0
votes
2answers
70 views

Does every free $R$-module have a maximal proper submodule?

Let $R$ be a commutative ring with $1$. We know that every finitely generated $R$-module has a maximal proper submodule. Is it true for any free $R$-module? In particular, can we do the following: ...
1
vote
1answer
46 views

Example of a module such that every proper submodule is finitely generated but the module is not.

Let $R$ be a ring with 1 and $M$ an $R$-module. What is an example such that $M$ is infinitely generated but every proper submodule is finitely generated.
0
votes
0answers
39 views

Bass numbers of minimax modules are finite?

Let $R$ be a commutative Noetherian ring, and $M$ be a minimax $R$-module. Are the Bass numbers of $M$ are finite? (An $R$-module $M$ is called minimax, if there is a finite submodule $N$ of $M$, such ...
0
votes
1answer
69 views

Modules with finite support in $\mathrm{Max}(R)$

Let $R$ be a commutative Noetherian ring, and $M$ be an $R$-module. Is the following statement true? If $\mathrm{Supp}_R(M)$ (support of $M$) is a finite subset of $\mathrm{Max}(R)$ (the set of all ...
3
votes
1answer
38 views

Injective hull of $\mathbb{ Z}_n$ [duplicate]

What is the injective hull of $\mathbb Z_n$? I know that in case $n=p$ is prime, the injective hull would be isomorphic to $\mathbb Z_{p^∞}$, but in general case, I have no idea. Can anyone be of ...
1
vote
2answers
30 views

Cardinality of minimal generating set of a module is constant

Let $R$ be a commutative ring with unity and $M$ be a finitely presented module over $R$. Then how to show that for any minimal generating set $S$, the cardinality is same? Edit: Thanks to Martin to ...
2
votes
2answers
22 views

A direct limit concerning some homomorphisms

In an algebra text there is the following argument I am stuck in the last part of which: "Let $f:B→C$ be an epimorphism in the category of $R$-modules, and $D=∑_{n=1}^∞c_nR$ be a countably generated ...
1
vote
1answer
62 views

Are $\mathbb{Q}$ or $\mathbb{Z}$ flat modules?

I have the following three question about flat modules. Why is not $\mathbb{Z}$ a flat $\mathbb{Z}$-module. Why is $\mathbb{Q}$ a flat module $\mathbb{Z}$-module. I need an example of a module which ...
2
votes
0answers
69 views

If $R$ is a domain and $M$ a finitely generated $R$-module, is it true that $\bigcap_{f\in M^{*}}\ker{f}=\operatorname{Tor}M$?

Let $R$ be a domain and $M$ a finitely generated $R$-module. Let $M^{*}=\hom_{R}(M,R)$. Let Tor$M$ be the torsion submodule of $M$. It it true that $$\displaystyle\bigcap_{f\in ...
2
votes
3answers
125 views

Isomorphic quotient of a module over Noetherian commutative ring

I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve ...
0
votes
0answers
14 views

$f_I(R)$ and $ f_I(M)$

The question is special case of this question. So, for background, see it. Is there a relation between $f_I(R)$ and $ f_I(M)$ ? Any hint, reference will be helpful. Thanks.
1
vote
0answers
56 views

Tensor product of free modules over free algebra

Suppose $M$ and $N$ are modules over a (commutative, unital) ring $S$. Let $R$ be a subring of $S$ such that $S,M$ and $N$ are all free, finitely generated modules over $R$. Question: Under what ...
0
votes
1answer
62 views

Rank of a module when the base ring is not a domain

Suppose $R$ is a commutative Noetherian local ring with $1$, which is not a domain. Let $M$ be a (non-free) finite $R$-module. What is meant by rank of $M$ in this case?
1
vote
0answers
61 views

$\operatorname{supp}(M) \subseteq \operatorname{supp}(N) \iff f_I(M)\subseteq f_I(N) $?

Let $ R $ be a commutative unital ring, $ I $ an ideal of $ R $, and $ M $ an $ R $-module. It has proven (here) that if $\operatorname{supp}(M) \subseteq \operatorname{supp}(N)$ then ...
1
vote
1answer
52 views

A Direct Sum of Members of a Certain Class of Modules

Let $S$ be a class of $R$-modules and let an $R$-module $M$ be countably generated. Suppose that, for every direct summand $K$ of $M$, each element of $K$ belongs to a direct summand of $K$ that is ...
1
vote
0answers
60 views

Surjectivity implies injectivity of finitely generated modules, localization?

The following problem is canonical: Suppose $A$ is a commutative unitary ring, and $M$ is a finitely generated module over $A$. If an endomorphism $f\colon M\to M$ is surjective, then it's also ...
-1
votes
1answer
37 views

Submodules and quotients of free modules over Noetherian local rings

Let $R$ be a Noetherian local commutative ring, $F$ a finitely generated free $R$-module and $A,B$ some arbitrary $R$-modules. Consider a short exact sequence $0 \to A \to F \to B \to 0$. In [Bruns, ...
1
vote
1answer
83 views

Example of strict inclusion for the localization of associated primes

Let $A$ be a commutative ring and $M$ an $A$-module. It is well known that $$\operatorname{Ass} M\cap\operatorname{Spec}S^{-1}A\subset\operatorname{Ass}S^{-1}M,$$ and that equality holds if $A$ (or ...
0
votes
0answers
41 views

Exercise 7.10 Atiyah, $M[x] $ is a noetherian $A[x] $-module [duplicate]

The exercise is: Let $M$ be a noetherian $A$-module. Then $M[x] $ is a noetherian $A[x] $ module. The action of $A[x] $ on $M[x] $ is the obvious one. In a previous exercise it was shown that ...
0
votes
2answers
42 views

$R^{(I)} \cong K \oplus H$ where $R^{(I)}$ is free but $K$ is not free

Let $R$ be a commutative ring with unit. Is there an example of a direct sum of $R$-modules $$R^{(I)} \cong K \oplus H$$ where $R^{(I)}$ is free but $K$ is not free ? Clearly $R$ can't be a PID.
1
vote
0answers
83 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)) = ...
2
votes
0answers
71 views

Automorphism of certain f.g. free modules

This is a quick question from Frohlich and Taylor's Algebraic Number Theory, II.4, p 94. Let $R$ be a Dedekind domain with quotient field $K$, $\mathfrak p$ is a non-zero prime ideal of $R$ and ...
1
vote
1answer
33 views

Basis for the completion of a free module

This (or similar) question might have been asked before- apologies for any duplication. I've got a Dedekind domain $R$, a non-zero prime ideal $P$ of $R$ and the completion $\widehat{R}$ of $R$ wrt ...
2
votes
1answer
74 views

Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?

The question is contained in title, I'm working with $A$-modules $M$ and $N$. I feel like Yoneda's lemma is what I'm looking for but it applies to functors into the category of sets, whereas ...
3
votes
1answer
75 views

How to prove this innocent looking isomorphism

I've got a Dedekind domain $R$ with quotient field $K$, a non-zero prime ideal $P$ of $R$. I form the completion $\widehat{K}$ of $K$ wrt the valuation $v_P$ associated to $P$. Let $\widehat{R}$ be ...
2
votes
0answers
70 views

Direct product of direct sum of a flat module

I have a problem concerning flat modules: Let $M$ be an $R$-module such that the direct product $M^A$ is flat for all sets $A$. I want to prove that $(M^{(B)})^A$ is also flat for any sets $A$ ...
4
votes
2answers
125 views

Proving that tensoring a projective module with a flat module gives a projective module?

If $P$ is a projective module and $M$ is a flat module, both over some commutative ring $R$, then how do you prove that $M\otimes_R P$ is flat? I tried using the fact that $P$ is the direct ...
2
votes
2answers
48 views

Relation between $\operatorname{Coker}(f)$ and $\operatorname{Coker}(f \otimes 1_P)$

Let $M,N,P$ be $R$-modules ($R$ commutative ring with $1$) and let $f:M\to N$ be a $R$-module homormorphism. Let tensor the homomorphism to get $ f \otimes 1_P : M \otimes P \to N \otimes P $. I ...
2
votes
1answer
63 views

Prove that if the induced homomorphism $M/\mathfrak aM \to N/\mathfrak aN$ is surjective, then $f$ it's surjective.

This problem it's from Atiyah and Macdonald, Chapter 2. Let $A$ be a commutative ring with $1 \ne 0$ and let $\mathfrak a$ be an ideal of $A$ contained in the Jacobson radical. Let $M$ be an ...
2
votes
1answer
59 views

Possible examples where the Zero Divisor Conjecture does not hold

Given a ring $R$ with a nonzero zero divisor $x$, it is easy to show that if $M$ is a nonzero $R$-module, then there exists $y\in R-\{0\}$ such that $ym=0$ for some $m\in M-\{0\}$. I was ...
0
votes
1answer
45 views

tensor product of R-algebra and f.g module [closed]

$R$ is a commutative noetherian ring. If $S$ is an $R$-algebra, and $M$ a finitely generated $R$-module, is $M\otimes_RS$ finitely generated $S$-module? I only need a hint. Thanks!
2
votes
1answer
52 views

Maximal linearly independent sets in a f.g. module

Suppose $M$ is a finitely generated module over a commutative unital ring $R$. Is it true that every maximal linearly independent set in $M$ has the same size? What is the most general condition ...
0
votes
1answer
53 views

Characterization of the kernel and cokernel of the natural homomorphism between a module and its double dual. [closed]

Let $R$ be a Noetherian ring and $M$ a finite $R$-module. Suppose $$ G \overset{\varphi}{\rightarrow} F \to M \to 0$$ is exact where $F,G$ are finite free modules. Suppose ...
0
votes
1answer
60 views

A finite module over a Noetherian ring is torsionless if and only if it is a submodule of a finite free module

Let $R$ be a Noetherian ring, and $M$ a finite $R$-module. Then $M$ is torsionless if and only if it is a submodule of a finite free module, where torsionless is defined here. (Bruns and Herzog, ...
5
votes
2answers
82 views

If $\{M_i\}_{i \in I}$ is a family of $R$-modules free, then the product $\prod_{i \in I}M_i$ is free?

If $\{M_i\}_{i \in I}$ is a family of free $R$-modules, then $\bigoplus_{i \in I}M_i$ is free. Is this true for the product $\prod_{i \in I}M_i$ too?
1
vote
1answer
56 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, ...
5
votes
1answer
77 views

Modules over local artinian rings

What is known about the structure of finitely generated modules over local artinian commutative rings $R$? Any information is appreciated. Let us denote by $\mathfrak{m}$ the maximal ideal and by $k$ ...
-2
votes
1answer
62 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
52 views

$M_{\mathfrak{p}} \otimes_{R_{\mathfrak{p}}} N_{\mathfrak{p}} = 0$ implies $M_{\mathfrak{p}} = 0$ or $N_{\mathfrak{p}} = 0$ [duplicate]

Studying commutative algebra I've found this statement: If $M$ and $N$ are finitely generated $R$-modules, with $R$ a commutative ring, and $\mathfrak{p} \subset R $ is a prime ideal, then ...
1
vote
1answer
42 views

Associated primes and Noetherian condition

Let $A$ be a commutative ring, $M$ an $A$-module, and $N\subset M$ a submodule. Consider the following two sets: $$\Omega:=\{\mathfrak{p}\in\operatorname{Spec}A \ | \ \mathfrak{p}=(N:m) \ \mbox{for ...
0
votes
2answers
50 views

“Finitely generated as an $R$-module”

Please could somebody explain to me what it means for something to be finitely generated as an $R$-module? I can't seem to find a definition anywhere! Thanks!
2
votes
1answer
74 views

Jacobson radical of a ring finitely generated over $\mathbb Z$

If a commutative ring $R$ with $1$ is finitely generated over $\mathbb Z$ could one deduce that the Jacobson radical of $R$ is nilpotent? I am aware of the well-known fact that when $R$ is ...
1
vote
0answers
29 views

An inverse limit of a certain inverse system

Let $∆$ be a directed set and $(N_i,f_{ji})_{i∈∆}$ be an inverse system of $R$-modules. Fix $α \in∆$ and consider $(M_i,g_{ji})_{i\in∆}$ as follows: $M_i=N_i$ for $i≥α$, $M_i=0$ for $i<α$, and ...
2
votes
1answer
75 views

Integral domain with a finitely generated non-zero injective module is a field

Suppose that $R$ is a integral domain. Suppose that there exists a non-zero finitely generated injective module $M$. How can I prove that $R$ is field?
0
votes
0answers
32 views

Suggest a good book or reference on graded modules over polynomial rings

I am looking for reference books or papers on graded modules over the polynomial ring $k[x_0, \ldots, x_n]$. Any good commutative algebra text like Eisenbud's Commutative Algebra already contains a ...
3
votes
1answer
62 views

Direct product of Cohen-Macaulay rings/Eisenbud, Exercise 18.6

Somehow I believe (or doubt (!)) that direct product of two Cohen-Macaulay (C-M) rings may not be C-M. Can anybody give me an example verifying this? I would be grateful to him/her.
0
votes
0answers
57 views

Sufficient conditions for quotient ring to be Cohen-Macaulay

We know that every Noetherian integral domain with (Krull) dimension $1$ is Cohen-Macaulay (CM). In a commutative algebra text the author have presented the following problem: "Let $(R,m)$ be a CM ...
2
votes
1answer
27 views

Irredundant intersection of submodules

Let $A$ be a commutative ring, $M$ an $A$-module, and $N_\alpha\subset M$ a family of submodules. Consider the intersection $$\bigcap_\alpha N_\alpha.$$ We say that the intersection is irredundant if ...
0
votes
1answer
25 views

$Hom_{A-Alg}(Sym(M),B)\cong Hom_{A-mod}(M,B)$

Let A be an ring. M be an A-module. Let $Sym (M)$ be the Symmetric algebra of M over A. Let B be an A-algebra. Why is $Hom_{A-Alg}(Sym(M),B)\cong Hom_{A-mod}(M,B)$ One way is clear - If we have a ...
0
votes
0answers
41 views

Direct limits commute with $\mathrm{Tor}$ functor

How one could prove that direct limits commute with the functor $\mathrm{Tor}$? Of course, I know that $\mathrm{Tor}$ with its first $0$ index is the same as tensor product which does commute ...