'_1\cap \dots \cap Q'_s$$ be reduced primary decompositions of $N$. Then $r=s$. The set of primes ...
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Further explanation on proof that associated primes are precisely those belonging to primary modules in reduced decomposition of $0$.

Consider the following theorem in Chapter X Noetherian Rings and Modules from Lang's Algebra (page 423, third edition): Theorem 3.5. Let $A$ and $M \neq 0$ be Noetherian. The associated primes of $...
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1answer
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ACC on chains of finitely generated submodules

Noetherian rings are those having ascending chain condition on ideals. There is also literature concerning ACC on n-generated (i.e., generated by n elements) ideals; see e.g, Commutative Rings with ...
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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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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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1answer
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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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Locally Noetherian Domain With Finitely Many Prime Ideals

Let R be a domain with finitely many prime ideals such that the localization at each prime, $R_{\mathfrak p}$, is Noetherian. Then is $R$ necessarily Noetherian?
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A short proof for $\dim(R[T])=\dim(R)+1$?

If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial ...
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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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depth $R_{\mathfrak p}=0$ implies $\mathfrak pR_{\mathfrak p}$ is associated to zero

Let $R$ be a commutative Noetherian ring with unity. I want to prove the fact that depth $R_{\mathfrak p}=0$ implies $\mathfrak pR_{\mathfrak p}$ is associated to zero. Since depth $R_{\mathfrak p}=...
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Let $(A, +, \cdot )$ be a ring such that $A/<x>$ is finite, for every $0 \neq x \in A$. Show that $A$ is noetherian.

I tried to show that every ideal $I$ of $A$ is finite. Let $0 \neq x \in I$. By hypothesis there is $x_1, x_2, ..., x_k \in A$ such that $$A/<x> = <x_1+<x>,..., x_k+<x>>$$ I ...
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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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Noetherian local ring is Artinian iff maximal ideal is nilpotent

I am browsing through some old lecture notes, and I am trying to prove the following: Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Show that the following are equivalent: ...
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Noetherian module does not contain a submodule $N$ which is a direct sum of $n$ simple modules

The question: Let $R$ be a ring and $M$ be a Noetherian module. Prove there is $k \in \mathbb{N}$ such for all $n > k$, $M$ does not contain a submodule $N$ which is a direct sum of $n$ simple ...
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Noetherian ring with infinite Krull dimension (Nagata's example).

I just started to read about the Krull dimension (definition and basic theory), at first when I thought about the Krull dimension of a noetherian ring my idea was that it must be finite, however this ...
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Non-Noetherian Domain Which is Locally Noetherian

Let $R$ be an integral domain such that the localization, $R_{\mathfrak p}$, at each prime ideal, $\mathfrak p \le R$ is Noetherian. Then is $R$ necessarily Noetherian? In the case of $R$ not ...
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Ring of Laurent Polynomials is Noetherian Implies Polynomial Ring is Noetherian?

Suppose $R[t, t^{-1}]$ is Noetherian. Is $R[t]$ necessarily Noetherian? (Here, $R$ is a commutative ring with an identity.) I am able to show that the converse is true, but am a bit stuck with this ...
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If $R$ is a right-Noetherian ring and the Jacobson radical satisfies the right Artin-Rees property, then $\bigcap^{\infty}_{n=1}\text{Jac}(R)^n = 0$

A (two-sided) ideal $I$ of a ring with identity $R$ has the right Artin-Rees property if for any right-ideal $E$ of $R$, there exists an integer $n\geq1$ such that $E\cap I^n\subseteq EI$. If $R$ ...
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Hartshorne Exercise III 3.2: $X$ is affine iff every component is affine

I'm trying to solve the following exercise frome Hartshorne's Algebraic Geometry: Exercise III 3.2. Let $X$ be a reduced noetherian scheme. Show that $X$ is affine if and only if each irreducible ...
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In a commutative local Noetherian ring $R$ with maximal ideal $J$, if $J$ is not nilpotent then $R$ is an integral domain.

We've just proved this result: Let $R$ be a commutative, local, Noetherian ring. Suppose that $J$ (the maximal ideal) is principal. Then every nonzero ideal of $R$ is a power of $J$. And now we ...
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Why noetherian ring satisfies the maximal condition?

"maximal condition" means if any non-empty collection of ideals in R has a maximal element (under set inclusion). And we define noetherian ring to be a ring such that any ascending chain of ideals is ...
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Noetherian ring has an associated prime ideal

Question Let $R$ be a Noetherian ring. Show that there is a prime ideal $\mathfrak{p}$ and a monomorphism of $R$-modules $f:R/\mathfrak{p}\to R$. Attempt I considered the set of ideals $\{I< R: \...
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Is Axiom of Dependence Choice necessary for two definitions of Noetherian to be equivalent?

There are two definitions of Noetherian R-module: That all ascending chains of submodules stop, and that all submodules are finitely generated. Typically people prove them equivalent with Axiom of ...
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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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Intersection of all symbolic powers of a prime ideal

This is Exercise 8.37 in R.Y.Sharp's Steps in Commutative Algebra: Let $P$ be a prime ideal of the commutative Noetherian ring $R$. Prove that $$\bigcap_{n=1}^{\infty} P^{(n)}=\{ r\in R \mid \...
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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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$0 \to L \to M \to N \to 0$ is a short exact sequence. $M$ is Noetherian $\iff$ $L$ and $N$ are Noetherian

Let $$0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0$$ be a short exact sequence of $A$-modules ($A$ is a commutative ring). Prove $$M \text{ is Noetherian} \iff L \text{ ...
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Non-unital ring $(2\mathbb{Z})[X]$ is not Noetherian

Let $R = 2\mathbb{Z}$. Then $R[x]$ is not a noetherian ring. I do not understand why this is so, because Hilbert's basis theorem says: If R Noetherian ring, then R[X] a is Noetherian ring (from wiki)....
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How to show the following ring is not Noetherian

$R \subset \mathbb{Q}[x]$ be the subring consisting of polynomials $f = a_0 + a_1 x+\cdots+a_n x^n$ s.t. $a_0\in \mathbb{Z}$. Show that $R$ is not a Noetherian ring. Given a subgroup $A\subset \...
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On finiteness of $\cup_{n\ge 1 } \operatorname{Ass}_R (R/I^n)$

Let $I$ be an ideal of a commutative noetherian ring $R$. How to prove that $\bigcup_{n\ge 1 }\operatorname{Ass}_R (R/I^n)$ is finite ? I am aware of Brodmann's result about Asymptotic stability ...
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Show that any non-zero prime ideal of $R$ is invertible.

Theorem $:$ Let $R$ be an integral domain such that $R$ is Noetherian, integrally closed and every non-zero prime ideal of $R$ is maximal with quotient field $K.$ Then every non-zero prime ...
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In an integrally closed, Noetherian, local, integral domain of dimension $1$, the maximal ideal $P$ is eventually principal

Let $R$ be an integrally closed, Noetherian, local, integral domain of dimension 1 with unique maximal ideal $P$. Take an element $a \in P$ that is non zero. Show that for some $n$, $P^n$ is ...
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If finite product of maximal ideals of ring $R$ is zero, then $R$ is Noetherian$\iff R$ is Artinian

I'm studying commutative algebra and now I am struggling to understand the following proof: Proposition. Let $R$ be a commutative ring with $1_R$, $t\in \Bbb{Z}^+$ and $P_1,\dots,P_t \in \mathrm{...
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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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A Noetherian integral domain is a UFD iff $(f):(g)$ is principal

Let $R$ be a Noetherian integral domain. For $f, g \in R$, define $(f):(g)=\{h \in R \mid hg \in (f) \}$. Sow that $R$ is a UFD if and only if $(f):(g)$ is principal for all $f,g \in R$. It is easy ...
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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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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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$R_{\mathfrak{p}}$ 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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Is any Finitely generated module over an Artinian ring , Noetherian?

In order to prove a statement, I need to prove the following claim : If $R $ is an (not necessarily commutative) Artinian ring and $M $ is a finitely generated $R $-module, then $M$ is Noetherian. ...
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1answer
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Confusion about Exercise II .5.15 in Hartshorne

I'm a bit confused about Exercise II 5.15 in Hartshorne's Algebraic Geometry, especially part (b) and (c) which are (b) Let $X$ be an affine noetherian scheme, $U$ an open subset, and $\mathscr{F}...
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If $R$ is a Noetherian ring then $R[[x]]$ is also Noetherian

Let $R$ be a Noetherian ring. How can one prove that the ring of the formal power series $R[[x]]$ is again a Noetherian ring? It is well-known that the ring of polynomials $R[x]$ is Noetherian. I ...
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When is a tensor product of two commutative rings noetherian?

In particular, I'm told if $k$ is commutative (ring), $R$ and $S$ are commutative $k$-algebras such that $R$ is noetherian, and $S$ is a finitely generated $k$-algebra, then the tensor product $R\...
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Ideals with equal squares in a Noetherian UFD of dimension 2

Let $I$ and $J$ be two ideals in $\mathbb C[X,Y]$ such that $\mu(I)=\mu(J) \le 3$ and $I^2=J^2$ . Then is it necessarily true that $I=J$ ? If not, then is it at least true if we assume $I,J$ are ...
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Ideal in polynomial ring which contains no non-zero prime ideal

Let $J$ be a non-zero ideal in $\mathbb C[X,Y]$ such that $J$ contains no non-zero prime ideal. Then is it true that $J$ has height $1$ ? Possible approach: Since $\mathrm{ht}(J^n)=\mathrm{ht}(J)$ ...