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0
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1answer
8 views

If $M/N$ and $N$ are noetherian $R$-modules then so is $M$

Let $M$ be an $R$-module. I want to show only using the definition of noetherian that if $N$ is a noetherian submodule of $M$ such that $M/N$ is noetherian, then $M$ is noetherian. I know that if $M_1 ...
0
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0answers
24 views

Every non-Noetherian module has a submodule maximal with respect to being not finitely generated. [duplicate]

Let $M$ be a module. Show that if $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated whenever $N<A\leq M$. The question is related to this one, which gives ...
1
vote
1answer
49 views

Tensor product of Hom-module and another ring

Let $A$ be a local noetherian ring, $B$ and $C$ are finitely generated $A$-algebras and $M$ is a finitely generated $B$-module. Is the natural morphism $\mathrm{Hom}_B(M,B) \otimes_A C \to ...
0
votes
0answers
18 views

Example where the closure of a point does not contain a closed point in a Nonnoetherian Scheme

Let $x\in X$ be a point of a Noetherian scheme. Then the closure of $x$ contains a closed point. Is there an example in non-Noetherian schemes where this property does not hold?
2
votes
1answer
48 views

Endomorphism of a finitely generated module and finite length of $\operatorname{Coker}f$

Let $M$ be a finitely generated module over a noetherian ring $R$ and suppose that $\dim R≤1$. And let be a one to one homomorphism $f:M\to M$. It is true that $\operatorname{Coker}f$ has finite ...
0
votes
1answer
54 views

Finite length module over Noetherian ring

Let $M$ be a finitely generated module over a noetherian ring $R$ and suppose that $\dim R\leq 1$. Then $M$ has finite length. A preliminary lemma: There exists an chain $\{0\}=M_0\subsetneq ...
1
vote
1answer
37 views

Unusual proof that $M$ is Noetherian if and only if $N$ and $M/N$ are Noetherian

Let $M$ be an $R$-module for a ring $R$. Let $N$ be a submodule of $M$. I read that one can prove that $M$ is Noetherian if and only if $N$ and $M/N$ are Noetherian using these two results: 1) If $M$ ...
0
votes
0answers
21 views

Can one add reasonable assumptions that we have $depth\ R \geq depth\ M$ for every $R$-module $M$?

Let $(R,m)$ be a commutative Noetherian local ring which is not CM. Let $M$ be a finite $R$-module. Here, Hanno shows that one can have any inequality between $depth\ R$ and $depth\ M.$ Still, the ...
0
votes
2answers
56 views

If $0 \rightarrow M' \rightarrow M \xrightarrow[]{f} M'' \rightarrow 0$ is exact then $M$ is Noetherian iff $M',M''$ are. [duplicate]

If $0 \rightarrow M' \rightarrow M \xrightarrow[]{f} M'' \rightarrow 0$ is exact then $M$ is Noetherian iff $M',M''$ are. For an infinite chain $M_i$ in $M$ we have chains $M'_i = M_i \cap M'$ ...
4
votes
1answer
81 views

Ore extensions of noetherian $k$-algebras are again noetherian (proof explanation)

Let $k$ be a field. All occuring $k$-algebras are required to be associative and unital. By noetherian I always mean left noetherian. In a lecture I’m currently taking the notion of an Ore ...
-1
votes
1answer
20 views

Converse Noetherian Relation

I have looked around and cannot find the answer so I ask here. It is well known that if $R$ is noetherian then $R[X]$ is too, but what about the converse? If $R[X]$ is noetherian can we say $R$ is? My ...
4
votes
2answers
39 views

$A$ noetherian, $A$-endomorphism not injective for all invariant submodules is nilpotent

Let $A$ be a noetherian ring, $M$ a finitely generated $A$-module, $T: M \to M$ an endomorphism. Assume that for all $T$-invariant proper submodules $N$ of $M$, the induced endomorphism $\overline ...
1
vote
1answer
29 views

Uncountably many left ideals?

Let $R$ be a following subring of $M_2(\mathbb{C}):$ \begin{equation*} R = \left\{ \begin{bmatrix} a & r \\ 0 & s \end{bmatrix} ~:~ a\in \mathbb{Q} ...
3
votes
1answer
24 views

Extension of idempotent ideals

Let $R$ be a Noetherian commutative ring with $1$. If $R[[x]]$ denotes the ring of formal power series over $R$ and $I$ is an idempotent ideal of $R$ I want to know whether the extension of $I$ in ...
2
votes
0answers
69 views

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 ...
3
votes
1answer
38 views

Verify proof that if $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian

I have to prove that If $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian We let $S$ be a non-finitely generated submodule of $M\otimes_R ...
3
votes
2answers
52 views

Proper ideal $I \implies \exists $ prime ideals $P_i$ such that $P_1 \cdots P_n \subset I$.

Let the below ideals be in a commutative Noetherian ring $R$. Corollary 22. (3) There are prime ideals $P_1, \dots, P_n$ (not necc. distinct) $\supset I$ such that $P_1\cdots P_n \subset I$. (Out ...
1
vote
2answers
46 views

If $R$ is a noetherian ring, then minimal primes of $R[x]$ are exactly the ideals $P[x]$ of $R[x]$ where $P$ is a minimal prime of $R$

Show that if $R$ is a noetherian ring, then minimal primes of $R[x]$ are exactly the ideals $P[x]$ of $R[x]$ where $P$ is a minimal prime of $R$. Definition of a minimal prime ideal of a ring ...
1
vote
1answer
22 views

Normal Noetherian rings of dimension at least 1

We want to pick $I_g$ to be "maximal" but what is the partial ordering to which it is maximal? For two $f,g \in A' - A$, I don't see how $I_f$ and $I_g$ would be related via subsets. How can we ...
0
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0answers
28 views

If $R$ is a Noetherian ring, why is $R[[x]]$ also Noetherian? [duplicate]

Prove that if $R$ is noetherian, then the formal power series ring $R[[x]]$ is noetherian. An element of the formal power series ring doesn’t have a highest degree term, so you have to use something ...
2
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0answers
72 views

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 ...
1
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0answers
28 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 ...
1
vote
1answer
38 views

matrix ring Noetherian, Artinian, semisimple?

Let $k$ be a field and $\Lambda=\begin{bmatrix} k & 0 \\ k^2 & k[x]/(x^2) \end{bmatrix}$. This ring is an algebra over $k$. (a) What is $\dim_k \Lambda$? (b) Is $\Lambda$ a left ...
2
votes
2answers
65 views

Is this ring R both noetherian and artinian?

Let the $R=\Big\{\begin{pmatrix}a & 0 \\ b & c \end{pmatrix} : a,b,c \in \mathbb{R}\Big\}$ be a ring. Is it: 1) Artinian? 2) Noetherian? If a ring $R$ is noetherian (artinian), then every ...
0
votes
1answer
77 views

The ring of real convergent sequences is Noetherian [closed]

Is it true that the ring of real convergent sequences is a Noetherian ring?
2
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1answer
41 views

Noetherian local domain $A$ with a prime $P$ so that $\operatorname{ht}P+\dim A/P<\dim A$

Is there a noetherian local domain $A$ with a prime $P$ so that $\operatorname{ht}P+\dim A/P<\dim A$? This is a follow up question to: Does codimension behave weirdly even in local rings?
2
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0answers
38 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 $C$ ...
0
votes
0answers
45 views

Can anybody show how primary decomposition of ideals is useful. Any application?

So far I have been able to only see some standard counter examples or examples of Lasker-Noether decomposition theorem ( primary decomposition ). Why should someone care to learn it? ( I am being ...
2
votes
1answer
98 views

$x$ is integral over $R$ if and only if for every minimal prime $\mathfrak q$ of $S$, $x$ is integral over the residue domains

I've made some progress on the following problem: Let $R$ be a Noetherian ring, $R \subseteq S$ an extension of rings, and $x \in S$. Show that $x$ is integral over $R$ if for every minimal prime ...
2
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0answers
34 views

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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0answers
76 views

Total quotient ring and localization

Let $R$ be a Noetherian normal ring, and let $\mathfrak p$ be a prime ideal of $R$, and let $Q(R)$ be the total quotient ring of $R$. For the tensor product $Q(R) \otimes_R R_{\mathfrak p}$, ...
2
votes
1answer
65 views

Goldbach property for polynomials with base ring a PID with infinitely many maximal ideals

Is it true that if $R$ is a PID with infinitely many maximal ideals, then every element of $R[x]$ of degree $n\ge1$ is a sum of two irreducible polynomials in $R[x]$? Even if this is not true in ...
1
vote
1answer
66 views

Noetherian criterion and principal ideals.

I struggle proving the following statement : Let $A$ be a ring. If for any $a\in A\setminus\{0\}$, $A/aA$ is noetherian, then $A$ is a noetherian ring. I first noticed that because $0\to aA\to ...
1
vote
3answers
213 views

Examples of Noetherian Domains of Dimension One [closed]

I am writing an assignment on Noetherian domains of dimension 1. I have been researching online left and right looking for concrete examples of this domain, but could not find any. I would therefore ...
2
votes
2answers
48 views

Localization of a one-dimensional noetherian integral domain

Let $o$ be a one-dimensional noetherian integral domain, and $a$ be a non-zero ideal of $o$. Suppose $p$ be a prime ideal containing $a$. Show that $o/o \cap ao_p= o_p/ao_p$. (Here $o_p$ is the ...
0
votes
1answer
42 views

Direct sum of noetherian rings

If $R_{1}, R_{2}, ..., R_{n}$ is a family of noetherian rings, is $$R = R_{1} \oplus R_{2} \oplus ... \oplus R_{n}$$ a noetherian ring? From a theorem I have an equivalence that says if a ring $R$ is ...
0
votes
2answers
47 views

What does this field notation mean?

I am reading a problem and its solution posted online here that says: Problem 3: Give an example of a Noetherian ring R that contains a subring that is not Noetherian. And then, Solution: ...
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0answers
15 views

$A \oplus M$ is Noetherian iff $A$ is Noetherian and $M$ is a f.g. $A$-module

Let $A$ be a commutative ring and $M$ an $A$-module. Define a multiplication on the additive group $B:=A \oplus M$ by $$(a,x)(b,y) := (ab,bx+ay) \ \ \ (a,b \in A,\ x,y \in M).$$ An easy computation ...
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vote
1answer
28 views

Zerodivisors and grade of an ideal

Assume $R$ is a commutative Noetherian ring and $M$ is a nonzero finitely generated $R$-module and $I\subseteq J\subseteq R$ are ideals so that $JM\neq M$. I'm trying to get a grasp of the notion of ...
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2answers
100 views

A field is always noetherian

I learned from this a recent posting here that a field is always notherian since a field has only $\{0\}$ and the field itself as ideals. However in a 3-year old posting here, it says $\mathbb Q$ as ...
3
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0answers
63 views

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.
0
votes
1answer
59 views

Prime ideals $\mathfrak{p} \supset \mathfrak{a}$ are finite in one-dimensional Noetherian domain

Let $A$ be a one-dimensional Noetherian domain. Let $\mathfrak{a} \neq 0$ be an ideal of $A$. How do I prove that prime ideals $\mathfrak{p} \supset \mathfrak{a}$ are finite? Thanks.
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vote
1answer
60 views

Is Noetherian required to conclude that radical of an ideal is a finite intersection of prime ideals?

Consider the collection $C$ of all ideals $I$ of $R$ such that $\sqrt{I}$ is not a finite intersection of prime ideals. Now using Zorn's lemma $C$ has a maximal element and that won't be prime. ...
4
votes
1answer
139 views

Knapp (Basic Algebra) Prop 8.52, error?

The above proposition says: Let $R$ be a Noetherian ring and let $I$ and $P$ be ideals of $R$ where $P$ is a prime ideal. If $IP=I$, then $I=0$. I feel that this is false. After passing to ...
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0answers
60 views

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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0answers
67 views

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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0answers
138 views

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 ...
3
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0answers
57 views

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 ...
2
votes
1answer
51 views

noetherian quotients and a finitely generated ideal

Suppose $A$ is a ring and $a$ and $b$ are ideals such that one of them is finitely generated. Also suppose $A/a$ and $A/b$ are noetherian. Apparently this implies $A/ab$ is notherian. I was ...
1
vote
1answer
79 views

A noetherian topological space is compact

Have to prove that every noetherian topological space $(X,\mathcal{T})$ is also compact. I don't understand the proof I found: Let $\{\mathcal{U}_\alpha\}_{\alpha\in\Lambda}$ be an open cover of ...