Questions tagged [noetherian]

For questions on Noetherian rings, Noetherian modules and related notions.

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257 views

$R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ Noetherian?

Let $R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective , then is $R$ a Noetherian ring ?
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2answers
136 views

Artinian - Noetherian rings and modules suggest study guide

What text or any document that has gathered this part of Algebra theory. Thanks. Pd: I seek on variety's book of commutative algebra but the subject is partially dealt
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1answer
51 views

Question about proof of Krull principal ideal theorem

How can we explain the following step in the proof of Krull principal ideal theorem: $l\{ ((z):x^n)/(z) \}$ or $l\{ ((x^n):z)/(x^n) \}$ is finite? $l(M)$ - length of module.
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1answer
140 views

About a short proof of Krull principal ideal theorem

How from this theorem I can get a proof of Krull principal ideal theorem? I understand that w.l.g. we can prove it for a Noetherian local ring. But why we can consider that $(x)$ is $M$-primary?
2
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1answer
115 views

If $A$ is a commutative, unitary ring and $I$ an ideal of $A$ such that $I$ and $A/I$ are Noetherian rings, then $A$ is Noetherian?

If $A$ is a commutative, unitary ring and $I$ an ideal of $A$ such that $I$ and $A/I$ are Noetherian rings, then $A$ is Noetherian ? I know just that if $A$ is Noetherian then $A/I$ is Noetherian....
1
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1answer
385 views

Noether's normalization lemma in practice (example)

I would like to know how to use the Noether's normalization lemma in practice. Noether's normalization lemma Let $k$ an infinite field, and $k[a_1,\dots ,a_n]$ be a finite $k$-algebra. There ...
2
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2answers
1k views

A Bézout UFD is a PID. [duplicate]

Let $R$ be an integral domain and a Noetherian U.F.D. with the following property: for each couple $a,b\in R$ that are not both $0$, and that have no common prime divisor, there are elements $u,v\in ...
3
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1answer
578 views

Noetherian local ring is Artinian iff maximal ideal is nilpotent

I am browsing through some old lecture notes, and I am trying to prove the following: Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Show that the following are equivalent: ...
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0answers
273 views

Proof of commutative Artinian ring is Noetherian

I think that I have a proof, but it seems much simpler than all proofs that I can find on the internet. Hence I suppose that there must be a mistake in my proof. The commutative ring $R$ is ...
0
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1answer
220 views

Let $R$ be a ring and $I$ be a finitely generated nilpotent ideal. If $R/I$ is noetherian (resp. Artinian) then $R$ is so.

Let $R$ be a ring and $I$ be a finitely generated nilpotent ideal. If $R/I$ is noetherian (resp. Artinian) then $R$ is so. In between step is $I^j/I^{j+1}$ is noetherian (artinian) $\forall j$. I am ...
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1answer
214 views

Example of a non noetherian module $M$ s.t. $M/IM$ is noetherian

What is an example of a non noetherian module $M$ s.t. $M/IM$ is noetherian? What is an example of a non artinian module $M$ s.t. $M/IM$ is artinian? What I was thinking that $k[x_1,...,x_n,...]$ is ...
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1answer
144 views

Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring. [duplicate]

Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring. I was trying to show that $R/I$ is a submodule of $M^n$ if $M=\langle f_1,...,f_n\rangle$. Now $M$ is an ...
2
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1answer
79 views

Noetherian Lemma Contradiction

There is a noetherian version of Higman's Lemma which says If $X$ is a noetherian poset, then $X^*$ is noetherian. Now I was thinking, given $X=\{1\}$ we get $X^*=\{1,11,111,\dots\}$ by the ...
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1answer
161 views

Can every element of a Noetherian ring be expressed as a product of irreducible elements?

Let $X$ be a Noetherian ring. Question: Can every element of $X$ be expressed as product of irreducible elements? I'm not assuming that $X$ is a Noetherian integral domain, only $X$ is Notherian. ...
3
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1answer
2k views

Noetherian rings have only finitely many minimal prime ideals. [duplicate]

We say that $p$ is minimal prime if It does not contain any other prime. Assume that $A$ is Noetherian ring Question: $A$ has only finitely many minimal primes. any suggestions please
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2answers
68 views

A module over a noetherian ring and its dual

Let $R$ be a noetherian ring, $M$ an $R$-module and $M^*$ be its dual. Is it true that if $M$ is noetherian then $f(x) = 0$ for all $f \in M^*$ implies $x =0$? How about the other way round? Ok, so ...
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2answers
109 views

Noetherian submodules and isomorphisms

Suppose $M$ is an $A$-Module, and $N$ is a submodule of $M$. Let $f:N\to M$ be an $A$-module epimorphism. How could I show that if $N$ is noetherian, then $f$ is an isomorphism? Thanks
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3answers
341 views

the ideal contains some power of the unique prime ideal containing it

Let $R$ be a Noetherian ring, and $I$ an ideal such that there exists a unique prime ideal $\mathfrak{p}$ containing $I$. Show that $I$ contains some power of $\mathfrak{p}.$ In this question, I ...
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2answers
77 views

Subgroups of $\mathbb Z^5$

Is there a nice way to see that any subgroup of $\mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z$ can have at most 5 generators? I know that $\mathbb Z$ is Noetherian, so ...
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1answer
142 views

Chains of ideals contained in maximal ideal in non-Noetherian commutative ring

Given a maximal ideal $M$ in a non-Noetherian commutative ring $R$, I'm trying to determine whether or not there can exist infinite strictly ascending chains of ideals of $R$ contained in $M$. I know ...
5
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1answer
622 views

A Noetherian integral domain is a UFD iff $(f):(g)$ is principal

Let $R$ be a Noetherian integral domain. For $f, g \in R$, define $(f):(g)=\{h \in R \mid hg \in (f) \}$. Sow that $R$ is a UFD if and only if $(f):(g)$ is principal for all $f,g \in R$. It is easy ...
2
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1answer
611 views

Polynomial ring in infinitely many variables over a noetherian ring is coherent

If $R$ is noetherian, show that the polynomial ring of infinite variables $R[x_1,x_2,...]$ is coherent, i.e. every finitely generated ideal is finitely presented. I don't really know how to get ...
5
votes
1answer
134 views

$R\subset A\subset R[X]$, $A$ is Noetherian. Is $R$ Noetherian?

Let $R\subset A\subset R[X]$ be commutative rings and suppose $A$ is Noetherian. Is $R$ Noetherian? I guess the answer is yes. Can we say from this relation that $A[X]=R[X]$? If yes, then by Hilbert ...
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0answers
30 views

The ring of polynomials in $X, Y$ all of whose partial derivatives with respect to $X$ vanish for $Y=0$ is Noetherian? [duplicate]

The ring of polynomials in $X,Y$ all of whose partial derivatives with respect to $X$ vanish for $Y=0$ is Noetherian ?
3
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3answers
239 views

$R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ , $R/I$ is finite ; then is $R$ a PID or at least Noetherian?

Let $R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ of $R$ , $R/I$ is finite; then is $R$ a PID or at least Noetherian ? I can only prove that $R$ must be an ...
2
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1answer
462 views

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

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 $...
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0answers
54 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 If $M$ isn't ...
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1answer
235 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 \mathrm{Hom}...
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0answers
31 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?
3
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1answer
97 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
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1answer
331 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 M_1\...
4
votes
1answer
729 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 $M$ ...
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1answer
47 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 ...
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2answers
198 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 $M'_i = M_i \cap M'$ and $ ...
4
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1answer
153 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 extension ...
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1answer
64 views

Converse Noetherian Relation [duplicate]

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
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2answers
132 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 T : ...
1
vote
1answer
45 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} ~\mbox{...
3
votes
1answer
47 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 $R[[...
2
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0answers
84 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 $\...
2
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1answer
131 views

ACC on chains of finitely generated submodules

Noetherian rings are those having ascending chain condition on ideals. There is also literature concerning ACC on n-generated (i.e., generated by n elements) ideals; see e.g, Commutative Rings with ...
3
votes
1answer
162 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 $M\otimes_R M$....
5
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2answers
211 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 ...
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2answers
157 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 $R$: ...
1
vote
1answer
70 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 see ...
2
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0answers
427 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 ...
2
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0answers
405 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
551 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 ...
3
votes
2answers
371 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 ...
2
votes
1answer
85 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?