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1answer
63 views

Question on a property of $\mathrm{Ass}(M)$ for modules over noetherian rings

I got stuck reading a proof of the following lemma (Lemma 0.19 in this file): Lemma Suppose that $M$ is a module over a commutative noetherian ring $R$ and let $m\neq 0 \in M$. Let $S$ be a ...
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0answers
46 views

Is the ring of germs of $C^\infty$ functions at $0$ Noetherian?

I'm considering the property of the ring $R:=C^\infty(\mathbb R)/I$, where $I$ is the ideal of all smooth functions that vanish at a neighborhood of $0$. I find that $R$ is a local ring of which the ...
0
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1answer
55 views

If every maximal ideal is finitely generated is the ring Noetherian? [duplicate]

$R$ is a commutative ring with $1$. Suppose every maximal ideal is finitely generated. Is this ring Noetherian? Equivalently, is every prime ideal finitely generated?
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1answer
48 views

Subring of a commutative Noetherian ring

We know that it's possible subring of the commutative Noetherian ring become not Noetherian (for example: Subring of a Noetherian ring need not be Noetherian?). But if $S$ be a subring of ...
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1answer
35 views

Show that if $M$ is a semisimple artinian module then $M$ is finitely generated.

The exercise is as follows: Show that for a semisimple module $M$ over any ring, the following conditions are equivalent: $(1)$ $M$ is finitely generated; $(2)$ $M$ is Noetherian; $(3)$ $M$ is ...
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1answer
56 views

If R is a Noetherian ring, is it true that there exists an integer N such that every ideal of R is generated by at most N elements? [duplicate]

I think, as R is an ideal of itself, and R is Noetherian thus R is finite generated, assume it is generated by N elements, then every ideal is generated by at most N elements. But my conclusion is ...
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0answers
48 views

Fact on Iwasawa module

The following fact falls under the category of Iwasawa modules. Let $M$ be a torsion free finitely generated module over the non commutative noetherian ring $\Bbb{Z}_p[[G]]$, (where $G$ is a p ...
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1answer
26 views

Conceptual problem in Noetherian rings and Zorn Lemma

I've got a problem, and it's that I can't see the difference of one of the definitions of Noetherian ring, and supposing the zorn lemma for that ring. I mean, if I use the definition of the maximal ...
0
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1answer
49 views

Primary Ideal and Associated Primes [duplicate]

I'm trying to understand the proof of the following statement: If $R$ is Noetherian, then an ideal $Q$ is $P$-primary for a prime $P$ $\Leftrightarrow$ $Ass(R/Q)=\lbrace P \rbrace$. I can show the ...
2
votes
1answer
27 views

Non-commutative noetherian integral domain-Ore condition

Let $R$ be a non-commutative integral domain with unity which is also a right Noetherian ring. By integral domain I mean that the product of nonzero elements is always nonzero. I am trying to show ...
5
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1answer
51 views

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 ...
0
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1answer
23 views

Non-Noetherian ring with Noetherian quotient

Let $R$ be a commutative ring with $1 \ne 0$. I'm trying to prove that if $R$ contains an ideal $I$ that is not finitely generated, then $R$ contains a proper ideal $J$ such that $R/J$ is ...
0
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0answers
47 views

$\mathrm{Ext}^i(-,A/\mathfrak{m})$ in $(A,\mathfrak{m})$ noetherian regular local ring

Dealing with $\mathrm{Ext}^i(\mathcal{F},k(x))$ on a smooth variety over a field $k$, with $\mathcal{F}$ coherent and $k(x)$ skyscraper sheaf of a closed point I foundin a proof that for $i=2,3$ (and ...
0
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1answer
45 views

Prove or Disprove: Finitely generated Artinian module is Noetherian.

I think it is true and I am trying to prove it. I am considering reducing the case to Artinian rings. Say $M$ is finitely generated Artinian $R$-module. Then $R/Ann(M)$ is an Artinian $R$-module. Thus ...
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1answer
36 views

Nontrivial map between finite modules over local noetherian ring

I have just read that over a noetherian local ring $(A,\mathfrak{m})$, there is always a nontrivial map between two finitely generated modules such that their support is exatly $\mathfrak{m}$. The ...
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0answers
43 views

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 ...
3
votes
1answer
74 views

A stronger definition of locally free modules

Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, Section 4.6, Exercise 4.12 (a) tells us if $M$ is a finitely presented $R$-module, then $M$ is projective if and only if $M$ is ...
2
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1answer
81 views

Is the intersection of two Noetherian rings Noetherian?

Is the intersection of two Noetherian rings also Noetherian? If yes, could you please give me the idea of proof. If not, give me an counterexample.
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1answer
40 views

Ring contained in a R-module finitely generated

Let $R$ be a Noetherian domain with quotient field $K$ and let $b_1,\ldots,b_n\in K$. Suppose that $R'$ is a integral domain, $R\subseteq R'$ and $$R'\subseteq \sum_j Rb_j.$$ Remark: It is ...
4
votes
1answer
75 views

$\varinjlim\operatorname{Hom}_R(N,M_i) = \operatorname{Hom}_R(N, \varinjlim M_i)$

Show that $\varinjlim \operatorname{Hom}_R(N,M_i) = \operatorname{Hom}_R(N, \varinjlim M_i)$ is true when $N$ is finitely generated and $R$ is noetherian. Do you think the noetherian condition is ...
0
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1answer
19 views

When is a D-module holonomic?

Suppose that I have a small family of partial differential equations such that $y(x)$ must be a solution to $$ P_i(x,D) y(x)=0,\ \ \ \ \ \ i=1,...,m $$ for all $i=1,...,m$, where I am using ...
0
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0answers
38 views

Show if $R$ is Noetherian, then $R_S$ is Noetherian [duplicate]

Show if $R$ is Noetherian, then $R_S$ is Noetherian. here is what I have read from somewhere else. Suppose $R$ is Noetherian and $J$ is an ideal $R_S$. Then $J=IR_S$ for some ideal $I$ of $R$. Since ...
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1answer
59 views

If M is an artinian module then the m-adic completion of M is complete? [closed]

Let $(R,m)$ be a noetherian local ring. If $M$ is an artinian module then the $m$-adic completion of $M$ is complete. Is this correct?
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1answer
52 views

$ M_i$ is injective, for all i, then $\bigoplus M_i$ is injective? [closed]

for every nonempty family of $\{M_i \}_i \in I$ of arbitrary $R$-module, if $\bigoplus M_i$ is injective for all $i$, $M_i$ is injective is the converse true? why?
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0answers
60 views

If $\Omega:M\to M$ is onto and $M$ is finitely generated then it is injective. [duplicate]

Suppose that $R$ is a commutative ring and that $M$ is an finitely generated $R$ module. If $\Omega : M \to M$ is an $R$-homomorphism that is onto, show that $\Omega$ is one one. If $M$ is a ...
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0answers
40 views

$IM=mM$. can we say that $I$ is a reduction ideal of $m$

Definition. Let $R$ be a Noetherian ring􀀀, $I$ a proper ideal,􀀀 and $M$ a finite $R$-module. An ideal $J\subset I$ is called a reduction ideal of $I$ with respect to $M$ if $JI^nM = I^{n+1}􀀀M$ for ...
2
votes
0answers
55 views

Algebraic characterization of commutative rings with Krull dimension=1,2, or 3

A commutative ring $R$ (with $1$) is $0$-dimensional if and only if $R/\sqrt 0$ is von Neumann regular. Besides this result, there is a wealth of information about zero-dimensional rings. I could not ...
0
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1answer
46 views

Is it possible that a finitely generated ring has an ideal that is not finitely generated

Sorry if this is duplicated. I couldn't find an exact answer of my question. One definition of Noetherian ring is: A ring $R$ is Noetherian if all its ideals are finitely generated. I know there are ...
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1answer
47 views

Question related to commutative ring being Noetherian

Let $A$ be a commutative ring with $1$, and $A = (f_1, \ldots, f_n)$. I want to prove the following: If $A$ is a Noetherian ring, then so is $A_{f_i}$ (which is the ring $A$ localized at $f_i \in ...
1
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1answer
30 views

Finitely Generated Sets

Let $F$ be a field and suppose $I$ is an ideal of $F[x_1, \ldots , x_n]$ generated by a (possibly infinite) set $S$ of polynomials. Prove that a finite subset of polynomials in $S$ is sufficient to ...
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2answers
71 views

Total ring of fractions of a Noetherian reduced ring is artinian

I'm doing the preparation to an exam, and I'm stuck in the following: If $R$ is a Noetherian ring with zero nilradical ($N(R) = 0$), and $S$ is the set of regular elements of $R$ ($r \in S$ if $rs ...
4
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1answer
67 views

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 ...
1
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1answer
50 views

Artinian module and short exact sequence

Consider the short exact sequence of modules $$0\rightarrow M\rightarrow N\rightarrow K\rightarrow0$$ Let $K$ be an Artinian (Noetherian) module, can we get $N$ is an Artinian (Noetherian) module?
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0answers
32 views

$I$ is a $J$-primary ideal of $R$ iff $I/L$ is a $J/L$-primary ideal of $R/L$

Let $R$ be a commutative unitary ring and $I$, $J$, $L$ be ideals of $R$ with $L$ proper, $L \subseteq I$ and $L \subseteq J$. A homework question asks to prove that if $R$ is noetherian then $I$ is ...
4
votes
1answer
105 views

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
38 views

what inequalities can one have between $depth\ R$ and $depth\ M$? when $depth\ R \geq depth\ M$

Let $(R,m)$ be a commutative Noetherian local ring which is not CM. Let $M$ be a finite $R$-module. what inequalities can one have between $depth\ R$ and $depth\ M$? Obviously there are ...
2
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1answer
103 views

Show that the ring $R$ is Noetherian [closed]

Let $$R = \{ \frac{f(z)}{g(z)}: f, g \in \mathbb{C}[z], g(z) \neq 0 \text{ for } |z| = 1 \}.$$ Prove that $R$ is a Noetherian ring. Note : $\mathbb C$ is the set of the complex numbers, $z$ is a ...
3
votes
1answer
82 views

$R$ noetherian, $I$ injective $R$-module $\Rightarrow$ $S^{-1}I$ is injective over $S^{-1}R$

I am trying to prove that if $R$ is a noetherian ring, $S$ a multiplicative part and $I$ an injective $R$-module, then $S^{-1}I$ is an injective $S^{-1}R$-module. So far I thought: I reduce to check ...
0
votes
1answer
121 views

Prove that some canonical homomorphism is injective.

Let $A \not= \{0 \}$ be a Noetherian commutative ring and let $M$ be an $A$-module. Prove that the canonical homomorphism $$M \to \bigoplus_{P \in \text{Ass}(M)} M_p$$ is injective. My question is, ...
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1answer
74 views

judge if nilradical equals jacobson radical

judge if nilradical equals jacobson radical 1)a noetherian ring that is not a artin ring. 2)a local integral domain that is not a field. 3)a integral domain with only finite number of ...
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1answer
68 views

Example of non noetherian ring and noetherian $\Bbb Z$-module

a non Noetherian ring that is a Noetherian $\Bbb Z$-module a Noetherian ring that is a non Noetherian $\Bbb Z$-module I have no idea in 1, and I'm not sure if $\mathbf{Q}$ is right for 2? ...
0
votes
1answer
95 views

Why field of fractions of $k[x_1,x_2,…]$ is Noetherian? [closed]

the classical counterexample of a subring of a noetherian rings that is not noetherian is $k[x_1,x_2,...]$, which is not noetherian, but the field of fractions of $k[x_1,x_2,...]$ is, can anyone ...
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0answers
26 views

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 ...
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vote
2answers
90 views

On a proof that left artinian implies left noetherian

Questions: [Refer to below] Could one elaborate on $\rm\color{#c00}{(a)}$, $\rm\color{#c00}{(b)}$ and $\rm\color{#c00}{(c)}$ ? My thoughts : $\rm\color{#c00}{(a)}$ For $r+J\in R/J$ and ...
1
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1answer
41 views

Noetherian ring and radical

Show that in a Noetherian ring $I$ and $J$ have the same radical if and only if there is a positive integer $N$ such that $I^N \subset J$ and $J^N \subset I$. [Hint: for the ``if'' direction, use a ...
1
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1answer
50 views

Noetherian ring not descending chain condition

Analogous to the ascending chain condition we can define a descending chain condition: if $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$ is a descending chain of ideals then there exists a ...
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1answer
56 views

Every ideal contains a power of it's radical in a Noetherian ring.

Let $I$ be an ideal in a Noetherian ring $R$. Prove that there exists a positive integer $N$ such that $\text{rad}(I)^N ⊂ I$. [Hint:Let $\text{rad}(I)=⟨g_1,\ldots,g_k⟩$,and suppose $g_i^{n_i} \in ...
2
votes
1answer
44 views

Exact sequence and Noetherian modules

Let $R$ be a ring, $X,Y,Z$ and $T$ four $R-$ modules such that there exists a short exact sequence $$0 \rightarrow X \xrightarrow{f_1} Y \xrightarrow{f_2} Z \xrightarrow{f_3} T \rightarrow 0 $$ Prove ...
2
votes
3answers
127 views

Commutative artinian ring is noetherian

Suppose R is a commutative Artinian ring then R is Noetherian. I am aware of the proof which uses the idea of filtration. But I would like to prove this fact without that idea but haven't got far ...
0
votes
1answer
30 views

Module and Noetherian/Artinian Rings

I am trying to prove that: Every finitely generated $F$-module $M$ is both Noetherian and Artinian where $F$ is a field. For this I am looking at the submodules of $F$ and saying that they are in ...