Questions tagged [noetherian]
For questions on Noetherian rings, Noetherian modules and related notions.
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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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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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1answer
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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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1answer
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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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1answer
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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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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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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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1answer
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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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1answer
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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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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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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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1answer
50 views
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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1answer
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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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1answer
50 views
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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0answers
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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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1answer
32 views
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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1answer
71 views
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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1answer
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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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1answer
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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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1answer
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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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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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0answers
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Is every commutative ring a limit of noetherian rings?
Let $\mathsf{Noeth}$ be the category of noetherian rings, viewed as a full subcategory of the category $\mathsf{CRing}$ of commutative rings with one.
Let $A$ be in $\mathsf{CRing}$.
Question 1. ...
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1answer
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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 $\mathrm{ht}(J^n)=\mathrm{ht}(J)$ ...
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0answers
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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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Existence of ideal in Cohen-Macaulay ring, going modulo which still gives Cohen-Macaulay [closed]
Let $R$ be a local Cohen-Macaulay ring of dimension $\le 2$. Does there necessarily exist an ideal $J$ of $R$ such that $\sqrt J$ is a minimal prime ideal of $R$ and $R/J$ is Cohen-Macaulay ?
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$K[X^2,X^3]\subset K[X]$ is a Noetherian domain and all its prime ideals are maximal
Consider $K$ field and consider the ring $R=K[X^2,X^3]\subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But ...
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1answer
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Exercise VIII.3.9 in Hungerford's Algebra
I am working on the following exercise (Exercise VIII.3.9 in Hungerford's Algebra).
Let $R$ be Noetherian and let $B$ be an $R$-module.
If $P$ is a prime ideal such that $P=\text{ann }x$ for ...
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2answers
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If a ring is noetherian, then so is its subring
Suppose $A\subset R$ are rings so that $R$ is also an $A$-module. Assume further that there is an $A$-module homomorphism $R\to A$ that is identity on $A$. How to deduce that $A$ is noetherian if $R$ ...
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0answers
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Noether Equation on Artin and Milgram Galois Theory book
On Section K Chapter 2 of Galois Book by Artin and Milgram it discusses Noether Equation, but I'm not sure what Noether Equation this is, I know about Noether Equation for symmetry in physics but not ...
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1answer
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$\mathbb{R} ^ \mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.
let $R=\mathbb{R}^ \mathbb{R}$ (all the functions like $f:\mathbb{R} \rightarrow \mathbb{R}$). For each $f, g \in R$ and $a \in R$:
$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$
I want to show that $...
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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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1answer
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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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1answer
39 views
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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1answer
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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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1answer
156 views
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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2answers
56 views
$\bigcap_{n\in\mathbb{N}} I^n = (0)$ 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 ...
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0answers
12 views
Support of a graded module of a ring concentrated in non-negative dimensions
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 ...
2
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1answer
33 views
If the completion of a module is trivial, must the module be trivial?
I want to prove the following lemma as a step in solving an exercise from Atiyah-MacDonald's Commutative Algebra.
Claim: Let $A$ be a Noetherian ring, and $M$ a finitely generated $A$-module. Let $I \...
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1answer
56 views
Noetherian domain whose fraction field is such that some specific proper submodules are projective
Let $R$ be a Noetherian domain (which is not a field) with fraction field $K$. Suppose every proper $R$-submodule of $K$ of the form $R[1/a]$, where $a\in R$, is projective as an $R$-module. Then, I ...
3
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1answer
58 views
How could I construct a module $M$ that has exactly $n$ composition series?
How could I construct a module $M$ that has exactly $n$ composition series?
I can't seem to find a series of submodules where each have exactly $n \in \mathbb{N}$ composition series. I don't know if ...
