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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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245 views

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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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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118 views

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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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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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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Ideals generated by regular sequences are generated by regular sequences in any order/Eisenbud, Exercise 17.6

I have a question regarding an exercise I found in Eisenbud's Commutative Algebra with a view towards Algebraic Geometry: Exercise 17.6: Any ideal of a Noetherian ring generated by a regular ...
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141 views

In an integral extension $A\subseteq B$, when does the Noetherian ness of $B$ imply that of $A$?

Let $A\subseteq B$ be an integral extension of commutative rings. If $B$ is Noetherian ring and finitely generated as $A$-module, the $A$ is Noetherian ring (we don't even need integral hypothesis ...
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102 views

Commutative local, artinian ring a homomorphic image of Noetherian (local) domain?

All rings are commutative with unity. Is every local, Artinian ring a homomorphic image of a Noetherian local domain? If this is not true, then, at least, Is every local, Artinian ring a ...
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A question regarding isomorphism of group rings

For a ring with unity $R$ and a group $G$ let $R[G]$ denote the group ring. Now let $R$ be a commutative Noetherian ring with unity such that $R[\mathbb Z_m] \cong R[\mathbb Z_n]$ (isomorphic as rings)...
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Integral domains $R$ such that any ring $R \subset S \subset K$ is Noetherian

Which integral domains $R$ with fraction field $K$ have the property that for any intermediate ring $R \subset S \subset K$, $S$ is Noetherian? I'm not really good at naming things, but let's call ...
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When does an integral group ring have finite global dimension?

Let $G$ be a finite group and $R=\mathbb{Z}[G]$ the integral group ring. If $G$ is such that $R$ is Noetherian (so $G$ polycyclic-by-finite) when does $R$ have finite global dimension? Another way of ...
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Proof of Krull's intersection theorem with Taylor expansion

I took a commutative algebra course last semester (with Kaplansky's book), and I've learned about Krull's intersection theorem. In the course, we proved it without using Artin-Rees Lemma. I heard that ...
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Can We Prove Cohen's Theorem for Modules by Using Cohen's Theorem for Rings?

Let $R$ be a ring. We have: Cohen's theorem for Rings. If all prime ideals of $R$ are finitely generated, then $R$ is Noetherian. Now let $M$ be an $R$-module. We have Cohen's Theorem for ...
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Corollary 4.19 from “Homological methods in commutative algebra”

I would like to show the following result: for a noetherian local ring $A$, we have $\mathrm{gl.dim}_A=\mathrm{hd}_A (k)$. Notice that the left side term of the equality is the global dimension of $A$,...
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42 views

Proving a ring $A$, generated by Noetherian subring and nilpotent element, is Noetherian again.

I am studying some algebra during my spare time. In particular I am learning about Noetherian rings. A friend sent me the following excersise, and I am not able to solve it. Suppose that a ring $A$ ...
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torsion free modules $M$ over Noetherian domain of dimension $1$ for which $l(M/aM) \le (\dim_K K \otimes_R M) \cdot l(R/aR), \forall 0 \ne a \in R$

Let $R$ be a Noetherian domain of Krull-dimension $1$ (i.e. every non-zero prime ideal maximal). Let $M$ be a torsion free $R$-module . Let $K$ be the fraction-field of $R$ and let $r=\dim_K S^{-1}M=\...
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Simplicity of Noetherian $B$, $A \subseteq B\subseteq C$, where $A$ and $C$ are simple Noetherian domains

After receiving important comments, which show that my original question has already been asked and answered, I now change my question to the following non-commutative setting: Let $A \subseteq B \...
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$\mathfrak q$ a $\mathfrak p$-primary ideal, show that every maximal chain from $\mathfrak q$ to $\mathfrak p$ has same length.

Let $A$ be a noetherian ring, and $\mathfrak q$ a $\mathfrak p$-primary ideal in $A$, then every chain of primary ideals from $\mathfrak q$ to $\mathfrak p$ has a finit length, the set of the length ...
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When is $RS^{-1}$ a local ring?

Suppose we have a noncommutative ring $R$ and multiplicatively closed set that is both right Ore, and right reversible, i.e. it is a right denominator set. Now, we can localize $R$ at $S$ to form $RS^{...
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$\mathbb{Z}[\sqrt{10}]$ is noetherian

How can we prove that $\mathbb{Z}[\sqrt{10}]$ is noetherian except by using Hilbert basis theorem? How can we find a sequence of ideals that satisfy the ACC?
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Local Noetherian ring and its invariants

Let $R$ be a local Noetherian ring and $G$ a finite group. Is $R$ finitely generated as a module over its invariants $R^G$? Thank you.
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Checking that a set is a finitely generated ideal

The exercise asks us to prove that $I = \{ f \in \Bbb R[X,Y,Z] \mid f(a,b,c) = 0, ~\forall\,(a,b,c)\in \Bbb S^2 \}$ is a finitely generated ideal of $\Bbb R[X,Y,Z]$. Well, clearly $I$ is an ideal of $...
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On $A\otimes_k B$ being noetherian

Let $k$ be a field, and let $A, B$ be commutative noetherian $k$-algebras. If either $A$ or $B$ is a localization of a finite type $k$-algebra, then clearly $A\otimes_k B$ is noetherian. Assume $A=B$...
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A question about the proof of Hilbert's Basis Theorem

I have a question regarding the proof of Hilbert's Basis Theorem. Say $I=(f_1,f_2,f_3,\dots)$ is an ideal in $A[x]$, where A is a Noetherian ring. Say we take the leading coefficients $a_i$ of all ...
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When are all (prime) ideals of an $R$-algebra, extensions of (prime) ideals of $R$?

Let $f:R\rightarrow R'$ be a homomorphism of commutative noetherian rings. When are all (prime) ideals of $R'$ extensions of (prime) ideals of $R$? Is it true for the case $R'$ is $R$-flat?
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Associated primes and finite base change

Let $R$ be an integrally closed commutative Noetherian integral domain. Let $R \subseteq S$ be a ring extension such that $S$ is also an integral domain and is finite as an $R$-module. Let $I$ be an ...
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383 views

Infinitesimal thickening of a smooth closed subscheme

Let $A$ be a noetherian ring (if it is useful I can assume that $A$ is an algebra of essentially finite type over a field) and $I \subset A$ is an ideal s.t. $A/I$ is smooth. Is it true that extension ...
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567 views

Coherent sheaves of finite length

Where can I find a treatment of coherent sheaves of finite length over, say, Noetherian schemes? Just things as basic as their definition and elementary facts about them. I am familiar with modules of ...
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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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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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Two definitions of an associated prime of an $R$-module

I have come across two definitions of an associated prime for an $R$-module $M$, one of which specifies that $R$ is Noetherian, however, I can't see the reason they would coincide. First one: A ...
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More aesthetically pleasing proof of lifting.

Say you have a complete Noetherian local ring $(A,\mathfrak m_A,k_A)$ and a complete Noetherian local $A$-algebra $B$ with residue field $A$. Furthermore, suppose you can find a surjective local ...
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Dimension of Quotient of Noetherian local ring

I have a question regarding the following exercise: Let $(A,\mathfrak{m})$ be a Noetherian local ring of dimension $d$. Let $k:=A/\mathfrak{m}$ be its residue field. $f_1, \ldots , f_r \in \...
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118 views

There exist an ideal whose associated primes are given set of non minimal prime ideals.

I was trying to understand a construction in the exercise of Atiyah and MacDonald, Introduction to Commutative Algebra, in chapter $4$ (exercise $19$, pg. $57$) and I stuck at the last line of that. I ...
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Serre's conditions and unmixedness

I'm trying to show the following claim: Let $R$ be a Noetherian ring. $R$ satisfies Serre's condition $S_k$ if and only if whenever $I$ is an ideal of the principal class (i.e., $I$ can be ...
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$R$ right Noetherian. Is it true that $R(x)\otimes_{R[x]}R(x)\cong R(x)$?

Let $R$ be a right Noetherian ring (actually it is left Noetherian as well) and $S=R[x]$ the polynomial ring in one (commuting) variable. If $X$ is the set of all monic polynomials then $X$ is a right ...
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When is every prime element of a Noetherian UFD irreducible in a simple algebraic ring extension

Let $D$ be a Noetherian UFD over a field $k$ of characteristic zero. Let $w$ be an algebraic, non-integral element over $D$, and denote its minimal polynomial over $D$ by $h(T)=d_nT^n+\cdots+d_1T+...
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Proving that $A$ is a Noetherian ring when $M$ is a faithful Noetherian $A$-module

Let $M$ be a Noetherian $A$-module (where $A$ is a commutative ring). I want to show that $M$ is faithful implies $A$ is Noetherian. I am very aware that this question is fully answered in both of ...
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Proof that $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a finitely-generated left $A_1(\mathbb{C})$-module

I want to prove that $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a finitely-generated left $A_1(\mathbb{C})$-module where $\lambda \in \mathbb{C}$ and $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a subring of $\...
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Equivalence of definitions of Noetherian Ring, another proof.

Let $R$ be a commutative ring with unity, then the following are equivalent -1. Every ideal in $R$ is finitely generated -3. Every nonempty collection of ideals of $R$ has a maximal element I will ...
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390 views

Intersection of height one prime ideals

Let $A$ be a commutative, integrally closed, noetherian ring and as $\mathfrak p$ ranges over height one prime ideals, we have: $$A = \bigcap_\mathfrak p A_{\mathfrak p}.$$ The proof I have seen ...
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141 views

For finitely generated $B$ all modules in exact sequence are finitely generated in PID

Let there be an exact sequence of $R$-modules: $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ where $R$ is a principal ideal domain and $B$ is finitely generated. Are $A$ and $C$ ...
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Characterization of Noetherian module

I guess this is a simple question but I don't see the answer. Definition: $M$ is Noetherian if every chain of submodules stabilize. Theorem: $M$ is Noetherian module iff every nonempty set $S$ of ...
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When a two-generated ideal of a noetherian integral domain have a finite projective resolution?

Let $R$ be a noetherian integral domain, and $I$ a non-zero ideal of $R$ which can be generated by two elements. (We do not know if $I$, considered as an $R$-module, is $R$-projective; maybe yes maybe ...
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149 views

Flat base change preserves the property of being non-degenerate

We say a homomorphism $f:A\rightarrow B$ of noetherian rings is non-degenerate if the induced map $f^*:{\rm Spec}(B) \rightarrow {\rm Spec}(A)$ maps ${\rm Ass}(B)$ into ${\rm Ass}(A)$. Let $f:A \...
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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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51 views

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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47 views

$R_{\mathfrak{p}}$ 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 ...