# 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 ...
1 vote
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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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1 vote
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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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1 vote
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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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### 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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### 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 $M_i$ ...
1 vote
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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 ...
### $\varphi'\in Hom_ R(M, N)$ 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 $\hat R$(...