For questions about modules over rings, concerning either their properties in general or regarding specific cases.

learn more… | top users | synonyms

-2
votes
1answer
84 views

A question about modules. [closed]

My Algebra book says that a module is the following map $$R\to End(M)$$ This implies that if $a,b\in M, a\neq b$, then $ra\neq rb$ for some $r\in R$. I dont see why that is.
1
vote
1answer
16 views

ideal,ring,flat module,modules over R

Is there a characterization of modules (AND equivalent characterizations of rings R) over integral domains R with the property that each left ideal in R is flat?When all left ideals are ...
1
vote
0answers
31 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 ...
0
votes
0answers
16 views

A condition of equivalence of flatness and projectiveness

This is a problem in "Foundations of Module and Ring Theory" of Wisbauer: " Let $R$ be a subring of the ring $S$ containing the unit of $S$. Show that a flat $R$-module $N$ is projective if and only ...
1
vote
0answers
49 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 ...
15
votes
5answers
1k views

Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$ Can you tell ...
5
votes
3answers
152 views

When is an element of a free module over a principal ideal domain contained in a basis?

I'm trying to show the following: Let $R$ be a principal ideal domain and let $M$ be free $R$-module of rank $n$. Let $Y=\{y_1,\ldots,y_n\}$ be a basis of $M$ and $x\in M$ with ...
1
vote
1answer
21 views

Simple $M_n(D)$-module with $D$ a division ring

Define $D$ to be a division algebra over a field $k$ and $R=M_n(D)$ the $n\times n$ matrix ring over $D$. A simple $R$-module $M$ is the quotient of $R$. I can write $R=\bigoplus_j I_j$ where $I_j$ is ...
0
votes
1answer
49 views

Confused about proof in Basic abstract algebra by Bhattacharya

On page 264 , 2nd edition. Theorem 5.1 It says Let M be a free $R$-module with "a basis" $\{e_1,\dots,e_n\}$ Then $M$ is $R$-isomorphic to $R^n$. Above he is defining the standard basis as the ...
1
vote
0answers
35 views

Simple Cogenerator for the category of left $R$-modules

I want to prove that if the category of left $R$-modules has a simple cogenerator $M$, then $R$ is a simple ring. (Here, $R$ is an arbitrary ring with $1$.) My try is to take an ideal $I$ of $R$ ...
0
votes
1answer
74 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 ...
2
votes
0answers
26 views

How Can I prove that this module is isomorphic to the generators with relations…

Let $G$ be a finite group of order $n$. Let $R=\mathbb{Z}G$ and $N=\sum_{g\in G} g$, observe that $gN=Ng=N$, $N^2=nN$. Let $r\in \mathbb{Z}$ be prime to $n$ and let $P_r$ be the ideal of $R$ generated ...
1
vote
1answer
28 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, ...
8
votes
1answer
354 views

Associate prime ideals and exact sequences of $R$-modules

Let $R$ be a commutative unitary ring, consider the exact sequence of $R$ - modules $$ 0\rightarrow N\rightarrow M\rightarrow L\rightarrow0. $$ We know that $$\text{Ass}(M)\subseteq \text{Ass}(N)\cup ...
3
votes
0answers
28 views

Tensor product of $A_n$ modules/ localisation at ring of differentials

I'm working through Coutinho's "A Primer of Algebraic D-Modules" and I've gotten stuck on the following question: Let $p \in K[x_1, \ldots ,x_n]$ be non-zero, and let $A_n$ be the Weyl Algebra. Show ...
2
votes
1answer
33 views

Submodules of semi-simple modules

Let $R$ be a ring (with unity, not necessarily commutative) and let $P$ be an irreducible $R$-module. Let $$M=\bigoplus_{i=1}^r P$$ be a direct sum of $r$ copies of $P$, for some $r\geq 1$. Then, $M$ ...
1
vote
1answer
55 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} ...
0
votes
0answers
39 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
74 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
1answer
42 views

Kahler forms of a smooth affine algebra vanish eventually?

If $k$ is a Noetherian ring, then do the Kahler forms of a smooth affine $k$-algebra of dimension $d$ vanish above $d$? I mean is: $\Omega^{d+1}_{A|k}\cong 0$?
0
votes
0answers
20 views

Classification of separable algebras over a commutative ring

A separable algebra over a field can be classified as a finite product of matrix algebras over division algebras whose centers are finite dimensional separable field extensions of the field. (See ...
1
vote
0answers
139 views

Associated prime ideals of Hom (Bruns and Herzog, exercise 1.2.27)

Let $R$ be a Noetherian ring and $M,N$ finitely generated modules. I want to show that $$\mathrm{Ass}_R(\mathrm{Hom}_R(M,N)) = \mathrm{Ass}_R(N) \cap \mathrm{Supp}(M).$$ I don't understand what ...
1
vote
0answers
21 views

Decomposition of finately generated graded modules over PID

I found this decomposition theorem used in a paper I'm reading, but it isn't referenced and I can't seem to find it in any of the books I have: Every graded module M over a graded PID decomposes ...
3
votes
0answers
70 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 ...
3
votes
4answers
71 views

Proving that P/PJ is a projective right module over R/J

If P is a projective right module over a ring R and J is a two sided ideal of R. Prove that P/PJ is a projective right module over R/J . My idea was trying to proof that " $M$ is an $R$-module ...
1
vote
1answer
30 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
54 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
71 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 ...
3
votes
2answers
53 views

If $N\cap rM=rN$ for all $r\in R$, then is $M=N\oplus K$ for some $K$?

Suppose $M$ is a finitely generated free module over a principal ideal domain $R$, and $N$ a submodule. Why does the condition $N\cap rM=rN$ for all $r\in R$ implies that $M=N\oplus K$ for some ...
1
vote
2answers
29 views

Why does tensoring a projective resolution with a flat module give another projective resolution?

This question came up in this thread: Proving that tensoring a projective module with a flat module gives a projective module? Assume $\left\{P_i\right\}$ is a projective resolution of an $R$-module ...
2
votes
0answers
66 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 $A$ ...
3
votes
1answer
56 views

Direct product commutes with direct sum?

Do direct products commute with the direct sums of vector spaces? Basically is $\underset{i \in I}{\prod} \underset{j \in J}{\bigoplus}M_{i,j} \cong \underset{j \in J}{\bigoplus}\underset{i \in ...
5
votes
1answer
39 views

If $0$ is the zero-object $ \Longrightarrow F(0) $ is the zero object when $F$ additive

Let $$ F : \text{A-Mod} \to \text{A-mod} $$ be an additive functor. Then if $0$ is the zero-object $F(0) $ is the zero object. Why this is true ? The definition of additive functor that I know is ...
1
vote
2answers
41 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 ...
4
votes
2answers
118 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
vote
2answers
186 views

Why is there only one way to make $M$ a $\mathbb{Q}$-module, if possible?

Suppose $M$ is a $\mathbb{Q}$-module. Why is the given action of $\mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $\mathbb{Q}$-module? I know that an abelian group can only be made ...
0
votes
2answers
56 views

Show that Q, as a Z module, is a direct summand in a direct product of copies of Q/Z.

Prove:Q, as a Z module, is a direct summand in a direct product of copies of Q/Z. This is a problem from P.J.Hilton&Stammbach's Homological Algebra. If this is true, then there exists a ...
2
votes
1answer
44 views

Showing that $\bigoplus_{i\in\mathbb N}\mathbb Z/2\mathbb Z$ is not a direct summand of $\prod_{i\in\mathbb N}\mathbb Z/2\mathbb Z$

I know that if the direct sum of countably many copies of $\mathbb Z/2\mathbb Z$, $I$ were a direct summand of the direct product of countably many copies, $R$ (direct summand as $R$-modules), then ...
2
votes
1answer
56 views

Linear dual of vector fields

Suppose that $M$ is a smooth manifold and $\mathfrak{X}(M)$ is the set of smooth vector fields on $M$. There are basically two different linear structures on $\mathfrak{X}(M)$: 1.) $\mathfrak{X}(M)$ ...
2
votes
0answers
73 views

Torsion free flat module over a discrete valuation ring is Cohen-Macaulay

Why a torsion free flat module over a discrete valuation ring is Cohen-Macaulay?
2
votes
0answers
142 views

“M is reflexive” implies “M is MCM”. Is the converse true?

Let $R$ be a local integrally closed domain of dimension $2$. Let $M$ be a nonzero finitely generated $R$-module. We know that "$M$ is reflexive" implies "$M$ is MCM". Is the converse true? (By MCM I ...
-2
votes
0answers
22 views

bases in free associative algebras

I need an example and more details for the below notion: If X={x_1,x_2,...} be a set and by X^* we denote the set of all words named w of elements of X. Let F be a field and F be the vector space ...
1
vote
1answer
23 views

Tensor product and free $R$-module construction

I am trying to understand an exposition on Tensor products by Keith Conrad. In the proof of Theorem 3.2 on page 7 it considers the free $R$-module on the set $M \times N$: $$F_R(M \times N) = ...
2
votes
1answer
108 views

The Gorenstein dimension of a ring

I'm studying on these notes. I have a question about page 64, the remark. A local ring is Gorenstein if and only if the Gorenstein dimension of the residue field is finite. Of course if the ...
2
votes
1answer
121 views

A Gorenstein domain that is not a complete intersection

Could you give me an example (with proof) of a Gorenstein domain that is not a complete intersection?
-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.
1
vote
1answer
57 views

Completion of a torsion-free module

Let $R$ be a Dedekind domain, $K$ its field of fractions, $P$ a non-zero prime ideal of $R$. Let $\hat R_P$ be the completion of $R$ wrt to the valuation $v_P$ induced by $P$ and let $L$ be a ...
1
vote
1answer
42 views

Localization of a torsion-free module

I have a Dedekind domain $R$ with field of fractions $K$ and a non-zero prime ideal of $P$ of $R$. Let $L$ be a torsion-free $R$-module. How can I show that $R_P\otimes_R L$ is torsion-free as an ...
5
votes
1answer
89 views

On the indecomposable decomposition of the reduction of an integral representation

Here is a problem I have been grappling with all day. I started out thinking it might be true but am now inclined to believe it is false, and would like to see a counterexample. Suppose $G$ is a ...