2
votes
2answers
31 views

Sub-modules of free modules

I'm going back through basic module theory notes, and I've come across a paragraph explaining that a sub-module of a finitely generated free module may not itself be free. In my course a free module ...
0
votes
0answers
13 views

show that exist a left quasi-inverse of any element in J(R)

I studying for math. but this problem, I don't understand. J(R): the radical of R, written as J(R),is the set of all elements of R which annihilate all the irreducible R-module. and show that exist a ...
0
votes
0answers
44 views

Tensoring over a direct sum of R

Is tensoring over a ring R same as tensoring over a finite direct sum of R. I mean is $A \otimes_R B$ isomorphic to $A \otimes_{R^n} B$? And if it doesnt hold for any ring $R$, Does it hold for ...
1
vote
2answers
32 views

The definition “module of finite type”.

I know the definition of "module of finite type" Is that different from finitely generated module ? Thanks.
1
vote
2answers
41 views

Can a the field of fractions or quotient field F of an integral domain R be free over some set as an R-module?

We know any integral domain R when extended to a quotient field F, then F is free as an F-module on the set {1}. Can this field be free over some set as an R-module.
0
votes
1answer
32 views

Show that the image or the kernel are submodule of R-module.

Let $R$ be Commutative ring and $M$ be an $R$-module. Show that $im(H)$ or $ker(H)$ are submodule of $R$-module $M$, where $$H\in Hom(M,-)$$ First, I think by lemma: If $M$ is an $R$-module and $N$ is ...
-1
votes
1answer
39 views

ring and module problem

Let $$F=\mathbb{R}$$ $$V=\mathbb{R}^{4}$$ consider two matrices $$S1=\begin{vmatrix} 0&-1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& -1\\ 0&0 & 1 & 0 ...
0
votes
0answers
15 views

Identifying the abelian group with a presentation matrix

I am doing problems from Artin: \begin{bmatrix} 2 \\ 1\\ \end{bmatrix} and \begin{bmatrix} 2 & 4\\ 1 & 4\\ \end{bmatrix} For the First one after manipulating rows I ...
0
votes
1answer
30 views

How to show Not a Free Module

Let $\mathbb K$ be a field, $A= \mathbb K [x,y]$ and $ M = Ax + Ay$. prove that $M$ is NOT a free module!
0
votes
0answers
19 views

Divisible Direct Sum or Direct Product

We know that direct sum and direct product of divisible R-modules are divisible when R is a domain. Does there exist non-divisible modules with their product or sum being divisible? Or, if R is not a ...
1
vote
1answer
32 views

About the definition of the tensor product of modules

I was reading something about tensor product (balanced product) of modules (over an arbitrary ring $R$), but I cannot realize why we need a left and a right module. Would it be the same with two right ...
2
votes
1answer
47 views

Localization of a module as direct limit

Let $A$ be a commutative ring, $S \subset A $ a multiplicatively closed set and $M$ an $A$-module. For every $s \in S$ we denote by $M_{s}$ the localization of $M$ with respect to $\{ 1, s, s^2, ...
1
vote
0answers
52 views

Prove this morphism is a derivation

Let $A$ a ring, $B$ a $A$-algebra, $I=\ker (B\otimes_A B \to B)$ given by multiplication, i wanna see why is $$d:B \to I/I^2$$ $$b \mapsto b\otimes 1 -1 \otimes b$$ well defined (in fact, why mod ...
2
votes
2answers
53 views

Tensor product is o?

$K$ generated by $(ar,rb)$ for every $a\in A$, $b\in B$, $r\in R$. $F$ generated by $r(a,b)$ for every $a\in A$, $b\in B$, $r\in R$. $(a,rb)=(ar,b)=r(a,b)$,then $K=F$, $F/K=0$ Apparently I ...
1
vote
0answers
26 views

Tensoring and retaining projectiveness

Let $A$be a unital associative ring, If $A$ is not a projective $A$-bimodule, however $A\otimes A$ is a projective $A$-bimodule, can we conclude that $A\otimes A \otimes A$ is also projective?
5
votes
1answer
60 views

Can a ring isomorphism change the structure of a module?

Let $M$ be an $R$-module, where $R$ is a ring with unit. Given a ring automorphism $\phi: R \rightarrow R$, we can define a new $R$-module structure on $M$ by $r \cdot x = \phi(r) x$ for all $r \in ...
1
vote
2answers
34 views

Free $A$-module isomorphic to a direct sum of copies of $A$?

Does this proposition hold even if it's not finitely generated? I think it does, since $M$ isomorphic to the direct sum of $M_i$, $M_i$ isomorphic to the direct sum of ...
1
vote
1answer
26 views

Free modules and ideals

I am trying to show that an ideal I of R=$\mathbb{C}[x_1,x_2]$ generated by $x_1, x_2$ is free R-module. I am trying to show that I has a basis of the two generators given above. But I am not able to ...
0
votes
0answers
18 views

projective dimension of modules over triangular matrix rings

Let $R$ and $S$ be rings and $M$ be an $S$-$R$- bimodule. Then consider the triangular matrix ring constructed from these data (as, e.g., in Auslander-Reiten-Smalo). what one can say about the ...
2
votes
1answer
33 views

What is the difference between a module of finite rank and finitely generated module.

R is an integral domain and every module we talk about is an R-module. If a module is finitely generated then obviously every element of the module can be written as finite R-linear combination of the ...
0
votes
1answer
62 views

The meaning of a symbol in the proposition

The question is what's the meaning of the symbol $\phi$? If it just a mapping,what's the mean of the equation? I guess it's $\phi(x)$, $x$ is the element of $M$. Then $\phi(x)$ is the element of ...
1
vote
1answer
19 views

Show polynomials $I$ is not finitely generated as $R$-module

Let $R=\{a_0+a_1X+\cdots+a_nX^n\;|\;a_0\in\mathbb{Z},a_1,a_2,\cdots ,a_n\in\mathbb{Q}, n\in\mathbb{Z}_{\geq 0}\}$ and $I=\{a_1X+\cdots+a_nX^n\;|\;a_1,a_2,\cdots ,a_n\in\mathbb{Q}, n\in\mathbb{Z}^+\}$. ...
0
votes
1answer
20 views

Prove module $M$ is finitely generated if $N$ and $M/N$ are finitely generated

Let $R$ be a ring with $1$ and $N$ be a submodule of an $R$-module $M$. Prove that $M$ is finitely generated if $N$ and $M/N$ are finitely generated. There are two definitions of "finitely ...
1
vote
1answer
24 views

If $N$ and $M/N$ are free modules of finite rank, so is $M$

Definiton: A $R$-module $M$ is said to be free of finite rank if $M\cong R^k = R\times R\times\cdots \times R$ (k times). Let $R$ be a ring with $1$ and $N$ be a submodule of an $R$-module $M$. ...
2
votes
1answer
62 views

Is any right exact sequence of modules induced by free modules?

Let $R$ be a ring and let $M \to N \to K \to 0$ be an exact sequence of $R$-modules. Is there an exact sequence of free modules $A \to B \to C \to 0$ and a commutative diagram $$\begin{array}{c} M ...
0
votes
1answer
35 views

Tor of submodule

Let $R$ be a $CRing$. If $i:A \rightarrow B$ is the inclusion of a $R$-subalgebra A into an $R$-algebra $B$, then what is ther relationship between: $Tor_{A^e}$ and $Tor_{B^e}$?
1
vote
1answer
28 views

Natural map of extension groups

Let $\Lambda$ be a cocommutative Hopf algebra over a commutative ring $R$. For two left $\Lambda$-modules $M$ and $N$, interpret $\mathrm{Ext}_{\Lambda}^n(M,N)$ as the set of equivalence classes of ...
3
votes
1answer
36 views

$\operatorname{rank}(F) = \operatorname{dim}_{k}(\frac{F}{mF})$

Let $R$ be a commutative ring with unit; $m$ is a maximal ideal; $F$ a free $R$-module. We know that $\frac{F}{mF}$ is a vector space over $\frac{R}{m} = k$ . I have to prove that ...
10
votes
1answer
107 views

If the tensor power $M^{\otimes n} = 0$, is it possible that $M^{\otimes n-1}$ is nonzero?

Let $M$ be a module over a commutative ring $R$. It is possible that $M \otimes M = 0$ if $M$ is nonzero, for example when $R = \mathbb{Z}$ and $M = \mathbb{Q}/ \mathbb{Z}$. What about when higher ...
0
votes
1answer
21 views

Are A and Hom(A,Q/Z) isomorphic?

If A is a Z-module, does this imply that A and Hom(A,Q/Z) are isomorphic ?
0
votes
2answers
65 views

Possible difference between $\mathbb{Z}$-modules and vector spaces

Suppose $G$ is a free abelian groups, i.e. a free $\mathbb{Z}$-module; we have a set $S \subset G $ such that $S$ spans $G$. Can we conclude that the rank of $G$ as a $\mathbb{Z}$-module is $ \leq ...
2
votes
1answer
44 views

Relation Between Free Quotients and Modules

Here is my question: Let $M$ and $M'$ be $R$-modules, where $R$ is a commutative ring, and $N \subseteq M$ and $N' \subseteq M'$ submodules. Suppose that $N \cong N'$ and $M/N \cong M'/N'$. Determine ...
0
votes
1answer
21 views

$R$ is an injective $R$- module $ \Rightarrow $ field if $R$ is a domain

This is an exercise from Rotman: Prove that if $R$ is a domain and an injective $R$-module, then $R$ is a field. Any hint ?
1
vote
2answers
52 views

Definition/existence/uniqueness of a minimal projective resolution

I'm reading Dave Benson's book "Representations and Cohomology," Volume I, and I'm trying to understand the following discussion on page $32$ in which he introduces the notion of a minimal projective ...
4
votes
1answer
71 views

Don't understand a proposition and its proof

Proposition 5.1 from Commutative Algebra by Atiyah and Macdonald: $x∈B$ is integral over $A$,then $A[x]$ is a finitely generated $A$-module. The elements in $A[x]$ are the set of all the sum. If ...
2
votes
1answer
52 views

Understanding the Bockstein homomorphism in group cohomology

Let $k$ be an algebraically closed field of characteristic $p>0$. Let $G$ be a finite group. In group cohomology, the Bockstein homomorphism is the connecting homomorphism ...
2
votes
1answer
48 views

Maximal $\mathbb{Z}$-submodules in $\mathbb{Q}$

Is it true that $\mathbb{Q}$ viewed as $\mathbb{Z}$-module ( i.e. abelian group ) has not maximal $\mathbb{Z}$-submodules ? Why ?
1
vote
1answer
23 views

Hom($P$, $R$) $\neq 0 $ if $P$ is a nonzero projective left $R$-module (Rotman)

I've found this exercise, number $3.11$ from Introduction to homological algebra. Prove that $\operatorname{Hom}(P, R) \neq 0 $ if $P$ is a nonzero projective left $R$-module. Any hint?
0
votes
1answer
37 views

finitely generated right R-module which is not cyclic

Let $A$ be a finitely generated right $R$-module which is not cyclic. Prove that there exist $B \le A_R$ maximal with respect to the property that $A/B$ is not cyclic. help please and thank you for ...
3
votes
1answer
21 views

Example of a ring whose left modules are all free but has some non-free right modules

As the question says, I'm looking for a ring with all free left modules but some non-free right modules. I had thought about looking for a ring not isomorphic to its opposite and try and use that a ...
14
votes
2answers
142 views

If $M\otimes N=R^n$ need $M$ be projective?

So if over a commutive ring, $R$, we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ be finitely generated projective? We have finite generation, because if $M\otimes N$ is ...
1
vote
1answer
24 views

GCD for multivariable polynomial ring

I'm reading Lectures on Modules and Rings by T. Y. Lam. It's on page 32 of the book, example 2.19A. It reads: (2.19A) Example. Let $k$ be a field. Then in the commutative polynomial ring $R = ...
3
votes
2answers
70 views

Tensor powers of injective linear maps of free modules

This is a basic question on tensor products of linear maps. Let $R$ be a commutative ring and let $\varphi: M\to N$ be an injective linear map of finitely generated free $R$-modules. Question: Are ...
0
votes
2answers
40 views

$R$ -module homomorphism

Let $M,N$ be $R$-modules. Suppose there exists $R$-module homomorphism $\phi:M \to N$ and $\psi:N \to M$ such that $(\psi \ \circ \ \phi)(m) =m, \forall m \in M. $ Could anyone advise me on how to ...
2
votes
1answer
28 views

Find an epimorphism in Mod-Z

How can I prove that there exists an epimorphism between a direct sum of Z and the set Hom(A,Q/Z) in Mod-Z ?
0
votes
0answers
17 views

Question about product of cyclic modules over a Euclidean Domain

Let $R \neq 0$ be a Euclidean domain. Let $d_1, d_2 \in R$ such that $0 \neq d_1 | d_2 \nsim 1$. Let $M$ be the $R$-module: $$M = \frac{R}{(d_1)} \oplus \frac{R}{(d_2)} $$ with the standard ...
-2
votes
1answer
67 views

Surjective Implies Injective for R-Homomorphism on Finitely Generated Module [duplicate]

Let $M$ be a finitely generated module over a ring $R$, and let $f$ be an $R$-homomorphism from $M$ to itself. Does $f$ injective imply $f$ surjective? Does $f$ surjective imply $f$ injective? I have ...
1
vote
1answer
26 views

Elements of a free module written uniquely as a linear combination of basis elements

Let $R$ be a ring and $M$ a free $R$-module with basis $X$. Is it so that every $m \in M$ can be written uniquely as a linear combination of elements of $X$? If not, in which cases is that true? ...
2
votes
1answer
35 views

Example of a faithful irreducible module

Let $R$ be a non trivial simple ring. I am trying to show that there is a faithful irreducible left $R$-module. Is the ring $R$ considered as a left module over itself such a module? I think ...
1
vote
0answers
25 views

Composition series of finite length modules

Given a field $K$ of characteristic zero. Let $V$ be a finite dimensional $K$-vector space and let ${f} \in End_K(V)$. Then $V$ can be regarded as a $K[X]$-module by the action: ...