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3
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0answers
18 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
64 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
26 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
27 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
36 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
31 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 ...
1
vote
1answer
35 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 ...
1
vote
3answers
56 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
20 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 ...
3
votes
1answer
66 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 $Q$ prime ideal such that $0 \subsetneq Q \subsetneq P$. ...
0
votes
2answers
30 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 ...
0
votes
0answers
17 views

Krull dimension of the completion

Let $(R,\mathfrak{m})$ be a commutative noetherian local ring of Krull dimension $n$ with maximal ideal $\mathfrak{m}$. Setting $\hat{R}$ as the $\mathfrak{m}$-adic completion of $R$, what is the ...
1
vote
1answer
24 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 ...
0
votes
1answer
55 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
216 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
45 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
21 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
63 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
29 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 ...
1
vote
0answers
110 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
46 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
127 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
31 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
50 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
32 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 ...
0
votes
1answer
48 views

Integral ring extensions and finitely generated as a module

Let $A \subset B \subset C $ be rings. Suppose that $A$ is Noetherian and $C$ is finitely generated as an $A$-algebra. I want to show that $C$ is finitely generated as a $B$-module $ \iff $ $C$ is ...
0
votes
1answer
37 views

Is this ring Noetherian

If $R=\{\frac m n\in{\mathbb{Q}}: \mbox{7 does not divide $n$}\}$. Is $R$ Noetherian? What i think is the ideals of $R$ are finitely generated. But how do i prove this, i have no idea. Can anyone ...
2
votes
2answers
82 views

Minimal injective resolution of a module

Let $R$ be a commutative Noetherian ring and $M$ an $R$-module. Let $0\rightarrow M \rightarrow E^{\bullet}$ be a minimal injective resolution of $M$ and $0\rightarrow M\rightarrow I^{\bullet}$ be an ...
-1
votes
1answer
34 views

Does extension of scalars take Noetherian modules to Noetherian modules?

Suppose $A$ is a commutative ring with unity, and $B$ is an $A$-algebra. If $M$ is a Noetherian $A$-module, is $M \otimes_A B$ Noetherian as a $B$-module? Note that there are no finiteness conditions ...
0
votes
0answers
30 views

Number of prime ideals that contain a non zero ideal

In the proof of proposition (12.3) of Neukirchs Algebraic Number Theory, we use the fact that for a one-dimensional noetherian integral domain, there are only finitely many prime/maximal ideals that ...
1
vote
0answers
35 views

Almost Dedekind (not Noetherian) [duplicate]

A domain $R$ is a Dedekind domain iff it is integrally closed, Noetherian with Krull dimension at most one. I think they put Noetherian in the definition not only to make proofs easier but also ...
1
vote
1answer
77 views

Noetherian group rings

I'm asking for an example of a finitely generated amenable group $G$ and a field $K$, such that the group ring $K[G]$ is not Noetherian. Is it also possible to find a finitely generated amenable ...
0
votes
1answer
38 views

Characterization of right noetherian rings

Here's a quick question on noetherian rings. I know that for a ring $R$, the following are equivalent. $R$ is left noetherian Every finitely generated left $R$-module is noetherian Every submodule ...
3
votes
1answer
88 views

A ring with finite dimensional vector space structure is noetherian?

Let $K$ be a field and $R$ a ring with finite dimensional vector space structure over $K$. Is $R$ necessarily a Noetherian ring? If $K \subset R$, then any ideal in $R$ is also a subspace and, since ...
2
votes
1answer
53 views

A two sided ideal of a Noetherian ring

$R$ is a left Noetherian ring with a minimal left ideal. Consider the set of minimal left ideals of $R$ ordered by inclusion. Then there is a maximal element $\mathfrak b= \bigoplus_{i\in I} \mathfrak ...
4
votes
1answer
164 views

Direct product of finitely many Noetherian non-unital rings is Noetherian

Let $A_1, A_2,...,A_n$ be Noetherian rings (not necessarily unital). Is the direct product $A:=A_1×A_2×⋯×A_n$ necessarily a Noetherian ring? If $A_1, A_2,...,A_n$ are unital, then one can prove ...
1
vote
0answers
90 views

What is $\operatorname{Ass}\operatorname{Ext}^i(M,N)$?

This is exercise 1.2.27 of Bruns-Herzog: Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $N$ an arbitrary $R$-module. Deduce that $\operatorname{Ass}(\operatorname{Hom}_R(M,N)) = ...
4
votes
1answer
83 views

Help with a problem from Christian Peskine's book about Artinian rings

I am stuck with this problem from the book of Complex Projective Geometry. Let $A$ be a Noetherian ring. Assume that if $a \in A$ is neither invertible nor nilpotent, then there exist $b \in A$ such ...
-1
votes
1answer
30 views

Noether normalisation $A=\mathbb{C}[x,y]/(f)$ where $f=(x-a)y^2-(x-b)$ find a transcendence element

Noether normalisation $A=\mathbb{C}[x,y]/(f)$ where $f=(x-a)y^2-(x-a)$ $a , b \in \mathbb{C}$ find $z \in A$. transcendence over $\mathbb{C}$ such that $A$ is integral over $\mathbb{C}[z]$ any ...
2
votes
0answers
50 views

A question about the proof of Hilbert's Basis Theorem

I have a question regarding the proof of Hilbert's Basis Theorem. Say $I=(f_1,f_2,f_3,\dots)$ is an ideal in $A[x]$, where A is a Noetherian ring. Say we take the leading coefficients $a_i$ of all ...
-3
votes
1answer
66 views

One dimensional noetherian domain

Let $(R,m)$ be a one-dimensional Noetherian domain. Is $R$ a regular or a topical ring like Gorenstein or other kinds?
2
votes
1answer
106 views

Exercise from Kaplansky's Commutative Rings and Eakin-Nagata Theorem

Exercise 15 of section 2-1 of Kaplansky's Commutative Rings is to show that if $T$ is a Noetherian ring and is finitely generated module over a subring $R$ of $T$, then $R$ is Noetherian. Kaplansky ...
2
votes
1answer
47 views

For some finitely many nonzero prime ideals, the contraction and extension of their product is zero

I was reading P.M. Eakin's thesis paper, The converse to a well known theorem on Noetherian Rings. The following is taken from Theorem 2, page 281 of that paper, and that's where I'm stuck. Let ...
2
votes
1answer
176 views

Is this module noetherian?

Let $k$ be a field, and let $A$ be a commutative $k$-algebra. Assume that $A$ is a noetherian ring, and let $I\subseteq A$ be a proper ideal. Consider the ideal $I\otimes_k A \subseteq A\otimes_k ...
1
vote
1answer
64 views

Conditions for a quotient module to be Noetherian

I'm solving this problem from "Introduction to Commutative Algebra" of Atiyah and Macdonald. Here is the problem: Let $M$ be an $A$-module and let $N_1, N_2$ be submodules of $M$. If $M/N_1, ...
0
votes
1answer
48 views

Depth zero module and $R$-regular element

Let $(R,m)$ be a commutative Noetherian local ring with $\operatorname{depth}(R)>0$ and $M$ be a finitely generated $R$-module with $\operatorname{depth}(M)=0$. Then can we take an $R$-regular ...
-1
votes
1answer
68 views

Is $R[X]/(f)$ Cohen-Macaulay if $R$ is so?

Let $R$ be a commutative (Noetherian) Cohen-Macaulay ring, and $f \in R[X]$ be monic. I guess that $R[X]/(f)$ is also Cohen-Macaulay. Is my hunch valid? Thanks for any help.
2
votes
0answers
109 views

Flat base change preserves the property of being non-degenerate

We say a homomorphism $f:A\rightarrow B$ of noetherian rings is non-degenerate if the induced map $f^*:{\rm Spec}(B) \rightarrow {\rm Spec}(A)$ maps ${\rm Ass}(B)$ into ${\rm Ass}(A)$. Let $f:A ...
0
votes
1answer
60 views

Ring of linear transformations modulo finite rank transformations [closed]

Let $ K $ be a field and $ V $ be a vector space of countable dimension (infinite) over $ K $, and let $ L = L (V) $ be the vector space of $ K $-linear transformations on $ V $. Let $ I $ be the ...
2
votes
1answer
106 views

Extension of Noetherian rings [closed]

Let $A \subset B$ rings. If $B$ is noetherian then $A$ is noetherian. Is false or true?