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

learn more… | top users | synonyms

1
vote
2answers
40 views

A module over a noetherian ring and its dual

Let $R$ be a noetherian ring, $M$ an $R$-module and $M^*$ be its dual. Is it true that if $M$ is noetherian then $f(x) = 0$ for all $f \in M^*$ implies $x =0$? How about the other way round? Ok, ...
1
vote
1answer
92 views

Injective resolutions of a complex

Let $\mathcal{A}$ be an abelian category, $M\in\mathcal{A}$. An injective resolution of $M$ is a quasi-isomorphism $M\longrightarrow I$, where $I$ is a complex of injective objects. This can be made ...
2
votes
0answers
38 views

If $E$ is a finitely presented left $A$-module, then tensor product by E commutes with direct product.

Please can anyone help me to prove this statement: If $E$ is a finitely presented left $A$-module, then for every family $(F_i)$ of right $A$-modules the canonical homomorphism $\phi: ...
17
votes
2answers
2k views

Is $A \times B$ the same as $A \oplus B$?

When $A, B$ are $K$-modules, then $A \times B$ is the same as $A \oplus B$. Let $A, B$ be two $K$-algebras, where $K$ is a field. Is $A \times B$ the same as $A \oplus B$? Thank you very much. ...
1
vote
1answer
33 views

Existence of injective homomorhism and surjective homomorphism of free modules implies isomorphism [duplicate]

Given an integral domain $R$, we have that $R^n$ embeds in $R^{n+1}$ as free modules over $R$. Can we have a surjective $R$-homomorphism from $R^n$ to $R^{n+1}$? Or more so, given two free $R$-modules ...
0
votes
2answers
69 views

Noetherian submodules and isomorphisms

Suppose $M$ is an $A$-Module, and $N$ is a submodule of $M$. Let $f:N\to M$ be an $A$-module epimorphism. How could I show that if $N$ is noetherian, then $f$ is an isomorphism? Thanks
4
votes
1answer
34 views

Apparent Contradiction to Weyl's Theorem

Let $L$ be $sl(2)$, i.e., $L=span\{h,e,f\}$, where $[h,e]=2e$,$[h,f]=-2f$,$[e,f]=h$. This is semi-simple. Suppose I create a module $V=span\{v_1,v_2,v_3\}$ and define actions as follows: ...
1
vote
1answer
231 views

Question in the exactness of the induced sequence of an exact sequence of module homomorphisms using the representation functor

I am referring to Theorem 6.3 p. 143 from Lang's Algebra (but the question description will be as self-contained as possible). Let $A$ be a commutative ring and $M$, $W$, $V$, $U$ be $A$-modules. ...
1
vote
1answer
32 views

Wedderburn component of $\mathbb C[G]$ corresponding to a contragredient character

Let $G$ be a finite group and let $\phi$ be an irreducible character of $G$ over $\mathbb C$. How does the Wedderburn component of $\mathbb C[G]$ corresponding to the contragredient character of ...
-3
votes
1answer
32 views

Find all subspaces of the real vector space $\Bbb R^2$.

Find all subspaces of the real vector space $\Bbb R^2$. Is is true for any elements $u=(a, b)$ and $v=(c, d)$ of $\Bbb R^2$,there exists a non-trivial subspace $W$ of $\Bbb R^2$ such that $u, v \, € ...
4
votes
1answer
69 views

Existence of a non-semisimple ring such that every module over it has a simple submodule [closed]

Does there exists a ring $R$ which is not semisimple but every module over it has a simple submodule?
2
votes
2answers
433 views

Proving that the length of the sum and intersection of two finite length modules is finite.

I am trying to solve the following question: Let $M_1, M_2 \subset M$ be finite length submodules of a module $M$ (where $M$ need not be of finite length). Show that $ \ M_1 + M_2 = \{x_1 + x_2 \in M ...
2
votes
1answer
135 views

Lemma 1.3.4(b) in Bruns and Herzog

My question refers to the proof of the second of the following lemma given in Cohen-Macaulay rings by Bruns and Herzog. Lemma 1.3.4 (Bruns and Herzog): Let $(R,\mathfrak m,k)$ be a local ring, and ...
-1
votes
1answer
34 views

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

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 ...
6
votes
2answers
240 views

Does absolutely flat commutative ring imply all ideals are idempotent?

From reading Atiyah and MacDonald, I know of the result that a absolutely flat commutative ring has all principal ideals idempotent. Reading around on math reference, I think that if a commutative ...
2
votes
1answer
43 views

$\text{Hom}_k(M,N)\cong M^*\otimes_k N$ as Hopf-algebra modules.

I'm reading Representations and Cohomology by D.J. Benson. At the beginning of the third chapter the following is explained: Let $\Lambda$ be a bialgebra over $R$ and $M,N$ left $\Lambda$-modules. We ...
-1
votes
2answers
67 views

Are polynomial rings finitely generated modules over the base ring?

Let $R$ be a commutative ring. Consider $R[X_1,...,X_n]$. Clearly it is a natural $R$-module. Is it true that $R[X_1,...,X_n]$ is a finitely generated $R$-module? If it is, then every its ideal is ...
2
votes
1answer
48 views

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

Let $R=\{a_0+a_1X+\cdots+a_nX^n\;|\;a_0\in\mathbb{Z},a_1,a_2,\dots ,a_n\in\mathbb{Q}, n\in\mathbb{Z}_{\geq 0}\}$ and $I=\{a_1X+\cdots+a_nX^n\;|\;a_1,a_2,\dots ,a_n\in\mathbb{Q}, n\in\mathbb{Z}^+\}$. ...
0
votes
3answers
123 views

Integral Dependence & Finitely Generated Modules

How to prove $(3)\Rightarrow(1)$ of this theorem: Let $A\subseteq B$ be commutative rings. The following are equivalent: $(1)~~x\in B$ is integral over $A$; $(2)~A[x]$ is a finitely generated ...
1
vote
0answers
23 views

Injective $A$-homomorphism is also surjective? [duplicate]

Let $A$ be a ring and let $M$ be an Artinian module over $A$. Let $f: M \to M$ be an $A$-homomorphism. Assume $f$ is injective. Does it follow that $f$ is surjective? I'm inclined to think yes, and ...
2
votes
1answer
31 views

Maps involving direct sums of modules

This might seem like a strange question... I am curious: If $A$ is a ring and $M_1, M_2, M_3, M_4$ are $A$-modules, and we have a map $f: M_1\oplus M_2 \rightarrow M_3\oplus M_4$, then is there some ...
5
votes
2answers
52 views

S,R bimodules subcategory of the category of S+R modules?

If $R$ and $S$ are commutative rings, then does the category $R \oplus S$-modules encompase the category of $(S,R)$-bimodules? I was thinking we can accomplish this by defining the action to be: ...
3
votes
0answers
87 views

Geometric interpretation of algebraic property

Let $X \subset \mathbb{C}^n$ be an irreducible affine algebraic curve with coordinate ring $$\mathbb{C}[X] = \mathbb{C}[z_1, \ldots, z_n] / (f_1, \ldots, f_m ) $$ with each $f_i \in \mathbb{Z}[z_1, ...
0
votes
2answers
34 views

How to solve this problem about faithful module.

Let $R$ be a non-zero commutative ring with identity and $M$ a unital $R$-module. The $R$-module $M$ is called faithful if $rM=0$ for $r\in R$ implies $r=0$. Let $M$ be a finitely generated ...
0
votes
1answer
26 views

Can I conclude that the following map is surjective?

I have a module homomorphism $A \rightarrow B \oplus C $ whose projection onto the first factor $B$ is surjective. If the projection onto $C$ is surjective and $C$ is a simple module, can I conclude ...
1
vote
1answer
30 views

Projective modules over a direct product of rings

Let $R$ and $S$ be rings, and let $\text{proj}(R)$ denote the category of finitely generated projective modules. Is there an equivalence of categories between $\text{proj}(R \times S)$ and ...
0
votes
0answers
57 views

Showing that a submodule of $\mathbb Z$ is free

$\textbf{Definition:}$ A free module is a module that has a basis. If I wanted to show that a $\mathbb{Z}$-submodule of $\mathbb Z$ is free, what would be a way of doing so? Since a ...
2
votes
1answer
23 views

If $\{v_0,v_1\}$ and $\{v_1,v_2\}$ form a basis for $\mathbb Z^2$ then $v_2=-v_0+a_1v_1$

I am reading the section on non-singular surfaces from Fulton's book, Introduction to Toric varieties. There we have the following - Let $v_0,v_1,\dots,v_d=v_0$ be a sequence of lattice points in ...
0
votes
2answers
31 views

Embedding $\mathbb{Z}$ into a $\mathbb{Q}$-module

$\mathbb{Z}$ is not a $\mathbb{Q}$-module since if $\frac{1}{2}\cdot1=x\in \mathbb{Z}$, then $2x=1$, which is absurd. However, according to Dummit and Foote (3rd Ed, page 359), there exists an ...
2
votes
1answer
27 views

Defining a radical of a module axiomatically

In these notes by Richard Vale, he approaches the notion of a radical of a module axiomatically, by saying the following. A radical is an assignment, to each R-module $M$, of a submodule $\tau(M) ...
8
votes
2answers
214 views

Why is this statement about generators of groups true?

Let $G$ be free abelian of rank $n$ and $H \subseteq G$ a subgroup also of rank $n$. It is known that $G/H$ is finite, in fact a direct sum of at most $n$ cyclic groups. Thus we can write $$G/H = ...
1
vote
1answer
67 views

Is an $R$-module equivalent to an $R[X]$-module?

This is probably a really simple question but I'm a little confused. I'm reading through a proof that concerns a finitely generated $R$-module $M$. The very first line is: Consider $M$ as an ...
0
votes
1answer
25 views

on the splitting of a presentation

Let $P$ be an $R-$module and let $F$ be a free module (say $F$ isomorphic to $R^n$). If $P$ is a summand of $F$, clearly there is a presentation (namely a surjective $R-$homomorphism) $p:F\to P$. ...
0
votes
1answer
51 views

Homomorphism between free modules [closed]

Assume $n>m$. Is it possible to have an injective $R$-module homomorphism from $R^n$ to $R^m$? I feel this is possible, the homomorphism is determined by where we send each basis. So we just ...
1
vote
1answer
28 views

Modulo Arithmetic - Find smallest divider greater or equal to.

I'm looking for a nice solution to the following problem: x mod(y + d) = 0. Where x is a positive integer and y is a positive integer smaller or equal to x. I'm looking to find the smallest d ...
0
votes
1answer
50 views

How does one compute quotients of $\Bbb Z$-modules?

Doing homology computations inevitably involves computing quotients of $\Bbb Z$-modules, and I am not familiar with any way of doing this. Some examples I have been working with are: $$\frac{\Bbb Z ...
2
votes
0answers
24 views

Compute directly that the mapping cone of a homotopy equivalence is contractible

Let's consider the category $Ch_R$ of cochain complexes of modules over a commutative ring $R$. I'm trying to prove that if the chain map $\phi:M\rightarrow N$ is a homotopy equivalence then its ...
0
votes
1answer
28 views

a question about finitely generated modules

Let $M$ be an $R-$module. Let $R^n$ be the direct sum of $n$ copies of $R$. Prove that if there is a surjective $R-$homomorphism $\pi:R^n\to M$ then $M$ is finitely generated. Clearly $R^n$ is ...
1
vote
1answer
30 views

Show that for a Noetherian ring R, the R-module R$^{n}$ is Noetherian

I'm allowed to use this: An R-module M is Noetherian when a submodule N is Noetherian and their quotient M/N is Noetherian. In my particular problem, M = R$^{n}$ and N = R. I believe I will have ...
1
vote
1answer
38 views

Baer Sum notation requires clearence.

I am working on Baer sum and I have my book by Rotman, Introduction to Homology, and also MacLanes book Homology and they use notation I am puzzled on. I have understood baer sum of extensions ...
0
votes
2answers
55 views

Subgroups of $\mathbb Z^5$

Is there a nice way to see that any subgroup of $\mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z$ can have at most 5 generators? I know that $\mathbb Z$ is Noetherian, so ...
3
votes
1answer
42 views

Baer sum of $\mathbb{Z}_9$ and $\mathbb{Z}_9$

I am working on trying to figure out the third extension of $\mathbb{Z}_3$ by $\mathbb{Z}_3$, I know one is $\mathbb{Z}_9$ and the neutral element (with respect to baer sum) $\mathbb{Z}_3\oplus ...
0
votes
1answer
12 views

$M/aM\rightarrow N/aN$ surjective implies $N/N'/a(N/N')$ is zero

I found the following statement in a problem sheet which I don't know how to approach. This is the following: If $u:M \rightarrow N$ is an $A$-module homomorphism and we denote $N'$ the image of $u$. ...
2
votes
1answer
68 views

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 ...
1
vote
2answers
36 views

Atiyah and Macdonald's proof of the existence of the tensor product

I have a question regarding the proof of the proposition 2.12 in Atiyah and Macdonald's book. They say: " Let C denote the free A-module $A^{(M \times N)}$. The elements of C are formal linear ...
1
vote
1answer
200 views

Localization of modules; sums commute

Let $B_i$ be a family of $R$ submodules of a module $B$. Show $$S^{-1}(\sum_{i\in I}B_i) = \sum_{i\in I} S^{-1}B_i$$ as $S^{-1}R$ submodules of $S^{-1}B$. If it helps, we know that ...
1
vote
0answers
39 views

How to show localization commutes with finite products/arbitrary coproducts?

I'm trying to do the following exercise from Vakil's notes on Algebraic Geometry: 1.3.F. EXERCISE. Show that localization commutes with finite products, or equivalently, with finite direct ...
4
votes
1answer
55 views

Direct limit of completions of finitely generated submodules

Let $A$ be a noetherian, local, integral domain with maximal ideal $\mathfrak m$. Moreover let $M$ be an $A$-module; I'd like to know if there exists an explicit expression of the module: ...
0
votes
0answers
17 views

Finite generation and associated graded modules

My question is as follows: Let $R$ be a ring and let $M$ a right $R$-module. Suppose that $(F_i)_{i \geqslant 0}$ is a filtration of $R$ and $(M_i)_{i \geqslant 0}$ is a compatible filtration of ...
1
vote
1answer
43 views

Irreducible Module in $\mathbb Q$ and $\mathbb C$ [closed]

Consider $V=(x,y,z)$ in $\mathbb{Q}^3$ and $\mathbb Q[X]$ a polynomial ring. Define $\mathbb Q[X]$-module by $X(x,y,z)=(-5z+5y,x+z,z)$. How can I show that V is irreducible $\mathbb Q[X]$-module? Now ...