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

learn more… | top users | synonyms

3
votes
2answers
225 views

Is every submodule of a free module of finite rank over a PID a direct summand?

Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a submodule. Is $N$ always a direct summand of $F$? I think the answer is yes, $N$ is also free of rank $m\leq n$ since we are working ...
0
votes
1answer
20 views

Condition under which one can extend a linearly independent set in a free $\mathbb{Z}$-module to a basis?

I'm having difficulty answering the following simple (?) question: Let $X = \mathbb{Z}^n$, and let $M$ be a submodule of $X$ closed under division - by which I mean that if $x \in X$ has $n x \in M$ ...
6
votes
1answer
103 views

Tensor product of $\mathbb{Q}$ with an infinite product [duplicate]

How can I prove that the tensor $\mathbb{Q} \otimes \left( \prod_n \mathbb{Z}/n\mathbb{Z} \right)$, where the product is taken over all the positive integers $n$, is not trivial?
3
votes
2answers
82 views

Determinant from Paul Garrett's Definition of the Characteristic Polynomial.

$\DeclareMathOperator{\id}{id} \DeclareMathOperator{\End}{End}$ On pg. 390 of Paul Garrett's notes on Algebra, a definition for the characteristic polynomial is given, which I discuss here. Let $V$ ...
3
votes
0answers
32 views

Elementary divisors for chains of submodules

Given free modules $N \le M$ of finite rank over a PID $R$, it's well-known that there is a basis $\{x_1,\ldots,x_n\}$ of $M$ and there are $e_1,\ldots,e_n \in R$ such that $\{e_ix_i\mid e_i \neq ...
0
votes
2answers
21 views

Cosets of submodule

Suppose we have a ring $R$ and an ideal $J \subseteq R$. Let $M$ be an $R$-module and $N$ be a submodule of $M$. Assume for two elements $m$ and $m'$ of $M$ we have $$m+ JM=m'+JM.$$ Since $N ...
0
votes
2answers
31 views

Generators for a free submodule of a free module

Suppose we are given a finitely generated free module $M$ over a ring $R$. Assume $N \subseteq M$ is a free submodule of $M$. If $B$ is a basis for $M$, does it follow that there exists a subset $A ...
0
votes
0answers
33 views
+50

Understanding tensor products and $R$-algebras

Sorry if this question isn't quite precise. Anyways, I'm reviewing for my algebra final right now and two things I don't think I quite grasp as well as I'd like are tensor products and $R$-algebras. I ...
0
votes
1answer
22 views

A $R$-module over a ring $R$ with identity is free iff it is a direct sum of copies of $R_R$, where $R_R$ is $R$ considered as a module over itself

Let $R$ be a ring with identity and $M$ be an $R$-right module. Then $M$ is called free over $X \subseteq M$ if for every module $N$ with mapping $\alpha : X \to N$ we can extend it uniquely to a ...
0
votes
1answer
44 views

Poincare series and Hilbert polynomial of some graded modules [on hold]

Let $k$ be a field, and let $k[X, Y ]$ be the polynomial ring in two variables equipped with the usual grading such that $\deg(X) = \deg(Y ) = 1$. Consider the ideals $I = (X, Y^2)$ and $J = (X^2, ...
-1
votes
1answer
59 views

Example of submodule which has higher “rank” than the module

Given a ring $A$, and an $A$-module $M$. Let $x_1, . . . , x_n ∈ M$. Say $\operatorname{rank}(M) = n$ if for all $m ∈ M$ there exist unique elements $a_i ∈ A$, $i = 1, . . . , n,$ such that ...
0
votes
1answer
24 views

Exact Sequences of Modules and Rank

Suppose that: $0 \rightarrow M_{1} \rightarrow M \rightarrow M_{2} \rightarrow 0$ Is an exact sequence of $R$-modules (for $R$ commutative integral domain). Show that $\mathrm{rk}(M) = ...
5
votes
3answers
136 views

nilpotent endomorphism on finitely generated modules over a domain

If $R$ is a domain and $f: R^n \to R^n$ is an $R$-module endomorphism. Suppose $f^m = 0$ for some $m> 0$. Show that $f^n = 0$. The cases $ m \le n$ is trivial. When $m>n$, I don't have much ...
1
vote
0answers
16 views

Uniqueness submodule of cyclic module

I'm having some problems with an exercise from Hungerford's Book of Algebra. It states: Let $R$ be a PID, and $A$ a unitary $R$-module such that $A$ is cyclic of order $r$. a) Prove that ...
1
vote
1answer
13 views

Kernel of $M\to M[U^{-1}]$ and primary decomposition of $(0)$

I am working on exercise 3.12 from Eisenbud's Commutative Algebra and I am having trouble parsing the question. Let $M$ be a finitely generated module over the Noetherian ring $R$. Given any ...
0
votes
1answer
56 views

$\mathbb C$-dimension of vector space $\mathbb C\otimes_{\mathbb R}\mathbb C$

Let $\mathbb R$ be the field of real numbers, $\mathbb C$ be the field of complex numbers. Consider $\mathbb C\otimes_{\mathbb R}\mathbb C$ as a $\mathbb C$-vector space via $a(b\otimes c) := ab ...
1
vote
2answers
25 views

CG-modules: what does this notation mean?

I am trying to solve a question, but I do not know what the notation used means. If anyone could help me out that'd be great! I don't need help doing the proof, just what the notation means would be ...
0
votes
0answers
28 views

If $f : M\otimes_A A/m \to N\otimes_A A/m$ is surjective , so is $f : M \to N$. [on hold]

Let $A$ be a local ring with maximal ideal $m$. Let $f : M \to N$ be a morphism of $A$-modules, where $N$ is finitely generated. Show that if the map $f : M\otimes_A A/m \to N\otimes_A A/m,\quad ...
3
votes
2answers
151 views

Example of a Tensor Product of Modules with Non-Decomposable Elements

Given a ring $R$ and $R$-modules $A_R$ and $_{R}B$, we define the tensor product $A \otimes_R B$ as the free abelian group on $A \times B$ modded out by the subgroup generated by the elements of ...
6
votes
1answer
712 views

A question about a proof of Noetherian modules and exact sequences

I proved part (i) of the following: Proposition 6.3. Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then i) $M$ is Noetherian $\iff$ $M'$ ...
1
vote
0answers
34 views

Matrices representing a map between free modules of infinite rank and Fitting's Lemma (Eisenbud)

p.497 of Commutative Algebra with a View Toward Algebraic Geometry, Eisenbud: If $\phi: F \rightarrow G$ is a map of free modules, then $I_j\phi$ is the image of the map $$\Lambda^j F ...
0
votes
0answers
34 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 ...
1
vote
1answer
73 views

When is an ideal also a ring, and could then be anything said about its relation to the original ring

If $R$ is a ring with unity $1$, then $S \subseteq R$ is called a subring if it is itself a ring with $1 \in S$. A subset $I \subseteq R$ is called an ideal if it is a group with respect to addition ...
0
votes
2answers
29 views

what is the difference between finitely generated module and finitely generated free module?

I am still confused about the difference between free module and finitely generated module. For example, $Z/2Z$ is finitely generated module, but why it is not finitely generated free module? What is ...
0
votes
1answer
28 views

Ideals and submodules are the same [closed]

My teacher has told me that for an R-module, that I is an ideal of R if and only if I is an R-submodule of R. I know this is true but I was wondering why? IS there an official proof?
-1
votes
1answer
10 views

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

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 $M_1 ...
1
vote
0answers
82 views

Tensor product of two simple modules [duplicate]

Let $k$ be a field of arbitrary characteristic. Suppose that $A$ and $B$ are finite dimensional $k$-algebras. Let $S$ be a finite dimensional simple $A$-module and let $T$ be a finite dimensional ...
3
votes
2answers
130 views

If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated.

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 ...
2
votes
2answers
96 views

Prove that there's a unique morphism that completes the commutative diagram

I have to prove that there's a unique $\gamma : M'' \rightarrow N''$ that completes this diagram considering the rows are exact. $$\begin{array} MM' \stackrel{f_1}{\longrightarrow} & M & ...
1
vote
1answer
40 views

For a Noetherian ring $R$, we have $\text{Ass}_R(S^{-1}M)=\text{Ass}_R(M)\cap \{P:P\cap S=\emptyset\}$

$\DeclareMathOperator{\ass}{Ass}$ Let $R$ be a Noetherian ring and $S$ be a multiplicative set in $R$. Then show that $\ass_R(S^{-1}M)=\ass_R(M)\cap \{P\in \text{Spec}(R):P\cap S=\emptyset\}$ ...
1
vote
1answer
27 views

Universal Linearizer of Alternating Multi-$F[x]$-Linear Maps is Same as that of Multi-$F$-Linear Maps.

Let $V$ be a an $n$-dimensional vector space over a field $F$. Let $M=F[x]\otimes_F V$. We can consider $M$ as an $F[x]$-module by extending scalars using the inclusion $F\to F[x]$. Fact 1. There is ...
0
votes
1answer
24 views

Structure of induced module for subgroup $H \le G$, confused by mixing up notions of $FH$-submodule and $F$-subspace

The following is a question on a passage in the book A Course in the Theory of Groups by Derek Robinson. There he constructs the induced module $M^G$ from a $FH$-module for some subgroup $H \le G$ of ...
1
vote
1answer
62 views

If $P$ is projective and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$

I need to prove that if $P$ is a projective module over $\mathbb Z$ and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$. This in turn would imply if $A$ and $B$ are ...
0
votes
1answer
18 views

Computing generators for a finitely generated module

I came across this problem yesterday: Let $R$ be a ring and $M$ an $R-$module. $\varphi:R^n\to M$ is a surjective $R-$module homomorphism if and only if $M$ is finitely generated. Given the set of ...
3
votes
1answer
39 views

In proof of $|G| = n_1^2 + n_2^2 + \ldots + n_h^2$ how could equality of dimensions concluded from ring isomorphism

In Derek Robinson, A Course in the Theory of Groups on page 224 he proves: Let $G$ be a finite group and let $F$ be an algebraically closed field whose characteristic does not divide the order of ...
1
vote
2answers
51 views

motivation for the direct limit [closed]

I know just the very basics on Category Theory and that's why I'm going to ask a stupid question. I'm trying to get an intuition for direct limits for my course on Commutative Algebra. All the books ...
4
votes
2answers
90 views

Count the number of elements of ring [closed]

1/ How to count the number of elements of $\mathbb{Z}[i]/(1+2i)^n$? 2/ How to write $\mathbb{Z}[i]/(1+2i)^n$ as direct sum of cyclic groups (in view of the structure theorem of finite abelian ...
1
vote
1answer
23 views

Show ring isomorphisms $End_R{({S_1}^{n_1} \oplus \ldots \oplus {S_r}^{n_r} )} \cong End_R{({S_1}^{n_1})} \times \ldots \times End_R{({S_r}^{n_r})}$

I have been struggling with this Representation Theory question for the past week: Let $R$ be a ring and $S_1, \ldots, S_r$ simple $R$-modules with $S_i$ not isomorphic to $S_j$ whenever $i \neq j$ ...
1
vote
1answer
20 views

Show that the left ideal $(N_G) \subset F[G]$ is a simple submodule of $F[G]$, where $N_G = {\sum}_{g \in G} {g} \in F[G]$. [duplicate]

I am trying to solve this Representation Theory question: Let $F$ be a field and $G$ a finite group. Let $N_G = {\sum}_{g \in G} {g} \in F[G]$. Show that the left ideal $(N_G) \subset F[G]$ is a ...
0
votes
0answers
15 views

Show that all simple modules of ${M_n}{(D)}$ are isomorphic to $D^n$, where $D$ is a division ring. [duplicate]

I am trying to solve part (c) of the following Representation Theory question: Let $D$ be a division ring and let $n$ be a positive integer. For $ 1 \leq l \leq n $ let $$C_l= \{A = (a_{ij}) \in ...
1
vote
1answer
22 views

Poincare series and the Hilbert polynomial of $A = A_0[X_1,\dots , X_s]$

Let $A = A_0[X_1, \dots , X_s]$ be a polynomial ring in $s$ variables over an Artin ring $A_0$. This is a graded ring, and can be regarded as a graded module over itself. 1. What are the ...
1
vote
1answer
51 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 ...
0
votes
0answers
43 views

Question about the classification theorem for finitely generated graded $F[t]$-modules

I am in the beginnings on learning persistence homology, and as a start I'm studying Gunnar Carlsson's survey "Topology and Data". Theorem 2.10 states the following: "Suppose $M_{\star}$ is a ...
1
vote
1answer
26 views

Submodule of a free module with the same rank is a direct summand

Let $M$ and $N$ be finite rank free modules over a ring with invariant basis number such that $N \subseteq M$ is a submodule of $M$. Suppose $M$ and $N$ have the same rank $n$. Is it possible for ...
3
votes
0answers
98 views

Structure theorem of modules in the graded case [duplicate]

I ran into an exercise in D. Passman's book A Course in Ring Theory (page 130) asking me to prove the structure theorem of modules over PIDs in the graded case. Find all finitely generated graded ...
0
votes
0answers
67 views

Decomposition of a graded module

In an article I am reading I found the following statement: If $D$ is a PID, then every finitely generated $D$-module is isomorphic to a direct sum of cyclic $D$-modules. That is, it decomposes ...
0
votes
1answer
24 views

where do elements go under multiplication in a graded module?

Assume that $M = \bigoplus_{n = 0}^\infty M_n$ is a graded $A$-module, where $A = \bigoplus_{n = 0}^\infty A_n$ is a graded ring. We have by definition $A_m M_n \subset M_{m + n}$. Does this mean that ...
0
votes
0answers
17 views

Why is $(A \otimes_\mathbb{Z} V_p )\bigcap R(G) = V_p$

I am reading the proof of brauers theorem in Serres book Linear Representations of finite groups and I have trouble understanding Lemma 5(page 75). Let $G$ be a group of order $g$. First let $g=p^nl$ ...
0
votes
0answers
26 views

If F is a simple module, is $F^n $ semi simple?

If F is a simple module, is $F^n $ semi simple? I'm assuming this is true by the classification theorem
0
votes
0answers
34 views

Origin of the name “Cauchy sequence” in abstract algebra

I know the origin in analysis as it is derived from our good old chap Cauchy himself. However I have read about Cauchy sequences in algebra, I'll use groups for this one, let $G$ be a group and ...