Questions tagged [noetherian]

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

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Showing that $R/\operatorname{ann}(A)$ is artinian

Let $R$ be a noetherian ring. Let $A$ be an $R$-algebra finitely generated as $R$-module, which is an artinian ring. Then $R/\operatorname{ann}(A)$ is artinian? My first attempt is, Q.1. Since $A$ is ...
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When $\mathrm{Hom}$ functor commutes with colimits in a category of modules? [duplicate]

I was looking through this question, and there's a thing I don't understand why it holds. I mean the next statement in the answer by @Peter McNamara: Since $R$ is Noetherian, $I$ is finitely ...
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Unimodular element of a module

Let $A$ be a noetherian ring and $M$ an $A$-module. An element $z \in M$ is said to be unimodular if $Az$ is a direct summand of $M$ and $Ann(z) = \{r \in A | rz=0\} = 0$. The order ideal of $M$ is: $...
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Krull dimension of an $r$-image of a Noetherian Module

Let $R$ be a ring and $M$ be a right R-module. The Krull dimension of $M$, denoted by ${\rm Kdim} M$, is defined by transfinite recursion as follows: if $M = 0$, then ${\rm Kdim}M = −1$; if $\alpha$ ...
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Relation between height of an ideal and unimodular row [closed]

A row $[b_1,\dots,b_{d+1}]$ is unimodular if there is $[a_1,\dots,a_{d+1}]$ such that $a_1b_1+\cdots+a_{d+1}b_{d+1}=1$. Let $A$ be a Noetherian ring of dimension $d$ and $I$ be an ideal generated by $...
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Lifting Property of flips

I am trying to understand the concept of flips before learning the Bass Cancellation theorem. Let $R$ be a Noetherian ring and $P$ be a projective $R$ module. Let $p,q \in P$ , $\phi \in Hom(P,R)$ ...
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A question related to the dimension of a ring [duplicate]

Let $R$ be a Noetherian ring and consider $S=R[x,x^{-1}]$. I would like to show that $\mathrm{dim}(S)=\mathrm{dim}(R)+1$. I know that for a Noetherian ring $A$, $\mathrm{dim}(A[t])=\mathrm{dim}(A)+1$, ...
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$M/xM$ free over $R/xR$ implies $M$ is free over $R$ when $R$ Noetherian

The following is from https://stacks.math.columbia.edu/tag/00NS. I'm having some difficulty understanding some steps of the proof. Let $R$ be a Noetherian local ring. Let $x \in \mathfrak m$. Let $M$ ...
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Question On Serre's Splitting Theorem

I am learning the splitting theorem from the book F. Ischebeck and Ravi Rao. The statement is as follows: Let $A$ be a commutative Noetherian ring of finite Krull dimension. Let $P$ be a finitely ...
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Surjection from quotient of power series ring over a DVR

Let $\mathcal{O}$ be a discrete valuation ring, $\varpi$ a generator of its maximal ideal. I'm reading that there exists a surjection $\frac{\mathcal{O}[[x_1,...,x_n]]}{(\varpi x_1,...,\varpi x_n)} \...
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Can generators of Noether normalization be zero divisors?

Let $k$ be a field and let $A$ be a finitely generated $k$-algebra. By Noether normalization one can find algebraically independent elements $x_1, \dots, x_n \in A$ such that $k [x_1, \dots, x_n] \...
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If $A$ is Noetherian, and $x$ is a non-unit and non zero-divisor, then can $(xr)=(r)$ for any non-unit $r$?

Problem I am attempting to prove that if $A$ is a Noetherian commutative ring with unit, and if $x$ is a non-unit and not a zero divisor, then any minimal ideal of $(x)$ has height $1$. Attempt I have ...
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Why $\mathfrak{p}A_\mathfrak{p} = 0$, where $A_{\mathfrak p}$ is the localization at the kernel $\mathfrak p$ of a surjective ring homomorphism.

Let $A$ be a commutative, Noetherian, local ring, $\mathfrak{O}$ a discrete valuation ring and $\lambda : A \rightarrow \mathfrak{O}$ be an epimorphism. Let $\mathfrak{p}=\ker(\lambda)$, and consider ...
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Finitely generated modules over a noetherian local ring whose completions are isomorphic

This is part of an exercise in Eisenbud's commutative algebra, chapter 7, no. 5. Suppose $M$ and $N$ are finitely generated modules over a Noetherian local ring $R$ whose completions $\hat{M}$ and $\...
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When are fiber products of Noetherian rings Noetherian?

I am trying to solve the following problem as part of a question from the Fall 2021 UCLA algebra qualifying exam: is the fiber product $\mathbb{Q} \times_{\mathbb{Q}[x]/(x^2-x)} \mathbb{Q}[x]$ in the ...
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2 votes
1 answer
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Zerodivisors on quotient module [closed]

Let $R$ be a Noetherian local ring and $M$ be a finitely generated $R$-module. Suppose that two ideals $I$ and $(x)$ are consisting of zerodiviors on $M$. Given a nonzerodivisor $a\in (I,x)$ on $M$, ...
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What conditions are necessary for an Artinian principal ideal ring to be uniserial?

Let $R$ an artinian commutative ring such that every factor ring (including $R$) is QF (quasi-frobenius, i.e. noetherian and self-injective). I'm looking for conditions to reach that this is a ...
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How large a class of rings are the noetherian domains that have finite character?

A commutative ring $R$ has finite character if every nonzero element of $R$ is contained in at most finitely many maximal ideals. I am interested in the particular case in which $R$ is a Noetherian ...
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Injective modules and prime ideals

Let $R$ be a ring with $1$. We say that a right $R$-module $M$ is indecomposable if $M$ can not be decomposed as the internal direct sum $M=L \oplus K$ with $L$ and $K$ non-zero. Suppose that $R$ is a ...
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Is polynomial ring of infinite many variables Laskerian?

Let $R$ be a commutative ring, we say $R$ is a Lasker ring, if every ideal of $R$ has primary decomposition. By Hilbert basis theorem and Lasker-Noether theorem (every Noether ring is a Lasker ring), ...
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Noetherian domain with unique principal prime ideal that is not a DVR

The question is whether such a thing exists. Namely, a discrete valuation ring (DVR), in whatever way you define it, is quite obviously a domain, Noetherian, and has a unique prime element up to ...
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Finitely generated module over Noetherian ring, all whose localizations at associated primes of the ring is $0$, is a torsion module?

Let $M$ be a finitely generated module over a commutative Noetherian ring $R$ such that that $M_P=0$ for every associated prime $P$ of $R$. Then, is it true that for every $m\in M$, there exists a non-...
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If $\operatorname{Proj} S$ is noetherian, must $S$ be noetherian?

Suppose $S$ is a positively graded ring, and $S$ is generated by $S_1$ (the degree $1$ part of $S$). If we know $\operatorname{Proj}(S)$ is Noetherian, can we deduce $S$ is a Noetherian ring? Could ...
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Submodules of Localization

Let $A$ be a polynomial ring $K[x,y,z]$ (or however number of variables) with $K$ a field. And suppose $f$ is a non-constant polynomial in $A$. Then clearly $A_f$ is an $A$-module that is not finitely ...
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Modules over a DVR can be considered as modules over its congruence algebra?

Let $A$ be a commutative, Noetherian, local ring, $O$ a discrete valuation ring and $\lambda : A \rightarrow O$ be an epimorphism. Let $p:=\ker(\lambda), \Phi_A:= p/p^2, I:=Ann[p], \Psi_A:=O/\lambda(I)...
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Condition for conormal module of commutative, Noetherian, local ring to have finite length

Let $A$ be a commutative, Noetherian, local ring, $O$ a discrete valuation ring and $\lambda : A \rightarrow O$ be an epimorphism. Let $p=\ker(\lambda)$, and consider the conormal $A$-module $p/p^2$. ...
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$M$ finitely generated $\nRightarrow$ $N$ and $M/N$ finitely generated

In an algebra lecture we looked at the following lemma: Let $R$ be a ring, $M$ an $R$-module and $N \subset M$ a submodule. Then $N$ and $M/N$ are finitely generated $\implies$ $M$ is also finitely ...
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Applications of Noetherian approximation for commutative rings

I am seeking an example of how Noetherian approximation can be used to simplify a proof in the context of commutative algebra strictly. To begin, I am referring to the fact that any commutative ring ...
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3 votes
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SES of finitely generated modules over noetherian ring split, given existence of an isomorphism.

I recently came across the following statement, however I always fail to prove it. I'm also a bit unsure where the assumption "noetherian" is used: Let $R$ be a noetherian ring, and $M_i$ ...
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1 answer
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What are some simple examples of "ID-only" rings? integral domains, that are neither atomic nor Noether nor Prüfer...

Most common examples in the literature are rings of "higher virtues", having finite decompositions into irreducibles (maybe non-unique) -> atomic rings (Cohn), or Noether, or Prüfer, or ...
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Is there equivalence in Hilbert's theorem on noetherian Rings? [duplicate]

In a course about commutative algebra, I came across the following theorem: $$\text{If $R$ is a noetherian ring, then $R[X_1,\dots,X_n]$ is noetherian.}$$ I can't help but wonder, if the converse is ...
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Why is the irreducible component of an irreducible set is the set itself?

Perhaps this is obvious, but it really confuses me. It seems that I have to accept the fact: Let $X$ be a noetherian topological space, and let $Y$ be an irreducible closed subset in $X$. Then the ...
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2 votes
1 answer
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$\varphi'\in Hom_ R(M, N)$ differs from $\varphi$ by an element of $P Hom_ R( M, N)$, then $\varphi'$ is an epimorphism.

Suppose $M$ and $N$ are finitely generated modules over a Noetherian local ring $(R,P)$ whose completions $\hat M$ (that is the inverse limit of $M/P^iM$) and $\hat N$, are isomorphic over $\hat R$(...
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2 votes
1 answer
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Finding a Noetherian faithful module

Let $M$ be a faithful $A$-module with the property that the modules of the form $IM$ for some ideal $I\subseteq A$ satisfy the ascending chain condition; also, suppose that $\{m_1,\dots,m_r\}\subseteq ...
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1 answer
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Sufficient conditions for $\operatorname {Spec} f$ to be closed

Let $f:A\to B$ be a ring homomorphism and suppose that $\operatorname {Spec}B$ is a Noetherian space. Show that (1) the prime ideals of $B$ satisfy the ascending chain condition, and (2) $f$ has the ...
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6 votes
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Can the Noetherian topological spaces be distinguished by looking at the category of topological spaces only?

A Noetherian topological space is one where every infinite descending sequence of closed sets $X_0 \supset X_1 \supset X_2 \cdots$ is eventually constant (paraphrased from Hartshorne page 5). ...
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Non-finitely generated R-module

Currently studying for qualifying exams and came across the following problem: Give an explicit example of a ring $R$ (commutative with identity) and a surjection $\psi: M \rightarrow N$ of finitely ...
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A less constructive proof of acc + dcc implies the largest chain of modules have finite length

We know  If a module satisfies both ACC and DCC then the  largest chain is of finite length. Instead of defining largest chains we define maximal chains (which are more constructive and easier to work ...
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1 answer
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A ring that is left Noetherian but not right noetherian

I am studying Chapter 8 of Hungerford for my qualifying exam. In Chapter 8.1., he gives an exercise concerning the ring that is right Noetherian (resp. Artinian) but not left Noetherian (resp. ...
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2 answers
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Is a ring whose every descending chain of prime ideals stabilizes, necessary Noetherian?

It is know that the DCC on prime ideals holds in Noetherian rings (see e.g., This question). I ask whether the converse holds: Is a commutative ring whose every descending chain of prime ideals ...
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3 votes
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Exercise with Noetherian modules [duplicate]

Suppose that $A$ is a ring, $M$ is a Noetherian $A$-module, and $\operatorname{Ann}M=0$. I must prove that $A$ is Noetherian. I tried to prove it by contradiction, assuming that $A$ isn't Noetherian. ...
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2 votes
1 answer
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Regular Noetherian Local Ring is Integral Domain

I am currently reading a proof of the fact that every regular Noetherian local ring $R$ is an integral domain. The proof argues by induction on $d=\operatorname{dim}R$. The base case $d=0$ is clear. ...
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1 answer
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Is the ring of lower triangular matrices over a field both right and left Noetherian?

I am looking at the ring $\left\{\begin{bmatrix}a&0\\b &c\end{bmatrix}: a,b,c\in k\right\}$, where $k$ is a field. I think that it is both left and right Noethrian, but I do not know how to ...
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1 vote
1 answer
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Existence of System of Parameters

I am currently reading Algebraic Geometry and Commutative Algebra by S. Bosch and I have troubles proving the following Theorem: Let $R$ be a Noetherian local ring with maximal ideal $m$. Then there ...
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1 answer
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Do ample sheaves descend along limits / noetherian approximation?

Suppose $X \to S$ is a proper (projective) morphism of schemes, $S$ is quasi-compact and quasi-separated, and $\mathcal L$ is a relatively ample sheaf on $X$. By stacks, tag 01ZA, the scheme $S$ is a ...
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1 vote
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Is a quotient of $k[x,y]$, where $k$ is a field, a finitly generated module?

i'm trying to prove the claim in the title, my argument is: $k$ is a field so it's Noetherian, hence so is $k[x,y]$ by the Hilbert basis theorem, hence $k[x,y]$ is a Noetherian $k$-module, and a ...
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1 vote
1 answer
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Not Hilbert's basis theorem

Hilbert's basis theorem tells us that $R[x]$ is Noetherian if $R$ is Noetherian. So, if every ideal $I\subset R$ is finitely generated, then every ideal $J\subset R[x]$ is finitely generated. This led ...
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0 votes
2 answers
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Does $R[x]$ Noetherian imply $R[x,y]$ is Noetherian

I am studying Noetherian rings at the moment and I know by the Hilbert Bases Theorem that if a ring $R$ is Noetherian then $R[x]$ is also Noetherian. My question is: Does this imply $R[x,y]$ is ...
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2 votes
1 answer
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Noetherian modules in exact sequence

Assume that the complex $$\cdots \xrightarrow{c_{i+2}} L \xrightarrow{c_{i+1}} M \xrightarrow{c_i} N \xrightarrow{c_{i-1}} \cdots$$ is exact and $L$ and $M$ are Noetherian. Prove that $M$ is ...
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6 votes
1 answer
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Prove certain ring is noetherian.

Let $R$ be a commutative ring with the property that for each nonzero ideal $I$ and element $a ∈ I$ there exists a unique ideal $J$ such that $IJ = (a)$. Show that $R$ is Noetherian. I am looking for ...
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