Questions tagged [free-modules]

Use this tag for questions about free modules and related notions as projective modules or free abelian groups. This tag should be used together with the tags of abstract algebra and modules.

Filter by
Sorted by
Tagged with
0
votes
0answers
4 views

$V\dot Rad(R)\subseteq Rad(V)$ where equality holds if $V$ is free or projective.

Question: Let $Rad(R)$ denote the Jacobson Radical of $R$. Show that $V\dot Rad(R)\subseteq Rad(V)$ where equality holds if $V$ is free or projective. Show that equality fails when $R=\mathbb{Z}$ ...
0
votes
0answers
28 views

understanding the contradiction of $I$ is not free.

My professor gave us this example on a module that is not free : $R = k[x,y,z]$ where $k$ is a field. $I = xyR + yz R + xz R \subset R.$ take $u = xy, v = yz, w = xz$ then $I = uR + vR + wR.$ And so, ...
2
votes
1answer
112 views

Tensor product of two spaces one of them is free module

Let $V$ be a real vector space and let $S(V^*)$ be the symmetric algebra of $V^*$. Suppose $L$ and $N$ are two $S(V^*)$-modules. How to prove the statement in corollary 16 in the article by Kumar and ...
0
votes
0answers
27 views

Show quotient of free algebra is four-dimensional.

Let $k$ be a field and consider the free algebra $k\{t,x\}$ and consider the ideal $I = (t^2-1,x^2,xt + tx)$. Consider the algebra $H := k\{t,x\}/I$. I want to show that $H$ is four-dimensional as a ...
-1
votes
1answer
19 views

Understanding an example of a finitely generated projective module which is not free.

Here is the example I know: Consider the ring $R = \mathbb Z_2 \times \mathbb Z_2$ and the submodule $\mathbb Z_2 \times \{0\}.$ it is by construction a direct summand of $R$ but certainly not free....
0
votes
1answer
40 views

Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically?

Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically $$0 \rightarrow \ker \varphi \rightarrow F \xrightarrow{\varphi} P \rightarrow 0? $$ ...
0
votes
1answer
54 views

Why I should show that no element of $R$ divides both $2$ and $1 + \sqrt{-5}$? [duplicate]

Here is the question I want to answer: Let $R = \mathbb Z[\sqrt{-5}]$ and $ I = (2, 1 + \sqrt{-5})$ be an ideal of $R$. Show that $I$ is not a free $R-$module. But here in this question How to show ...
1
vote
1answer
29 views

What are the implications of a surjection between free modules?

My professor said the following in his lecture: Suppose $M$ and $P$ are free modules and $\alpha: M\rightarrow P$ is a surjection. Then $\alpha$ splits and the kernel of $\alpha$ is a summand of $M$. ...
0
votes
1answer
64 views

Showing that an ideal is not a free $R-$module.

Here is the question I want to answer: Let $R = \mathbb Z[\sqrt{-5}]$ Show that $I = (2, 1 + \sqrt{-5})$ is not a free $R-$module. Hint: It helps to view $R$ as $\mathbb Z[x]/(x^2 + 5).$ My questions ...
1
vote
1answer
42 views

If $F$ is free, then the functor $M \mapsto \operatorname{Hom}_A(F,M)$ is exact.

This is from Serge Lang's Algebra 3rd Edition (see the picture down below if you feel confused). My problem is, how to prove that if $F$ is free, then $M \mapsto \operatorname{Hom}_A(F,M)$ is exact. ...
1
vote
0answers
13 views

Taking quotients over isometric groups and determining the cokernel of transformations

We know that given group $A$ and subgroups $B_1,B_2$ such that $B_1\cong B_2$, it is not necessary that $A/B_1\cong A/B_2$. Now take for instance the following transformation $\varphi:\mathbb{Z}^3\...
0
votes
2answers
34 views

Prove that for every set there is a free module on that set

While studying the book "Module theory: an approach to linear algebra" by T. S. Blyth (the electronic edition of the book is free to download), I came across a proof that confused me. We ...
2
votes
1answer
38 views

How to prove that there is an isomorphism between a free module and a direct sum of copies of $R$?

The definition for a free $R$-module is the following: An $R-$module $F$ is free if there exists a linearly independent generating set $B=\{b_i\}_{i\in I}$, called a basis. Note that in this case $F$ ...
0
votes
0answers
25 views

Finitely generated modules over a polynomial ring

Let $M$ be a finitely generated module over $\mathbb{Q}[x_{1},\dots,x_{n}]$ but suppose that $M$ is not a finitely generated $\mathbb{Q}$-module. I would like to know if $M$ has a free direct summand ...
0
votes
1answer
39 views

$\mathbb{Z}^{\oplus \mathbb{N}} = \mathbb{Z}^{\oplus \mathbb{N}} \oplus \mathbb{Z}^{\oplus \mathbb{N}}$.

I am solving the following question from Aluffi chapter 0. I constructed the following counter example. I know by intuition that the following there exists an isomorphism $\mathbb{Z}^{\oplus \mathbb{N}...
1
vote
0answers
27 views

A free $A$-module $M$ which is also finitely-generated. [duplicate]

I have come upon with the following problem: Let $M$ be an $A$-module (assume that $A$ is commutative with identity) which is both free and finitely-generated. My question is whether there is an ...
0
votes
0answers
24 views

Is a bounded complex of projective $R$-modules of finite type quasi-isomorphic to a bounded complex of free $R$-modules of finite type?

The question is the one in the title. Let $(P,d)$ be a bounded complex of projective $R$-modules. My attempt is to consider, since projective is summand of free, for every integer $i$ a free module $F^...
1
vote
0answers
74 views

The structure of the quotients of free modules

Let $R$ be a commutative (unitary) ring. Every $R$-module is the quotient of a free $R$-module, and every free $R$-module with basis $B$ is isomorphic to $R^{\oplus B}:=\bigoplus_{i\in B}R$. That is, ...
1
vote
2answers
40 views

Confused about Tensor Product of R-modules

In Tu's book on Differential Geometry he first defines $Free(V\times W)$ as: $$\sum r_i(v_i, w_i), r_i \in R, (v_i, w_i) \in V \times W$$ where the sum is finite. The way I understand it is the above ...
0
votes
0answers
45 views

How to find a syzygy module is graded free module over R?

I am a bit new to commutative algebra. It is an algorithmic question rather than mathematical. Suppose that I have a graded polynomial ring $R = \oplus_{i \geq 0} R_i$ and a graded module $M=\oplus_{i ...
1
vote
1answer
43 views

How decomposes an $R[x]/(x^n)$-module which is free of finite rank over $R$?

Let $R$ be a commutative ring (with unit) and let $R'$ be the $R$-algebra $R[x]/(x^n)$. Let $r$ be a positive integer. Let $V$ be a finitely generated $R'$-module which is free of finite rank over $R$....
0
votes
1answer
69 views

If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ be a module homomorphism, then $\mathbb{Z}^m/\operatorname{im}(f)$ is a finite abelian group

If $f:\mathbb{Z}^m\to\mathbb{Z}^m$ is an injective group homomorphism, then is $\mathbb{Z}^m/\operatorname{im}(f)$ a finite abelian group? I think yes. Let the homomorphism be given by a $m\times m$ ...
0
votes
1answer
45 views

Smith normal form and elementary divisors

Say I have the matrix $\begin{pmatrix}16&16&8\\8&6&2\\3&4&2\end{pmatrix}$. To give a structure for a module homomorphism whose representation is given by this matrix, I must ...
4
votes
0answers
75 views

Prove, using category theory, that every base of free $R$-module ($R$ a conmutative ring) have the same cardinality

As said in the question, I'm trying to prove that if $R$ is a commutative ring (with unity) and $F$ is a free $R$-module, then every base of $F$ have the same cardinality. I know some algebraic proofs ...
2
votes
1answer
69 views

Is $\bigoplus_{x \in X} \mathbb{Z}$ reflexive?

Basically the title. I am trying to figure out whether or not $Hom(Hom(\bigoplus_{x \in X} \mathbb{Z}, \mathbb{Z})) \cong \bigoplus_{x \in X} \mathbb{Z}$ with the isomorphism $i(y)(f)=f(y)$. Here is ...
0
votes
2answers
39 views

Existence of epimorphism from a Ring to a Module

So the problem is the following, given a conmutative ring $R$ and a free $R$-module let's say $M$, such that $\dim(M)=2$. Prove there is no epimorphism from $R$ to $M$ I've been struggling for a while ...
0
votes
2answers
66 views

Index of submodule generated by multiplication by one element of free $\mathbb{Z}$-module

This question appears in an algebraic number theory notes with $M=\mathcal{O}_K$ for a number field $K$. Let $M$ be a free $\mathbb{Z}$-module of rank $n$ which is a ring itself. Let $x\in M$, then ...
0
votes
1answer
21 views

Higher Ext vanish iff module is free

Let $\Lambda$ be a commutative ring with $1$. Let $M$ be a $\Lambda$-module. Is it true that $\mathrm{Ext}^i_\Lambda(M,\Lambda)=0$ for all $i\ge1$ if and only if $M$ is free over $\Lambda$? The 'if' ...
1
vote
1answer
65 views

Free module is a right $R$-module with scaling operation.

Let $R$ be a ring and ruppose $F$ is a set of all functions from $S$ to $R$. Then, every element $u\in S$ can be viewed as a function from $S$ to $R$: $$u(t)=\begin{cases} 1 \hbox{ if } t=u\\ 0 \hbox{...
1
vote
0answers
38 views

Tensor product of free R modules is $R^{mn}$

Let R be any ring (not commutative maybe). Let $R^m$ and $R^n$ be considered as R-R bimodules. What is their tensor product as R-R bimodule? I know as abelian groups it is $R^{mn}$. Is it also the ...
0
votes
0answers
32 views

Misconception of extension of scalars of free $K$-algebras

I guess I have a bit of a misconception about the extension of scalars of free $K$-algebras where $K$ is a field. Consider the following situation: Let $K$ be a field and let $F$ be a reduced and free ...
2
votes
1answer
32 views

Kernel of surjective map from rank two module is free and rank 1.

I am having a bit of trouble with the following question 11(a) (section 2) in Waterhouse affine group schemes. It's really just a module theory question. It asks: Let $M$ be a free rank $2$ $R$-module....
0
votes
1answer
60 views

Why the canonical epimorphism $\pi:F\rightarrow F/IF$ is a one to one map?

The statement is the following: Let $R$ be a ring with identity, $I(\neq R)$ an ideal of $R$,$F$ a free $R$-module with basis $X$ and $\pi:F\rightarrow F/IF$ the canonical epimorphism. Then $F/IF$ is ...
0
votes
1answer
17 views

Extending a morphism among R-modules using a basis to make a diagram commute

let M,N R-modules, and L a free R-module. consider the morphism g:L⟶N,f:M⟶N, f epimorphism. Show that exist h:L⟶M such that f∘h=g I try to define h in some basis, with that i can use a theorem that ...
1
vote
1answer
37 views

If an ideal is free then it has rank 1.

Suppose we consider an ideal $I \subset R$ as an $R-$submodule. If $I \neq 0$ and it's free, then it's free of rank 1. I was wondering whether it's enough to say that $I$ is isomorphic to $R^n$, but ...
-1
votes
1answer
52 views

Linear independence of set in modules

Suppose $R$ be a commutative ring with unity and $N$ is a free $R$-module. m is a maximal ideal of $R$ and $m\in$ m. Then the set of elements $\{ x_1,x_2,...,x_{n-1}, m x_n \}$ which spans $N$ is not ...
2
votes
1answer
64 views

Lattices over modules

The standard construction of lattices in $\mathbb{R}^n$ can be generalised by taking any finite-dimensional vector space $V$ over a field $F$ and a subring $R$ of $F$, as per Wikipedia (https://en.m....
0
votes
1answer
46 views

Basis of quotient ring as a free module over a subring

Define the $k$-algebra $R=k[x,y]/(x^3-y^2)$. Fix some nonzero $m,n\in\mathbb{N}$. Set $s=x^my^n+(x^3-y^2)\in R$, and let $S=k[s]$. I want to show that $R$ is a free $S$-module with rank $2m+3n$. My ...
0
votes
1answer
87 views

Does $R/I \cong R$ implies $I=0$? [duplicate]

I think this is an obvious question, yet I'm stuck: Does $R/I \cong R$ implies $I=0$ where $R$ is a commutative ring with unity? Say there exists $I\neq 0$ s.t. $R/I\cong R$. Then since $R$ is free, ...
0
votes
2answers
56 views

$\operatorname{Hom}_R(F,R)$ is isomorphic to $F$ where $F$ is a free $R$-module of finite rank

[Dummit and Foote, Exercises 10.3, problem 13] Suppose $R$ is a commutative ring with $1$. Suppose $F$ is a free $R$-module of finite rank. Then show that $\operatorname{Hom}_R(F,R) \cong F$. Suppose ...
0
votes
1answer
36 views

A question in proof of a theorem in modules

While studying algebra from Thomas Hungerford I have a question in proof of a theorem on page: 186 . Its image: Question is in 3rd last line of the proof. How does $x\neq x' $ implies $1_{R} \in I$? I ...
1
vote
1answer
45 views

Do non-trivial kernels of maps between free modules always contain a basis element?

Let $R$ be a commutative ring with identity and $f: R^m \rightarrow R^n$ a homomorphism of $R$-modules with $m>n$. Then does there exist a basis $x_1,\cdots , x_m$ of $R^m$ such that $f(x_1)=0$? ...
0
votes
0answers
17 views

Then $bP$ is a projective $Ab$-module.

$A$ is a $k$-algebra and $b$ is an idempotent in $Z(A)$. Let $P$ be a projective $A$-module. Then $bP$ is a projective $Ab$-module. Here is my proof: $P$ is a direct summand of a free $A$-module and ...
0
votes
2answers
73 views

Let $k$ be a field. Is $k[x,x^{-1}]$ free over $k[x]$?

I see that $\{x, x^{-1}\}$ do not form a basis over $k[x]$ since for example $x^2 \cdot x^{-1} +(-1)\cdot x =0$. I'm not sure if this is enough to conclude that $k[x,x^{-1}]$ is not free over $k[x]$. ...
1
vote
0answers
53 views

$A$ is finitely generated free.

$R$ is left noetherian. $A$ is finitely generated $R$-module. Then $A$ is finitely generated free. I can't see why it is free. I know that $A$ is finitely presented, so there is an exact sequence: $0 \...
0
votes
0answers
22 views

Isomorphism on quotient vector spaces for modules over a complete local ring

Let $R$ be a complete local ring with maximal ideal $\mathfrak{m}$. Let $M$ be a separated module over $R$ and $S$ be a minimal generating set for $M$ (not necessarily finite). If $F = F(S)$ is the ...
2
votes
1answer
25 views

Determine whether $V_{T_1}$ and $V_{T_2}$ are isomorphic as $\mathbb{Q}[t]$-modules

Given a two-dimensional $\mathbb{Q}$-module, $V$ with basis $\{e_1,e_2\}$, define two $\mathbb{Q}[t]$-module structures as follows: $$T_1:= \begin{cases} te_1 \mapsto e_2\\ te_2 \mapsto ...
0
votes
0answers
24 views

Free modules of $F[x,y]$

I am trying to find out the free modules of $F[x,y]$. ($F$ is a field) Trivially, $F[x,y]$ is an $F[x,y]$ free module. Also , I can have $F[x,y,z]$ as an $F[x,y]$ free module(similar to: for a ...
2
votes
1answer
87 views

Two finitely generated module over $\mathbb{C}[[x_1, x_2, \dots, x_n]]$

I'm struggling with the following problem. Let $A = \mathbb{C}[[x_1, x_2, \dots, x_n]]$ be the ring of formal power series over $\mathbb{C}$. Show that if two finitely generated $A$-modules $M, N$ ...
1
vote
1answer
93 views

Clarification in a proof that free modules are flat

In section 10.5 of Dummit and Foote, we are given a proof that free modules are flat that goes like this: 1). Finitely generated free modules are flat (easy) 2). Suppose now that $F$ is an arbitrary ...

1
2 3 4 5
8