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Questions tagged [noetherian]

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

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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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The Ring Game on $K[x,y,z]$

I recently read about the Ring Game on MathOverflow, and have been trying to determine winning strategies for each player on various rings. The game has two players and begins with a commutative ...
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4answers
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Why are Noetherian Rings important?

I know they are important in abstract algebra, but why do people study them? Why are they so important to study? Do they make certain things easier to understand?
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1answer
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Is the ring of holomorphic functions on $S^1$ Noetherian?

Let $S^1={\{ z \in \Bbb{C} : |z|=1 \}}$ be the unit circle. Let $R= \mathcal{H}(S^1)$ be the ring of holomorphic functions on $S^1$, i.e. the ring of functions $f: S^1 \longrightarrow \Bbb{C}$ which ...
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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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1answer
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Subring of a finitely generated Noetherian ring need not be Noetherian? [duplicate]

A common example showing that a subring of a Noetherian ring is not necessarily Noetherian is to take a polynomial ring over a field $k$ in infinitely many indeterminates, $k[x_1,x_2,\dots]$. The ...
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Primary ideals of Noetherian rings which are not irreducible

It is known that all prime ideals are irreducible (meaning that they cannot be written as an finite intersection of ideals properly containing them). While for Noetherian rings an irreducible ideal is ...
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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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1answer
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Hilbert's Basis Theorem - Clever Proof?

So I am studying commutative algebra at the moment and I have come across the proof of the Hilbert Basis Theorem (the proof I have is the same as the one in Reid's Undergraduate Commutative Algebra). ...
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2answers
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Why are noetherian and artinian modules important?

As a TA I was recently asked to give the students an introduction to two (quite related) concepts that are new to me, noetherian and artinian modules. I intend to prove the characterisation theorem (i....
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1answer
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Ideals in $C[0,1]$ which are not finitely generated (From Atiyah- Macdonald )

I'm trying to solve the following problem from Atiyah-Macdonald: Is the ring of continuous function on $[0,1]$ is Noetherian ? Certainly not, here are two non terminating ascending chain of ideals:...
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Is noetherianity a local property?

Let $R$ be a ring with finitely many maximal ideals such that $R_{\mathfrak m}$ ($\mathfrak m$ maximal ideal) is noetherian ring for all $\mathfrak m$. Is $R$ noetherian? I think $R$ has to be ...
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2answers
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Converse to Hilbert basis theorem

Prove the converse to Hilbert basis theoren: If the polynomial ring $R[x]$ is Noetherian, then $R$ is noetherian.
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Exercise 5.5.M, Foundations of Algebraic Geometry by Ravi Vakil

In the following $A$ is a Noetherian ring. The basic definition here is: A prime $p\subsetneqq A$ is said to be associated to $M$ if $p$ is the annihilator of an element $m$ of $M$. The exercise ...
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What is an easy example of non-Noetherian domain?

Keep in mind, I'm strictly an amateur, though a very old one. I learned about imaginary numbers barely two years ago and ideals a year ago, and I'm still decidedly a novice in both topics. In the ...
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2answers
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Krull dimension of polynomial rings over noetherian rings

I want to prove the following theorem concerning Krull dimension: Theorem If $A$ is a noetherian ring then $$\dim(A[x_1,x_2, \dots , x_n]) = \dim(A) + n$$ where $\dim$ stands for the Krull ...
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Ring of formal power series over a principal ideal domain is a unique factorisation domain

An exercise in my algebra course book asks to prove that if $R$ is a PID, then $R[[x]]$ is a UFD, where $R[[x]]$ is the ring of formal power series over $R$. After some failed attempts at proving the ...
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1answer
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Grade of a maximal prime ideal in a Noetherian UFD

Here is an another problem in Commutative Rings by Kaplansky, p. 103, no. 15. Let $R$ be a Noetherian UFD. Let $(a,b) \not= R$ where $a,b \in R.$ Prove that any maximal prime of $(a,b)$ has grade ...
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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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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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Is this ring Noetherian?

The subring of $\mathbb{C}[x,y]$ consisting of all polynomials $f(x,y)$ whose gradient vanishes at the point $x=y=0$. Is this ring Noetherian?
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Noetherian ring with finitely many height $n$ primes

If $R$ is a Noetherian commutative ring with unity having finitely many height one prime ideals, one could derive from the "Principal Ideal Theorem", due to Krull, that $R$ has finitely many prime ...
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1answer
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Direct product of finitely many Noetherian non-unital rings is Noetherian

Let $A_1, A_2,...,A_n$ be Noetherian rings (not necessarily unital). Is the direct product $A:=A_1×A_2×⋯×A_n$ necessarily a Noetherian ring? If $A_1, A_2,...,A_n$ are unital, then one can prove that ...
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1answer
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Looking for a special class of ideals such that If every ascending chain of ideals from this class stabilizes, then $R$ is a Noetherian ring.

A commutative ring $R$ is called Noetherian if any one of the following holds: $1.$ Every ascending chain of ideals in $R$ stabilizes, that is, $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ ...
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Intersection of powers of an ideal in a Noetherian ring

Given a Noetherian ring $R$ and a proper ideal $I$ of it. Is it true that $$\bigcap_{n\ge 1} I^n=0$$ as $n$ varies over all natural numbers? If not, is it true if $I$ is a maximal ideal? If ...
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1answer
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The total ring of fractions of a reduced Noetherian ring is a direct product of fields

This is question 6.5 in Matsumura's "Commutative ring theory": How can I prove that the total ring of fractions of a reduced Noetherian ring is a direct product of fields?
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3answers
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Are Dummit and Foote making a mistake in proving Cohen's theorem?

Exercise 11 on page 669 (this is Chapter 15) wants to prove Cohen's theorem that if every prime ideal of a ring is f.g. then every ideal is f.g. that is the ring is noetherian. The highbrow (perhaps?) ...
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2answers
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Help with a problem about Artinian rings from Christian Peskine's book

I am stuck with problem 3, chapter 4, from the book of Peskine, An Algebraic Introduction to Complex Projective Geometry. Let $A$ be a Noetherian ring. Assume that if $a \in A$ is neither ...
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Why are the irreducible components of a subspace of Noetherian space just the intersection with the irreducible components?

Suppose you have a Noetherian topological space $X$, and its finitely many irreducible components are $X_1,\dots,X_n$. If $U\subseteq X$ is open, why are the irreducible components of $U$ precisely ...
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Noetherian rings and prime ideals

Let $R$ be a noetherian ring and $P\subset Q$ be prime ideals. I'm trying to prove that if there exists another prime ideal $P_1$ such that $P\subset P_1\subset Q$ and $P\ne P_1\ne Q$, then there are ...
8
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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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Integral domain over which every non-constant irreducible polynomial has degree 1

Let $R$ be an integral domain such that any polynomial $f(X) \in R[X]$ , which is irreducible in $R[X]$, has degree $1$. Then is it true that $R$ is a field ? If this is not true in general , What if ...
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Noetherian property for exact sequence

Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then $M$ is Noetherian is equivalent to $M'$ and $M''$ are Noetherian. For the ''$\Leftarrow$''...
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3answers
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Is the function ring $C^{\infty}(M)$ noetherian? [closed]

Let $M$ be a smooth manifold and $C^{\infty}(M)$ be its function ring. Is this a noetherian ring?
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3answers
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Nontrivial ideal of a Noetherian domain contains a finite product of nonzero prime ideals

If $R$ is a Noetherian domain and $ 0 < U < R$ an nontrivial ideal of $R$. How to prove that there exists nonzero prime ideals $p_1,...,p_n \subset R$ such that the product $ p_1 p_2 ...p_n \...
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1answer
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Is the global section ring of a Noetherian Scheme Noetherian as well?

As the title suggests, I am asked to prove that, given a Noetherian scheme $(X,\ \mathcal{O}_{X})$ and any open subset $U\subseteq X$, $\Gamma(U,\ \mathcal{O}_{X}):=\mathcal{O}_{X}(U)$ is a Noetherian ...
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1answer
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Non Noetherian subring of Z[X]

Let $\mathbb Z[X]$ be the ring of polynomials in one variable. It is a well-known fact that it is a Noetherian ring (because $\mathbb Z$ is a PID and therefore Noetherian and if $R$ is Noetherian then ...
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1answer
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Global dimension of quasi Frobenius ring

Let $R$ be a quasi-Frobenius ring (so $R$ is self-injective and left and right noetherian). I want to prove that $lD(R)=0$ or $\infty$, where $lD(R)$ denotes the left global dimension. I'm unsure ...
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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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Integral closure and $\bigcap \mathfrak{a}^n$

Let $R$ be a domain such that $\bigcap_{n=1}^\infty \mathfrak{a}^n=0$ holds for all proper ideals $\mathfrak{a}$ of $R$ (this holds, for example, if $R$ is Noetherian). Let $K$ be the quotient field ...
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Show that the ring of continuous functions $f:\mathbb R\to\mathbb R$ is not Noetherian

Prove that the ring of continuous functions $f:\mathbb R\to\mathbb R$ is not Noetherian. I know that to be Noetherian, every ideal is generated by finitely many elements or equivalently R satisfies ...
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Is a direct limit of Noetherian rings necessarily Noetherian?

Is the direct limit of Noetherian rings necessarily Noetherian? And if it is, how to prove this? If it is not, what is a counterexample? (I was thinking this question: if $A_{m}$ are Noetherian for $...
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1answer
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$R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ Noetherian?

Let $R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ a Noetherian ring ?
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A commutative noetherian ring in which each ideal $I$ is principal and $I^2=I$ must be a finite product of fields

PROBLEM A commutative noetherian ring $R$ in which each ideal $I$ is principal and $I^2=I$ must be a finite product of fields. I am lost with the condition $I^2=I$ and the desired result "a finite ...
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$I$-adic completion

Let $A$ be a commutative noetherian ring, and suppose that $A$ is $I$-adically complete with respect to some ideal $I\subseteq A$. Is it true that for any ideal $J\subseteq I$, the ring $A$ is also $J$...
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$\mathbb Z[\sqrt {-5}]$ is Noetherian

I'm trying to prove that $\mathbb Z[\sqrt {-5}]$ is Noetherian. I already know that $\mathbb Z[X]$ is Noetherian and I'm trying to find a surjective map $$\varphi: \mathbb Z[X]\to \mathbb Z[\sqrt{-5}...
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1answer
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Krull dimension of Noetherian local rings is finite

Does anyone know an "elementary" proof of the fact that a Noetherian local ring has finite Krull dimension? The one I know is from Atiyah&Macdonald's book Introduction to Commutative Algebra, ...
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1answer
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Is the integral closure of an integrally closed Noetherian domain in a finite extension field Noetherian?

Just as the title says. Let $R$ be a Noetherian integral domain, let $K$ be its field of fractions, let $L$ be a finite extension of $K$, and let $S$ be the integral closure of $R$ in $L$. Must $S$ ...
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2answers
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Proving a ring is Noetherian when all maximal ideals are principal generated by idempotents

Let $R$ be a commutative ring with unity such that all maximal ideals are of the form $(r)$ where $r\in R$ and $r^2=r$. I wish to show that $R$ is Noetherian. I know that if all prime (or primary) ...
6
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1answer
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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 ...