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3
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0answers
28 views

Litterature on noncommutative ring

I am looking for books or notes about non commutative rings with with a maximum of data exposed without the help of modules (because I have many references which deal with the subject but modules are ...
2
votes
0answers
49 views

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 ...
1
vote
0answers
49 views

Projectivity of a (prime) ideal in a noetherian integral domain

Assume $R$ is a noetherian integral domain (and assume $R \neq k[x_1,\ldots,x_n]$), $I$ is a non-zero ideal of $R$ ($I$ is finitely generated, since $R$ is noetherian), and $I$ is not necessarily ...
5
votes
1answer
51 views

Is the ring of Laurent polynomials in $n$ noncommuting variables Noetherian?

Suppose we have a Noetherian ring $R$. Is it true that the ring of Laurent polynomials $R\langle x_1,\,x_1^{-1},\ldots,\,x_n,\,x_n^{-1}\rangle$ in $n$ noncommuting variables is also Noetherian? If so, ...
0
votes
1answer
19 views

Example of noetherian module non-uniquely expressible as sum of indecomposable submodules

If $M$ is a noetherian module then it can be written as a finite sum of indecomposable submodules of $M$. The same can be concluded if we assume instead $M$ to be artinian. If we ask for both $M$ to ...
6
votes
3answers
85 views

A commutative noetherian ring in which each ideal $I$ is principal and $I^2=I$ must be a finite product of fields

PROBLEM A commutative noetherian ring $R$ in which each ideal $I$ is principal and $I^2=I$ must be a finite product of fields. I am lost with the condition $I^2=I$ and the desired result "a ...
0
votes
2answers
98 views

Noetherian ring under some conditions has at least two minimal prime ideals

Question is : Suppose $R$ is a noetherian ring. Prove that $R$ is either an integral domain, has nonzero nilpotent elements, or has at least two minimal prime ideals. [Use the previous exercise.] ...
0
votes
1answer
36 views

Noetherian ring of symmetric polynomials

I wish to show that $k[x_1,x_2,..,x_n]^{\Sigma_n}$, which is the ring of all symmetric polynomials, is Noetherian. I thought the easiest way to do this would be to show that every ideal is ...
1
vote
1answer
39 views

how to tell if a ring is noetherian

In general how would i tell if the rings $\mathbb{Z}[\sqrt d]$ and $\mathbb{Z}[\frac{x}{y}]$ are noetherian? I know that the ring $\mathbb{Z}$ is noetherian as all ideals are contained in a finite ...
1
vote
1answer
69 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 ...
2
votes
1answer
65 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
votes
1answer
70 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?
2
votes
1answer
56 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 ...
1
vote
1answer
39 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 ...
3
votes
1answer
69 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 ...
0
votes
0answers
51 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 ...
1
vote
1answer
29 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 ...
2
votes
1answer
38 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
votes
1answer
58 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
votes
1answer
25 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
votes
0answers
48 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
votes
1answer
59 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 ...
1
vote
1answer
37 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 ...
9
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0answers
59 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
78 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
votes
1answer
84 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.
1
vote
1answer
41 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
83 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
votes
1answer
21 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
votes
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 ...
0
votes
0answers
46 views

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

Definition. Let $R$ be a Noetherian ringτ€€€, $I$ a proper ideal,τ€€€ and $M$ a finite $R$-module. An ideal $J\subset I$ is called a reduction ideal of $I$ with respect to $M$ if $JI^nM = I^{n+1}τ€€€M$ for ...
2
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0answers
58 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
votes
1answer
49 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 ...
1
vote
1answer
49 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
vote
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 ...
1
vote
2answers
90 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
votes
1answer
75 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
vote
1answer
62 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?
0
votes
0answers
35 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
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1answer
157 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 ...
1
vote
1answer
39 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 ...
3
votes
1answer
107 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
89 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
136 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, ...
1
vote
1answer
84 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 ...
1
vote
1answer
75 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? ...
3
votes
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 ...
1
vote
2answers
99 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
vote
1answer
44 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
vote
1answer
52 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 ...