The tag has no wiki summary.

learn more… | top users | synonyms

1
vote
1answer
36 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
46 views

$\varinjlim Hom_R(N,M_i) = Hom_R(N, \varinjlim M_i)$

please help to show that $\varinjlim Hom_R(N,M_i) = Hom_R(N, \varinjlim M_i)$ is true when $N$ is finitely generated and $R$ is a noetherian $R$-module. Do you think the noetherian condition is ...
2
votes
0answers
32 views

Example for $f_m(S/I)\ge depth S/I+2$ [migrated]

Let $ R $ be a commutative unital noetherian ring, $ I $ an ideal of $ R $, and $ M $ a finite $R$-module with $\dim M\gt 0$. $f_I(M) = \inf\ \{i : H_I^i(M)$ is not finitely generated$\}$ is defined ...
0
votes
0answers
8 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
35 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 ...
-5
votes
1answer
52 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?
0
votes
1answer
46 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?
0
votes
0answers
59 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 ...
0
votes
0answers
34 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
46 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
40 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 ...
0
votes
0answers
20 views

Conditions equivalent to Noetherianness

Let $R$ be a left Noetherian ring. We know that any direct sum of injective left $R$-modules is again injective. Since any injective module is quasi-injective, we infer that (1):"any direct sum ...
1
vote
1answer
46 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
29 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
66 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
61 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
45 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
31 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
75 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
34 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
votes
1answer
97 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
78 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
118 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
58 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
62 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
84 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 ...
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
82 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
0answers
30 views

Noetherian local ring with maximal ideal $M$

Let $R$ be a Noetherian local ring with maximal ideal $M$. If the ideal $M/M^2$ in $R/M^2$ is generated by $\{ a_1+M^2, \dots, a_n +M^2\}$, then the ideal $M$ is generated in $R$ by $\{ a_1, \dots , ...
0
votes
0answers
29 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 ...
0
votes
0answers
41 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 ...
0
votes
0answers
34 views

noetherian Ring

Let $I$ be an ideal in a Noetherian ring $R$. Prove that there exists a positive integer $N$ such that $(rad(I))N ⊂ I$. [Hint:Let $rad(I)=⟨g_1,...,g_k⟩$,and suppose $g_i^{n_i} ∈I$.Use ...
2
votes
1answer
41 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
72 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
27 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 ...
4
votes
2answers
106 views

In a Noetherian integral domain, a principal prime ideal can't have proper non-zero prime ideals

Let $R$ be an integral domain and Noetherian. Let $P \subset R$ be a non zero prime ideal. Prove that if $P$ is principal then there is no prime ideal $Q$ such that $0 \subsetneq Q \subsetneq P$. ...
1
vote
2answers
34 views

Noetherian property for exact sequence

Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then $M$ is Noetherian is equivalent to $M'$ and $M''$ are Noetherian. For the ...
2
votes
1answer
29 views

Why are the irreducible components of a subspace of Noetherian space just the intersection with the irreducible components?

Suppose you have a Noetherian topological space $X$, and its finitely many irreducible components are $X_1,\dots,X_n$. If $U\subseteq X$ is open, why are the irreducible components of $U$ precisely ...
1
vote
1answer
58 views

Maximal among some ideals is prime

I am reading a lemma on noetherian integral domains but I am stuck, I am bring it up here hoping for help. The original passage is in one big fat paragraph but I broke it down here for your easy ...
8
votes
3answers
245 views

Are Dummit and Foote making a mistake in proving Cohen's theorem?

Exercise 11 on page 669 (this is Chapter 15) wants to prove Cohen's theorem that if every prime ideal of a ring is f.g. then every ideal is f.g. that is the ring is noetherian. The highbrow (perhaps?) ...
0
votes
1answer
50 views

Dedekind domain necessary for equivalence of flatness and torsion-free

It is well-known that for finitely generated modules over a Dedekind domain, flatness and torsion-free are equivalent. Is this true for general Noetherian rings? If not, where is the dimension one ...
2
votes
1answer
24 views

Intersection of nonzero ideals in a right Noetherian domain is nonzero

I've been asked to show that in a right Noetherian domain, the intersection of nonzero right ideals is nonzero. A hint is given, saying that if not, then any nonzero right ideal contains a direct sum ...
2
votes
2answers
77 views

Show that the ring of continuous functions $f:\mathbb R\to\mathbb R$ is not Noetherian

Prove that the ring of continuous functions $f:\mathbb R\to\mathbb R$ is not Noetherian. I know that to be Noetherian, every ideal is generated by finitely many elements or equivalently R ...
0
votes
0answers
33 views

If $\chi:Q\to \mathbb{R}$ is discrete (i.e. $Im(\chi)\simeq \mathbb{Z}$) and $Q$ is polycyclic group then $\mathbb{Z}Q_{\chi}$ is noetherian?

Where $\chi:Q\to \mathbb{R}$ group homomorphism, $\mathbb{R}$ is $(\mathbb{R}, +)$ group and $Q_{\chi}=\{q \in Q | \chi(q)\geq 0\}$. I know that if $Q$ is abelian finitely generated group and $\chi: Q ...
3
votes
0answers
193 views

Hilbert's Basis Theorem - Clever Proof?

So I am studying commutative algebra at the moment and I have come across the proof of the Hilbert Basis Theorem (the proof I have is the same as the one in Reid's "Undergraduate Commutative ...
0
votes
1answer
67 views

Closedness and going up property

Let $f: A\rightarrow B$ be a homomorphism of commutative unital rings. The problem is to show that if $f$ has going-up property and $\text{Spec }B$ is Noetherian topological space then $f^*: ...
3
votes
1answer
136 views

Is $\Bbb{R}[X,Y]/(X^2+Y^2)$ a UFD or Noetherian?

Hello everyone I would like to know if $R$:= $\Bbb{R}[X,Y]/(X^2+Y^2)$ is UFD or Noetherian. I'm not really confortable in seeing how $\Bbb{R}[X,Y]/(X^2+Y^2)$ looks like. From what i've ...
0
votes
1answer
35 views

Proving a submodule is Noetherian

I'm doing some independent study with a professor in Ring/Field theory (I'm an undergraduate) and I have been having a hell of a time wrapping my head around problems involving Noetherian rings and ...
1
vote
1answer
51 views

Noetherian toplogical space exercise

Let $X$ be a noetherian topological space. Prove the following statements: (a) If $F \subset X$ is closed, then there exist $n \in \mathbb N$ and irreducible closed subsets $F_1,\ldots,F_n \subset ...
1
vote
1answer
38 views

Equivalent conditions for a topological space to be Noetherian

Problem Show that the following statements are equivalent: (a) $X$ is a noetherian topological space (b) Every non-empty family of closed subsets of $X$ has a minimal element. (c) If $$U_1 \subset ...