Questions tagged [noetherian]

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

Filter by
Sorted by
Tagged with
1 vote
0 answers
73 views

Is The sub ring $\mathbb{C}\{z\}$ Noetherian?

I am attempting to prove (or disprove) that $\mathbb{C}\{z\}$ is Noetherian, where $\mathbb{C}\{z\}$ is the subring of $\mathbb{C}[[z]]$ consisting of functions holomorphic in a neighborhood of $0 \in ...
Santa-claus 's user avatar
0 votes
1 answer
105 views

Is the ring $\mathbb{R} [x]$ Noetherian?

I aim to assess the soundness of my approach. In an exercise, the question is posed regarding whether $\mathbb{R}[x]$ is Noetherian, and my response is negative. To support this, I employ the ...
Santa-claus 's user avatar
0 votes
1 answer
47 views

The ring of Laurent polynomials over a Noetherian ring is Noetherian

In the Wikipedia about Larent polynomial, there is this result: The ring of Laurent polynomials over a field is Noetherian. I was wondering if would it be enough ...
ghc1997's user avatar
  • 1,351
1 vote
2 answers
95 views

Let $R$ be a Noetherian commutative ring with zero nilradical and with any localization at a maximal ideal as a finite ring. Prove that $R$ is finite

Suppose that $A$ is a Noetherian commutative ring such that: (1) the nilradical (intersection of all prime ideals) of $A$ vanishes, and (2) localization at every maximal ideal is a finite ring. Prove ...
Squirrel-Power's user avatar
1 vote
1 answer
112 views

Small generating set for the unique minimal prime ideal of a finitely generated $\mathbb{C}$-algebra

Let $A$ be a $\mathbb{C}$-algebra, generated by $n$ elements ($n$ finite). Assume that $A$ has a unique minimal prime ideal $\mathfrak{p}$. Write $t$ for the minimal number of generators for the ideal ...
Object's user avatar
  • 285
1 vote
1 answer
50 views

ID's, PID's, Noetherian rings and valuation rings: implications amongst them

I am trying to establish some implications between being an ID, a PID, a Noetherian ring and a valuation ring. First of all, I know that PID $\Rightarrow$ Noetherian, because in a PID every ideal is ...
kubo's user avatar
  • 1,762
0 votes
1 answer
60 views

Algebraic properties of the ring of analytic functions on the complex plane

Let $r, R>0$ and $D_{R}=\{z\in\mathbb{C}: \|z\|<R\}$ and $D_{r,R}=\{z\in\mathbb{C}: r<\|z\|<R\}$. Consider the following $\mathcal{R}_{1}:=\mathcal{O}_{R}=\{f:D_{R}\to\mathbb{C}: f\hspace{...
user 987's user avatar
  • 591
-1 votes
1 answer
45 views

Show quotient of ideals is Noetherian

Not sure how to approach this problem: Let A be a commutative ring, I,J ideals of A such that A/I is a Noetherian ring. Show that J/(I ∩ J) is Noetherian as a A-module So far I only concluded that if ...
MischiefManaged's user avatar
0 votes
0 answers
32 views

Chain of kernels of maps induced from picking a basis of modules

Let $R$ be a Noetherian ring. Suppose that $M_1$ is a finitely generated $R$-module, so that after picking a list of $n_1$ generators, $M_1$ is isomorphic to $R^{n_1}/M_2$, where $M_2$ is another ...
Ben Lou's user avatar
  • 13
0 votes
1 answer
44 views

Is each component of a graded module over a $k$-algebra a finite-dimensional vector space?

I have some problems with an argument in a proof of a lemma: Let $M = \oplus_{-\infty}^{\infty} M_n$ be a finitely generated graded $A$-module and $A=\oplus_{n\geq 0} A_n$ a graded commutative ring ...
Heraklit's user avatar
  • 387
2 votes
0 answers
112 views

Proving that if $J=\bigcap_{n\geq0}I^n$ then $IJ = J$

I'm having some trouble with the following exercise: Let $R$ be a Noetherian commutative ring, $I$ and ideal and $J=\bigcap_{n\geq0}I^n$. Show that $IJ = J$. (Hint: Assume that $J\not\subseteq IJ$ ...
Eduardo Magalhães's user avatar
1 vote
1 answer
60 views

Ascending chain conditions between two rings

This question has been solved. I have a question about ascending chain between different rings. Here is my question: Assume that $f:S \rightarrow R$ be an injective ring homorphism and $I_1\subset I_2\...
fusheng's user avatar
  • 1,076
2 votes
0 answers
68 views

Are module finite algebras over Noetherian semiperfect rings again semiperfect?

Let $S$ be a Noetherian semiperfect ring (https://en.m.wikipedia.org/wiki/Perfect_ring). Let $R$ be a module finite associative $S$-algebra. Then, is $R$ also a semiperfect ring? (Clearly, $R$ is ...
uno's user avatar
  • 1,560
1 vote
1 answer
90 views

Length of maximal ideal in noetherian ring

Let $A$ be a noetherian commutative ring and $\mathfrak{m}$ a maximal ideal. Is it possible to have $ht(\mathfrak{m}) > 1$? Here $ht(\mathfrak{m})$ is the height of $\mathfrak{m}$, that is the ...
Analyse300's user avatar
3 votes
1 answer
141 views

Need help showing that the only submodules of $M$ are the ones in an ascending chain.

$\color{Green}{Background:}$ $\textbf{Assumed facts:}$ $\textbf{Theorem 1:}$ Let $R$ be a ring. Then the following conditions are equivalent: $(1)$ Every ideal of $R$ is finitely generated $(2)$ ...
Seth's user avatar
  • 3,219
0 votes
1 answer
87 views

Prove $\cap_{n\geq0} I^n = (0)$

Let $A$ be an integral domain and noetherian and let $I \subset A$ a proper ideal such $I*\cap_{n\geq0} I^n = \cap_{n\geq0} I^n$. Prove that $\cap_{n\geq0} I^n = (0)$ I'm trying to prove it by getting ...
Juan José Campos's user avatar
0 votes
0 answers
46 views

Dummit & Foote (3rd ed.) 15.1.11

I have some specific questions regarding this not answered here: Are Dummit and Foote making a mistake in proving Cohen's theorem? (for example). Note, I don´t claim the exposition below is a full ...
Ben123's user avatar
  • 868
2 votes
1 answer
45 views

On which rings must a finitely generated module be finitely presented? Is there an 'if and only if' characterization for such rings?

As is well known, if $R$ is a Noetherian ring, then a finitely generated module over $R$ must be finitely presented. However, this is not necessarily true for coherent rings. For example, consider $k$ ...
Liang Chen's user avatar
4 votes
1 answer
278 views

Why is the noetherian ring property not definable in first-order logic?

I am reading this paper on the connection between model theory and algebraic geometry. https://math.uchicago.edu/~may/REU2015/REUPapers/Zhang,Victor.pdf On page 9, I have trouble understanding Example ...
Y.X.'s user avatar
  • 3,935
0 votes
0 answers
39 views

Is the ideal of projective variety homogeneous when the projective variety is defined over a ring

We call the set $\mathbf{I}(V) = \{f \in K [X] | f(x)=0,\text{ for all } x \in V\}$ the homogeneous ideal of projective variety $V$. Indeed, if $f$ and $g$ both vanish on $V$, and $r$ is an arbitrary ...
Zirui Yan's user avatar
2 votes
0 answers
109 views

Show that the ring of formal power series in a commutative ring $R$, $R[[x]]$ is noetherian.

Yes, I am aware that this has been answered (If $R$ is a Noetherian ring then $R[[x]]$ is also Noetherian), but the answers given did not answer my specific question regarding this: In my notes from a ...
Ben123's user avatar
  • 868
0 votes
0 answers
105 views

Non-Noetherian ring that has an ideal $I$ containing a power of $\sqrt{I}$

We already knew that if $R$ is a Noetherian ring, then every ideal $I$ contains a power of its radical. Now suppose that $R$ is non-Noetherian, is there any example of $R$ such that one of its ideals, ...
Mystery girl's user avatar
2 votes
0 answers
52 views

Rings where finitely generated ideals are closed under countable intersection

Does there exist a characterization of those rings $R$ such that finitely generated left ideals are closed under countable intersection? For example, any noetherian ring has this property, since all ...
Piotr Pstrągowski's user avatar
1 vote
1 answer
18 views

counter-example for quotient of artinian and noetherian modules

Suppose I have an $R$-module $P$ and let $M, N\in P$ be submodules. Furthermore, suppose that both $M$ and $N$ are artinian and noetherian. Is it always true that the quotient $P/M\cap N$ is also ...
User666x's user avatar
  • 844
0 votes
0 answers
66 views

What are some standard generalisations of Noetherian rings?

Let $\textbf{CRing}$ be the category of commutative rings with $1$. Then, an immensely important subclass is that of the Noetherian rings, those rings which satisfy either of the equivalent conditions:...
legionwhale's user avatar
  • 2,336
0 votes
0 answers
46 views

Why does an injective $R$-module homomorphism $M\to N$ preserve submodules?

Let $R$ be a commutative ring with unity. Let $0\to M\xrightarrow{f} N\xrightarrow{g} P\to 0$ be a short exact sequence of $R$-modules. I am looking at a proof of the fact that $N$ is Noetherian $\iff$...
IAAW's user avatar
  • 1,466
0 votes
1 answer
122 views

Proving a quotient ring is Noetherian. [duplicate]

I am trying to prove that if $R$ is Noetherian, and $I$ is an ideal of $R$, then $R/I$ is Noetherian. I know that both $R$ and $I$ are finitely generated since they are both ideals of $R$. I’m having ...
user873295's user avatar
0 votes
0 answers
53 views

the case of Noetherian local ring

I have some question about the propositon(c) of thetheorem 3.16 in Qing Liu's book. the statement is: Let (A,m) be a Noetherian local ring, and $\hat{A}$ its m-adic completion. Let (B,n) be a local ...
梦尽缘灭's user avatar
2 votes
0 answers
45 views

Is Artinian assumption necessary here in Matsumura's book?

I quoted Theorem 13.2 below from Matsumura's book Commutative Ring Theory: Let $R=\bigoplus_{n\geq 0}R_n$ be a Noetherian graded ring with $R_0$ Artinian, and let M be a finitely generated graded $R$-...
William Sun's user avatar
  • 2,369
0 votes
1 answer
68 views

A principal ideal is prime in a coprime localization -- is it prime in the ring?

Let $R$ be a normal Noetherian domain. Suppose there is $x \in R$ such that the principal ideal $xR$ has only one associated minimal prime ideal $\mathfrak{p}$. Assume there exists a multiplicative ...
Grabovsky's user avatar
  • 151
3 votes
1 answer
44 views

Noetherian, self-injective ring $R$ with non-torsionless $R$-module

Let $R$ be an (associative, unital) ring. By an $R$-module we mean a left $R$-module. We call an $R$-module torsionless if it can be embedded into a direct product of the regular $R$-module $R$. I am ...
Margaret's user avatar
  • 1,729
1 vote
1 answer
37 views

On the property of a ring modulo its Jacobson radical being a division ring

Let $(R,\mathfrak m)$ be a commutative Noetherian local ring. Let $A$ be a finitely generated associative $R$-algebra. Let $x\in \mathfrak m$ be a non-zero-divisor on $A$ such that $xA\neq A$. If $A/J(...
uno's user avatar
  • 1,560
0 votes
1 answer
30 views

relation between support and submodules

Support of an $R$-module $M$ is defined as $\mathrm{Supp}_R(M)=\{\mathfrak{p}\in Spec(R); M_{\mathfrak{p}}\neq 0\}$. Let $M$ and $N$ two finitely generated modules over a Noetherian ring $R$ such that ...
Saeed Yazdani's user avatar
0 votes
1 answer
120 views

Colon of Two ideals in a ring [closed]

Is there any example of a commutative ring $R$ with unity and its proper ideal $I$ where $(I^{k+1} : I) = I^k $ does not hold for some value of k ?
Pritam 's user avatar
4 votes
1 answer
122 views

Geometric intuition behind Noetherian rings?

In Algebraic Geometry, there is this nice interpretation for the localization of rings and for quotient rings. Let $A$ be a ring. Then, localizations of $A$ at different elements correspond to the ...
Josu P. Z.'s user avatar
1 vote
0 answers
68 views

Are sober noetherian spaces sequential?

A sequential topological space $X$ has a few different equivalent definitions: $X$ is the quotient of a first-countable space $X$ is the quotient of a metric space Sequentially open subsets of $X$ ...
saolof's user avatar
  • 639
0 votes
1 answer
88 views

$k[x^{1/2},x^{1/4},x^{1/8},...]/(x)$ is not noetherian

Let $R$ be a non-trivial commutative ring. If $R$ has only one prime ideal $\mathfrak{p}$, is $R$ Noetherian? As pointed out in the comments, this affirmation is false. E.g., take $k[x^{1/2},x^{1/4},...
tomate's user avatar
  • 119
0 votes
1 answer
85 views

Show that the identity component $G^\circ\sqsubseteq G$ (i.e., is characteristic) in a linear algebraic group $G$.

This is Exercise 7.6.6 of Humphreys', "Linear Algebraic Groups". The Question: Show that the identity component $G^\circ$ of a linear algebraic group $G$ is a characteristic subgroup. The ...
Shaun's user avatar
  • 44.4k
0 votes
0 answers
40 views

Rank of commutative free monoid and Krull dimension of Monoid ring

We just introduced the notion of the Krull dimension of a ring in class and I was thinking about the following: Let A be a commutative, noetherian ring with unit and let $n=\text{dim}(A)$ be the Krull ...
Liva's user avatar
  • 61
0 votes
2 answers
91 views

Proving that $\mathbb{Z}$ is a Noetherian ring

So by definition I know that for all increasing chain of ideals: $I_1\subseteq I_2\subseteq \ldots I_k\subseteq I_{k+1}\subseteq \ldots$ in $\mathbb{Z}$, $\exists n\in \mathbb{N}, n \geq 1: I_n = I_{n+...
Tomas Rojas's user avatar
-1 votes
1 answer
37 views

Noetherian induction; closed sets that are not finite union of irreducibles

I am following an online course on algebraic geometry. On today's lesson, the lecturer stated the following theorem: Any Noetherian space is a finite union of irreducible subspaces. For the proof, he ...
kubo's user avatar
  • 1,762
0 votes
0 answers
64 views

Quotient of the polynomial ring over a local noetherian ring is finitely generated?

I am going through these lecture notes, and I am confused about the proof of Corollary 10.10 there. Corollary 10.10. Let $A$ be a local noetherian domain with maximal ideal $\mathfrak{p}$, let $g \in ...
Kyaw Shin Thant's user avatar
0 votes
0 answers
36 views

Smoothness criterion using completion.

Let $A\to B$ be a ring map of Noetherian commutative unital rings. Let $\mathfrak{p}\subseteq A$ and $\mathfrak{q}\subseteq B$ be primes such that $\mathfrak{q}\cap A=\mathfrak{p}$. I wonder whether ...
Display Name's user avatar
  • 1,373
0 votes
0 answers
68 views

Decomposition of bivariate polynomials over finite fields as a sum of univariate products

Let $p$ be a prime. Given a bivariate polynomial $f(X,Y)\in \mathbb{F}_p[X,Y]$ with degrees $d_1,d_2$ in $X,Y$ respectively, what is the lowest known upper bound on the smallest integer $k$ such that $...
Mathdropout's user avatar
0 votes
1 answer
97 views

Why is it called Hilbert's "basis" theorem? [duplicate]

The statement of the theorem doesn't seem to hint at any "basis" whatsoever, so why the "basis"? The theorem: If $A$ is a Noetherian ring, then $A[x]$ is also Noetherian.
Atom's user avatar
  • 3,791
0 votes
0 answers
29 views

What is an example for a noetherian connected graded algebra with infinite depth?

Let $k$ be a field, and $A=A_0\oplus A_1\oplus \dotsb$ be an $\mathbb{N}$-graded $k$-algebra which is left-noetherian and connected (i.e. $A_0=k$). Depth of a left $A$-module $M$ is defined as the ...
Noto_Ootori's user avatar
0 votes
0 answers
50 views

A Noetherian-as-a-module ring is Noetherian as a ring?

This SE answer claims [see the edit] (in the beginning paragraph) that if $A$, $B$ are rings with $B$ being Noetherian as an $A$-module, then $B$ is Noetherian as a ring because any chain of $B$-...
Atom's user avatar
  • 3,791
1 vote
0 answers
33 views

Hint about a problem with Noethenian rings and its ellements

Consider the following problem Let $R$ be an integral domain and $R^{\times}$its group of units. An element of $S=R \backslash\left(R^{\times} \cup\{0\}\right)$ is irreducible if it is not a product ...
Maths Wizzard's user avatar
2 votes
2 answers
124 views

What is this ring $\mathbb{Z}[x,1/x]$?

I can see that it is the ring of polynomials over $\mathbb{Z}$ with variables $x,1/x$, i.e. things like $x+1/x$. Is there a better way to visualize the elements of this ring? I am wondering if it is ...
quanticbolt's user avatar
  • 1,746
1 vote
1 answer
58 views

Annihilators of elements in the associated graded ring

Let $R$ be a Noetherian ring, $I\subset R$ an ideal, and $S\subset R$ a multiplicatively closed set. Let $T$ be the associated graded ring $\text{gr}_I(R):=R/I\oplus I/I^2\oplus I^2/I^3\oplus ...$ ...
Bromelain's user avatar
  • 421

1
2 3 4 5
20