Questions tagged [noetherian]

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

2
votes
2answers
177 views

What can be said of the lattices of submodules of a noetherian/finitely generated module?

I've asked several times about properties of the lattice of submodules/ideals of modules/rings with specific properties. This time I wonder about what interesting properties can one see in the ...
4
votes
4answers
386 views

Noetherian module and noetherian ring [closed]

If $R$ is a noetherian ring then also $R[x]$ is a noetherian ring, i.e. $R[x]$ is noetherian as $R[x]$-module. Is $R[x]$ also noetherian as $R$-module?
-1
votes
2answers
120 views

Prove that there is a finite subset of these polynomials whose zeros define the same locus.

I am attempting to solve Ch 14 Problem 6.1 from Artin's Algebra textbook. Let $V\subset\mathbb{C}^n$ be the locus of common zeros of an infinite set of polynomials $f_1, f_2, f_3, \cdots$ ...
2
votes
1answer
57 views

Does there exist a $1$-dimensional Noetherian domain whose set of all prime ideals has cardinality $\aleph_1$ ?

Does there exist a $1$-dimensional Noetherian domain whose set of all prime ideals has cardinality $\aleph_1$ ? What I know is that for $1$-dimensional Noetherian domain $R$, $\operatorname{max} \...
1
vote
0answers
55 views

Can the Artin-Rees lemma be derived from Krull Intersection theorem?

The Krull Intersection theorem states that : For a finitely generated module $M$ over a Noetherian ring $R$ and any ideal $I$ of $R$, we have $I(\cap_{k=1}^{\infty}I^k M)=\cap_{k=1}^{\infty}I^k M$ . ...
1
vote
0answers
72 views

If R has ACCP does R/I have ACCP? (Counter Example)

Let $R$ be an Integral Domain satisfying Ascending Chain Condition for Principal Ideals. If $I$ is a proper ideal of $R$, does $R/I$ have ACCP? This statement is False but I do not have a counter ...
4
votes
0answers
147 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 ...
0
votes
0answers
44 views

Looking for easy example of 2-dimensional Noetherian domain which is not Cohen-Macaulay

I am looking for an easy example , with proof, of a 2-dimensional Noetherian domain which is not Cohen-Macaulay . I know and it is easy to prove that such an example doesn't exist in dimension 1. ...
2
votes
1answer
52 views

Noether polynomial ring over $2\Bbb Z$

Given a definition of $R=2\Bbb Z$ prove that ring $R[x]$ is not noether. I assume that the proof should be based on the fact that a ring is noether if and only if any ideal is finitely generated. ...
3
votes
1answer
125 views

Non-unital ring $(2\mathbb{Z})[X]$ is not Noetherian

Let $R = 2\mathbb{Z}$. Then $R[x]$ is not a noetherian ring. I do not understand why this is so, because Hilbert's basis theorem says: If R Noetherian ring, then R[X] a is Noetherian ring (from wiki)....
1
vote
0answers
26 views

Noetherianity of $\mathbb{R}^+$

I have some confusion about some properties of noetherian posets. I am defining a poset $X$ to be $\textbf{noetherian}$ if every ideal of $X$ Is finitely generated. I am trying to show that this is ...
5
votes
1answer
94 views

Integral domain with Noetherian spectrum and algebraically closed fraction field

If $R$ is an integral domain satisfying acc on radical ideals (i.e. Noetherian spectrum) and if the fraction field of $R$ is algebraically closed, then is $R$ a field ? If $R$ is normal (integrally ...
8
votes
0answers
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 ...
0
votes
0answers
42 views

A question on the permutation of regular sequence of length two in commutative Noetherian ring

Let $R$ be a commutative ring. If $(x,y)$ is a regular sequence in $R$ then I can show that the image of $x$ in $R/yR$ is not a zero divisor and $x S=S$ where $S=ann_R(y)$. Hence if $R$ is Noetherian ...
0
votes
1answer
61 views

On permutation of regular sequence of length two in commutative Noetherian ring

If $(x,y)$ is a regular sequence in a commutative Noetherian ring , then is it true that $(y,x)$ is also a regular sequence ? Let $R$ be a commutative ring. If $(x,y)$ is a regular sequence in $R$, ...
0
votes
1answer
121 views

Commutative Noetherian prime ideals finitely generated

In Bosch's Algebra you're asked to prove that every commutative ring R is Noetherian iff every ideal is finitely generated I think I managed to prove the if part (I write it just to be more ...
1
vote
0answers
48 views

Associated primes to principal ideals in a Noetherian domain

Let $R$ be a Noetherian domain. Let $\mathcal S:=\{P \in\mathrm{Spec}(R) : P \in \mathrm{Ass}(R/xR)$, for some $ 0\ne x \in R\}$. Then how to show that $\cap_{P\in \mathcal S} R_P \subseteq R$ ?
1
vote
0answers
104 views

Checking if quotient ring is Noetherian or Artinian over a module

Given a quotient ring $k[t,w]/w^2$, and the lemma that if $A \subset B$ is a R-submodule then $B$ is Noetherian iff both $A$ and $B/A$ are Noetherian. Can we see if $k[t]$ is Noetherian over itself ...
-2
votes
1answer
259 views

Commutative Noetherian, local, reduced ring has only one minimal prime ideal?

Let $R$ be a commutative Noetherian, local, reduced (no non-zero nilpotent) ring; is it true that $R$ has only one minimal prime ideal? If $R$ is Noetherian and local, then I can show that $(0)$ is ...
3
votes
0answers
89 views

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=\...
1
vote
1answer
156 views

If M is a a left module over $M_n(D)$ where $D$ is a division ring, M is Noetherian iff Artinian

I was hoping for an elementary method of approaching this. My attempt: D is a division ring so Noetherian/ Artinian. Then $M_n(D)$ is Noetherian/ Artinian as a matrix ring over Noetherian/ Artinian ...
5
votes
1answer
398 views

Why noetherian ring satisfies the maximal condition?

"maximal condition" means if any non-empty collection of ideals in R has a maximal element (under set inclusion). And we define noetherian ring to be a ring such that any ascending chain of ideals is ...
3
votes
0answers
73 views

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 \...
1
vote
1answer
196 views

Finitely generated modules over noetherian rings

There is a question in here that has a proof which I can’t understand, (maybe they use higher level of algebraic definition and I’m new), could you please explain it? Let $R$ be a commutative ...
2
votes
3answers
118 views

Is $\mathbb{Z}[\frac{1}{2}]$ Noetherian?

I am a bit confused with the question wether $\mathbb{Z}[\frac{1}{2}]$ is a Noetherian ring or not. I would say that $\mathbb{Z}[\frac{1}{2}]$ is a Noetherian ring. Reason : $\mathbb{Z}[\frac{1}{2}...
2
votes
0answers
96 views

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 \...
1
vote
1answer
143 views

Noetherian local domain of dimension one

While revising for my exam in Commutative algebra, I came across the following statement that: Let $A$ be a Noetherian local domain of dimension one. Let $x,y \in A$ with $x \neq 0$ and $y \in \...
1
vote
1answer
59 views

Ascending chain of proper submodules in a module all whose proper submodules are Noetherian

Let $M$ be a module over a commutative ring $R$ such that every proper submodule of $M$ is Noetherian, then is it true that every ascending chain of proper submodules of $M$ terminate ?
3
votes
1answer
317 views

Prove that the field $k(x)$ of rational functions over $k$ in the variable $x$ is not a finitely generated $k$-algebra.

I am working through Chapter 15 of Dummit and Foote's Abstract Algebra text, and I am stumped on how to prove the following (Exercise 3): Prove that the field $k(x)$ of rational functions over $k$ in ...
4
votes
1answer
133 views

In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element? [closed]

In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element?
-2
votes
1answer
35 views

Do not understand a corollary of $N$,$N/M$ are Noetherian submodules of $M$ iff $M$ is noetherian,

Do not understand a corollary of: "$N$,$N/M$ are Noetherian submodules of $M$ iff $M$ is noetherian" My professor said as a corollary of the above statement that: "If $R$ is a left Noetherian ...
3
votes
1answer
177 views

A difficulty in solving a question in chapter 8 hungerford.

How can I prove that the ring of all $2*2$ matrices $$S=\begin{equation*} \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \qquad \end{equation*}$$ Such that $a$ is an integer and $b,c$ are ...
1
vote
0answers
71 views

Commutative rings over which every module $M$ satisfies $\mathrm{Ass}(R/\mathrm{Ann}\; M) \subseteq \mathrm{Ass}(M)$

Let $R$ be a commutative ring with unity. If $M$ is a Noetherian $R$-module, then we know that $\mathrm{Ass}(R/\mathrm{Ann}\; M) \subseteq \mathrm{Ass}(M)$. My question is: Can we characterize ...
1
vote
1answer
141 views

Artinian and Noetherian ring of matrices

I am trying to solve an exercise about Artinian an Noetherian rings of $2 \times 2$ matrices but I really can't get to a solution. The exercise is the following: Set $$ R = \left\{ \begin{pmatrix} ...
2
votes
1answer
44 views

Monomorphism of noetherian R-modules

Suppose $R$ is a noetherian ring. Then there exists a $R$-module monomorphism: $R/I \rightarrow R$ with $I$ being a prime ideal of $R$. I found a proof but I cannot really understand the intuition ...
1
vote
1answer
67 views

Noetherian module

Let's consider the ring $R = \begin{bmatrix}\Bbb{Q} & 0\\\Bbb{Q} & \Bbb{Z}\end{bmatrix} = \left\{\begin{bmatrix}q & 0\\p & z\end{bmatrix} {\Big|\,} q,p \in \Bbb{Q}, z \in \Bbb{Z}\right\...
4
votes
0answers
103 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 ...
3
votes
1answer
90 views

In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I),P=I:a$ for some $a\in R$.

Maybe this question is too obvious to everyone. In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I)$, then $P=I:a$ for some $a\in R$ where $\mathrm{Ass}(I)$ denotes associated primes ...
11
votes
0answers
246 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 ...
-1
votes
1answer
212 views

A ring is Noetherian if its prime ideals are finitely generated

I am working on Exercise 3.15 from Aluffi's book Algebra: Chapter 0. 3.15. Recall that a (commutative) ring $R$ is Noetherian if every ideal of $R$ is finitely generated. Assume the seemingly ...
1
vote
1answer
49 views

Show that, if $\dim_{R/m}(m/m^2) = 1$, then there are no ideals $I$ of $R$ such that $m^{k + 1} \subsetneq I \subsetneq m^k$ .

Let $R$ be a noetherian local ring and let $m$ be its maximal ideal. I have already proved that $m^{k}/m^{k+1}$ is an $R/m$-vector space and that if $a_1, \ldots, a_n \in m$ generate $m/m^2$ as a ...
1
vote
1answer
350 views

$R[X]$ noetherian with $R$ non noetherian [duplicate]

Let $R$ be a ring. If $R[X]$ is noetherian, is R necessarily noetherian ? I think that the answer is no, but could you show me the easiest example to understand ?
1
vote
1answer
81 views

Ideals and Tensor product by a finitely genrated module over Noetherian ring

I was trying to solve this problem: Let $R$ be a commutative Noetherian ring. Prove that a finitely generated $R$-module $M$ is flat if and only if $Tor_1(R/m,M)=0 $ for any maximal ideal $m$ of $R$. ...
1
vote
0answers
107 views

Example of noetherian ring that nilradical is prime and (0) is not primary

I have problems to give an example of noetherian ring $R$ whose nilradical is prime and $(0)$ is not primary. I can find a lot of examples when $(0)$ is not primary, for example $\mathbb{C}[x,y]/(x^...
2
votes
0answers
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 ...
4
votes
1answer
112 views

Are all connected and locally integral affine schemes globally integral?

In these notes, on p. 2 Section 4, Kedlaya claims an affine scheme is integral if and only if it is connected and every local ring is an integral domain. But elsewhere I have seen that this requires a ...
1
vote
0answers
75 views

Is $\mathbb R \otimes_{\mathbb Q} \mathbb R$ Noetherian ring?

Is $\mathbb R \otimes_{\mathbb Q} \mathbb R$ Noetherian ring ? I understand that $\mathbb R \otimes_{\mathbb Q} \mathbb R$ being an infinite dimensional vector space over $\mathbb Q$ is not ...
1
vote
0answers
37 views

Let $M$ be a Noetherian $A$-module. Show that $M[x]$ is a Noetherian $A[x]$-module. [duplicate]

Let $M$ be a Noetherian $A$-module. Show that $M[x]$ is a Noetherian $A[x]$-module. Comments: I was able to show that $M[x] \cong M\otimes_A A[x]$. It is possible to construct an isomorphism $\phi$ ...
3
votes
1answer
76 views

Can the infectiousness of being Noetherian / Artinian in exact sequences of modules be generalised to lattices (with extra structure)?

It is well known that for every exact sequence $$0 → M' → M → M'' → 0$$ of modules over some ring, $M$ is Noetherian / Artinian if and only if both $M'$ and $M''$ are. If the arrows in such a ...
-2
votes
1answer
302 views

Direct sum of injective modules

Let $M$, $(E_i)_{i \in I}$ be $A$-modules. 1) There exists a canonical injective morphism $\varphi :\bigoplus_{i \in I}$ Hom$(M,E_i) \to $ Hom$(M, \bigoplus_{i \in I}E_i)$ 2) If $M$ is finitely ...