1
vote
0answers
22 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
24 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 ...
3
votes
1answer
46 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
62 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
60 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
110 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 ...
-1
votes
0answers
81 views

Is $\operatorname{depth}(I,M)=\operatorname{depth}(S^{-1} I,S^{-1} M)?$ [closed]

Let $A$ be a ring, $I$ an ideal of $A$, and $M$ a finitely generated $A$-module such that $M\neq IM$. Show that there is a maximal ideal $\mathfrak m$ of $A$ such that ...
1
vote
2answers
40 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
56 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
43 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
30 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? If this is not true, what can i add to ...
1
vote
1answer
42 views

exact sequence problem

Let $B_1 \stackrel{f}{\to} B \stackrel{g}{\to} B_2 \to 0$ be an exact sequence. For any module $M$ the sequence $$0 \to \mbox{Hom}(B_2,M) \stackrel{g*}{\to} \mbox{Hom}(B,M) \stackrel{f*}{\to} ...
3
votes
1answer
46 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
49 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
52 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, ...
6
votes
2answers
58 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
55 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
69 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
58 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.
3
votes
0answers
51 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
39 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
60 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
27 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
62 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
29 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
58 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
51 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
26 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
24 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 ...
1
vote
0answers
37 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 ...
0
votes
1answer
28 views

$\hom_R(A,B)$ is finitely generated if $R$ is noetherian [duplicate]

This is part of an exercise I'm doing, from Rotman Introduction to homological algebra. Let $R$ be a commutative ring, and let $A$ and $B$ be finitely generated $R$-modules. Then if $R$ is ...
0
votes
1answer
26 views

Chains in modules

I'm answering this question: 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 but $A$ is finitely generated whenever ...
0
votes
0answers
44 views

Minimal number of generators and associated prime ideals

Let $R$ be a commutative Noetherian local ring, $M$ a finitely generated $R$-module with minimal number of generators $\nu_{R}(M)=s$, and let $Q$ be the total quotient ring of $R$. Then, ...
3
votes
1answer
25 views

Does there always exist finitely presented submodules?

Suppose $M\neq 0$ is an $A$-module where $A$ is a commutative ring with $1$. Is it always possible to find a finitely presented submodule $N\neq 0$ of $M$? It is not interesting when the ring ...
3
votes
0answers
58 views

Length of a composition series of a module

If $A=\mathbb{C}[x,y]_{(x,y)}$, then what is the length of $A$-module $$A/(x^3-x^2y^2+y^{100},x^3-y^{999})\ ?$$ Any suggestion ?
1
vote
1answer
47 views

Symmetric powers of ideal quotients in a local ring.

Let $R$ be a local ring and $I \subset R$ any ideal. When is it the case that $(I \: \backslash I^2)^n = I^n \: \backslash I^{n+1}$? Put another way, when is the natural map $\text{Sym}^n(I/I^2) ...
6
votes
0answers
72 views

When does the tensor product consist of elementary tensors only?

The question is: Assume that $R$ is a (commutative) ring. Under what conditions on $R$-modules $M,N$ does the tensor product $M\otimes_RN$ consist of elementary tensors only? That is, every ...
2
votes
1answer
51 views

Criterion for locally free modules of rank $1$

Let $R$ be a commutative ring and let $M$ be a finitely generated $R$-module such that the $R_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$ is free of rank $1$ for every prime ideal $\mathfrak{p}$. Can we ...
1
vote
1answer
55 views

A statement equivalent to flatness

If $R$ is a ring with identity and $P$ is a flat right $R$-module, it is a fact that any $R$-homomorphism $f$ from a finitely presented right $R$-module $M$ to $P$ factors through a finitely generated ...
2
votes
1answer
108 views

How to construct a nonzero homomorphism from a module to a proper submodule?

Let $M$ be a finitely generated module over a commutative ring and $N$ be a non zero proper submodule of $M$. Then is it always possible to have a non zero homomorphism $f$ from $M$ to $N$?
3
votes
0answers
57 views

Equations in the semiring of f.g. modules

Let $R$ be a commutative ring. Then we may consider the semiring $G(R)$ of isomorphism classes of finitely generated $R$-modules with $+ = $ direct sum, $* = $ tensor product, $0 = $ zero module, $1 = ...
3
votes
0answers
51 views

When is $N\otimes_A B \to N$ an isomorphism?

Let $A, B$ be commutative (unital) rings and $f\colon A \to B$ an $A$-algebra. There then exists a canonical functor $f_*\colon \mathbf{Mod}_B \to \mathbf{Mod}_A$ such that, for every morphism of ...
2
votes
2answers
53 views

The Relationship Between Cohomological Dimension and Support

Let $ R $ be a commutative unital ring, $ I $ an ideal of $ R $, and $ M $ an $ R $-module. The cohomological dimension of $ M $ with respect to $ I $ is defined as $$ \operatorname{cd}(I,M) ...
1
vote
1answer
54 views

Choice of ideal so that localisation of a module must be free?

If $R$ is a Noetherian (commutative) ring and $M$ is a finitely generated $R$-module, then what choice of nonzero ideal $I$ of $R$ is such that $M_P$ is a free $R_P$-module for any prime $P$ in $R$ ...
1
vote
1answer
41 views

Inequality amongst projective dimensions!?

Assume $φ : R\to S$ is a ring homomorphism between commutative rings sending unity to unity. Taking any $S$-module $M$ as an $R$-module, is it true that always $$\operatorname{pd}_R (M)\le ...
0
votes
2answers
39 views

Zero direct summand

I want a suggestion for the following: Let $M$ be a finitely generated $R$-module, where $R$ is a commutative Noetherian local ring with $1_R$, and $N$ a direct summand of $M$ such that $N⊆mM$, ...
1
vote
1answer
72 views

Projective dimension zero or infinity

Let $R$ be a commutative Noetherian local ring (having unity) with the maximal ideal $m$, which consists only of zero-divisors. Then for any finitely generated $R$-module $M$, the projective ...
5
votes
2answers
300 views

Exercise 2.27 Atiyah-Macdonald, absolute flatness

A commutative ring $R$ is absolutely flat if every $R$-module is flat. Prove that the following are equivalent: 1) $R$ is absolutely flat 2) Every principal ideal of $R$ is idempotent 3) Every ...
1
vote
0answers
54 views

Resolution of module over polynomial ring

The problem is: Let $F$ be a field, and let $R = F[x_1, \ldots, x_r]$, the polynomial ring over $F$. Consider the $R$-module $M = R/(x_1, \ldots, x_r) \cong F$. Find a resolution of $M$ by free ...