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

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6
votes
0answers
83 views

On describing a sort of “well-behaved” subgroups of a free abelian group.

I found this question when I tried to figure out what kind of subgroups of a free abelian group behave just as well as in the finitely generated case. Let $M$ be a free abelian group and $N$ a ...
2
votes
1answer
25 views

Definition of Torsion Submodule

If $R$ is an integral domain one defines an $R$-module $M$ to be torsion-free if the torsion submodule $T(M)=\{x\in M\mid l_R(x)≠ 0\} $ is zero, where $ l_R(x)$ is the left annihilator of $x$ in $R$. ...
1
vote
1answer
32 views

Let $R$ be an infinite comutative ring with unity, $M,N$ be $R$-modules, $f:M \to N$ be a surjective module homomorphism; then $|M|=|N ||\ker f|$?

Let $R$ be an infinite commutative ring with unity, $M,N$ be modules over $R$, let $f:M \to N$ be a surjective module homomorphism; then is it true that $|M|=|N || \ker f|$ ($M,N$ are not ...
1
vote
1answer
19 views

Indecomposable module, Algebra

Let R be $K[x]$, the the ring of polynomials over K, and $A \in M_n(K)$. Then $K^2$ is a R-modul by using the Matrix A, with $p(x) \circ v=p(A)v$, where $v \in K^2$ and $p \in K[x]$. Now I got show ...
5
votes
2answers
853 views

Isomorphism between quotient modules [closed]

Is it true for a commutative ring $R$ and its ideals $I$ and $J$ that if the quotient $R$-modules $R/I$ and $R/J$ are isomorphic then $I=J$?
0
votes
1answer
36 views

Isomorphism of modules [duplicate]

Are $\mathbb{C}[x,y]/(x,y)$ and $\mathbb{C}[x,y]/(x-1,y-1)$ isomorphic as $\mathbb{C}[x,y]$-modules? I think they are cyclic so they are isomorphic, but I'm not sure.
0
votes
0answers
34 views

Find an $R$-module homomorphism $f:R\longrightarrow M$ such that $r_0m=0$ implies $r_0f^{-1}(m)=0$ , $(m\in M)$

Let $R$ be a commutative ring with $r_0\in R$ a fixed element, and $M$ be an $R$-module. I search for an $R$-module homomorphism $f:R\longrightarrow M$ such that $r_0m=0$ implies $r_0f^{-1}(m)=0$ , $(...
1
vote
0answers
22 views

(B,N) pair and Steinberg idempotent

Let $q=p^f$ where $p$ is prime and $G$ be a finite group with a $(B,N)−$pair ($T=B\cap N$ and $W=N/T$), and assume that $B=UT$ with $U\triangleleft B$ and $U\cap T=1$. Define $$e=\dfrac{1}{[G:U]}\...
0
votes
0answers
10 views

Dual basis for lattices with non-degenerate pairing

Let $\Lambda \subseteq \Lambda'$ be fin. gen. free $\mathbb{Z}$-modules and let $(.,.): \Lambda' \times \Lambda \to \mathbb{Z}$ be a non-degenerate bilinear pairing that is symmetric on $\Lambda$. Let ...
2
votes
1answer
33 views

On selfinjectivity of Hopf algebras

Any group algebra $kG$ of a finite group is selfinjective. More generally Gentile proves that for a group ring $RG$ with $R$ commutative and torsion free as a $\Bbb Z$-module, $RG$ is selfinjective ...
0
votes
1answer
35 views

Let $M_1$, $M_2$ be Artinian modules over $R$. Then $M_1\times M_2$ is Artinian.

Using exact sequences, it's fairly easy to prove the converse, but I can't figure out how to prove this statement. Suppose we have a descending chain $N_1\supset N_2\supset\cdots$ of $R$-submodules ...
1
vote
1answer
40 views

Noetherian vector space is finite-dimensional

Given a field $k$, and a $k$-vector space $V$ which is noetherian as $k$-module, I want to show that $V$ is finite-dimensional. Is it correct that this follows because since $V$ is noetherian, every ...
1
vote
1answer
44 views

Modules over algebras vs Modules over Rings?

Since I've started module theory I was confused with a point. What is more general, the theory of modules over algebras or over arbitrary rings? I hope this is not a pointless question so let me try ...
0
votes
1answer
37 views

Find two generators of a $\Bbb{Q}[x]$-module $\Bbb{Q}^5$.

Let $T$ be a linear transformation on $\Bbb{Q}^5$ which is defined by $$T(v)= \begin{pmatrix} -1 & 0 & 0 & 0 & 3\\ 1 & 2 & 0 & -4 & 0\\ 3 & 1 & 2 & -4 &...
3
votes
3answers
103 views

Finitely generated projective modules over polynomial rings with integral coefficients

There is famous Quillen-Suslin theorem which states that every finitely generated projective module over a ring of polynomials $k[x_1,...,x_n]$, where $k$ is a field, is free. I have never carefully ...
1
vote
2answers
30 views

Modules as morphisms to endomorphism rings

An $A$-module $M$ may be thought of as a (surjective) ring homomorphism $f: A \to E(M)$, where $E(M)$ is a ring of group endomorphisms of $M$. Then $am = f(a)(m)$. Is there any more to this ...
1
vote
1answer
55 views

Tensor fields on a manifold

Let $M$ be an $n$-dimensional smooth manifold. It is easily shown that the modules $\Gamma(TM)$ (the real vector space of vector fields on $M$) and $\Gamma(T^\ast M)$ (the real vector space of $1$-...
0
votes
0answers
19 views

An example of a module that have no supplement.

We see that if R/J is not coclosed coprojective and J has a supplement then R/J is projective. Now we are looking for an example J has no supplement and also R/J is not coclosed coprojective. But we ...
0
votes
0answers
25 views

What properties are preserved by direct limits? [closed]

We know that direct limit of a directed family of flat $R$-modules is also flat ($R$ is a commutative ring with $1$ and all modules are unital). I am looking for other properties of modules which ...
1
vote
1answer
83 views

Show that $R$ is a field

Let $R$ be a commutative ring with unit. If $R\neq 0$ such that each finitely generated $R$-module is free then $R$ is a field. In my notes there is the following proof: We need to show that ...
1
vote
2answers
119 views

Specific basis of A-algebra B that is also a free A-module of finite rank.

I have a problem that seems (at least to me) harder then I initially thought. Let $B$ be an $A$-algebra that is also a free $A$-module of finite rank (if necessary we can assume that $B$ is ...
0
votes
1answer
52 views

Show that $\text{Hom}_R(R^n,M)\cong \prod_{i=1}^n\text{Hom}_R(R,M)$

Let $R$ be a commutative ring with unit and let $M$ be a $R$-module. It holds that $\text{Hom}_R(R^n,M)\cong \prod_{i=1}^n\text{Hom}_R(R,M)$, right? How could we prove this? Do we have to define ...
-2
votes
1answer
27 views

How can we find a generating set?

How can we find a generating set of the $\mathbb{Z}$-module $\mathbb{Z}$ that does not contain the basis which is $\{1\}$ ? I saw in my notes that such a set is the $\{2,3\}$. Why is this a ...
2
votes
2answers
81 views

$\text{Ext}_R^n(\bigoplus_{i\in I} M_i, N) = \prod_{i\in I}\text{Ext}_R^n(M_i,N)$

Let $(M_i)_{i\in I}$ be a collection of $R$-moduls. Show that for all $N\in \text{Ob}(_R\text{Mod})$ is $$ \text{Ext}_R^n(\bigoplus_{i\in I} M_i, N) = \prod_{i\in I}\text{Ext}_R^n(M_i,N). $$ My ...
0
votes
1answer
50 views

Finite type + integral = finite

Let $A \subseteq B$ be rings (comm. with unity). I am struggling to see why the following equivalence holds for $B$ interpreted as a $A$-Algebra: $A \rightarrow B$ is of finite type and $A\...
1
vote
1answer
64 views

Extension of Scalars is well-defined

The reason I'm asking this, is because as an exercise, I'm asked to prove the following: Let $A$, $B$ be rings, $f:A\to B$ a ring homomorphism inducing $A$-module structure on $B$, and $M$ a flat $A$-...
1
vote
1answer
20 views

Empty singular submodule

I search for a module $M$ with its singular submodule $Z(M)$ the empty set, i.e. for every element $m$ of $M$ the annihilator of $m$ in $R$ is not essential, say, as right $R$-module; or proving that ...
1
vote
1answer
32 views

groups as modules

Let $G$ a abelian group. Prove that $G$ is a $\mathbb{Z}$-module. Let $x\in G$ and $n\in\mathbb{Z}$ we define $xn$ as follows: If $n\geq 0$, then $x0=0$ and $x(n+1)=xn+x$. If $n<0$, then $xn=(-x)(...
0
votes
0answers
53 views

primary decomposition of injective envelope of a module

The Exercise A3.6 of Eisenbud's book, Commutative Algebra with a view Toward Algebraic Geometry, is: Assuming that $R$ is Noetherian, let $M$ be any finitely generated $R$-module. a. Let ...
0
votes
1answer
31 views

Prove that any finitely generated submodule of $R^+$ (the field of quotients) is free of rank $1$

I am working on the following problem: Let $R$ be a principal ideal domain and $R^+$ the field of quotients. Then $R^+$ is an $R$-module. Prove that any finitely generated submodule of $R^+$ is a ...
0
votes
0answers
29 views

Finding an essential submodule

Let $R$ be a commutative ring with unity, and let $M$ be a unitary faithful $R$-module. Assume the annihilator $A$ of $r\in R$ is essential in $R$ (as an $R$-module). I search for an essential ...
1
vote
1answer
22 views

Name for submodule killed by a right ideal

Let $\mathfrak a$ be a right ideal in a ring $R$. The set $N=\{m\in M: \mathfrak am=\mathfrak 0\}$ is a submodule of the left $R-$module $M$: If $m,n\in N$, $a\in \mathfrak a$, then $a(m-n) = am-an =...
1
vote
1answer
70 views

Show that $V=\mathbb KG\oplus\cdots\oplus \mathbb KG$.

Let $V$ a $\mathbb KG$-module such that the character of $V$ is such that $\chi_V(g)=0$ for all $g\in G\setminus \{1\}$. Show that there is an $m$ s.t. $$V=\underbrace{\mathbb KG\oplus\cdots\oplus \...
1
vote
1answer
45 views

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

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
70 views

Poincaré series and Hilbert polynomial of some graded modules [closed]

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, Y^2)...
0
votes
0answers
16 views

Let $M$ be a finitely generated module over a P.I.D.. If $M=A\oplus B$, $A$ is a torsion submodule and $B$ is free, then $A=M_{\text{tor}}$.

I am reading the Goodman's Algebra. There is a lemma. Let $M$ be a finitely generated module over a principal ideal domain $R$. (a) If $M=A\oplus B$, $A$ is a torsion submodule, and $B$ is free, ...
0
votes
0answers
36 views

projective resolution for an $I$-torsion $R$-module

Let $R$ be a commutative Noetherian ring with non-zero identity, $I$ be an ideal of $R$ and $M$ be an $I$-torsion $R$-module. We know that there exists an injective resolution of $M$ in which each ...
2
votes
0answers
23 views

$R$-module strcture on $M\oplus A$ so that it becomes a $R$-algebra?

Let $R$ be a commutative ring with identity, $A$ an associative $R$-algebra without identity and $M$ a $A$-bimodule. I read I can endow $M\oplus A$ with a structure of $R$-algebra with the product $$(...
1
vote
2answers
32 views

Reducing a marix to Smith Normal form

I am trying to reduce the following matrix to Smith Normal form - $$ A= \begin{pmatrix} 1&0&0\\ 1&2&0\\ 1&0&3 \end{pmatrix}$$ What ever row and column operations I try, I end ...
2
votes
1answer
58 views

A question on Auslander-Bridger transpose

I am learning Auslander-Reiten Theory. When I read the book Frobenius Algebras I. Basic Representation Theory, I have some problems. On page 236-237, there is the following Proposition 4.5. Let $...
26
votes
2answers
3k views

Tensor products commute with direct limits

This is Exercise 2.20 in Atiyah-Macdonald. How can we prove that $\varinjlim (M_i \otimes N) \cong (\varinjlim M_i) \otimes N$ ? Atiyah-Macdonald give a suggestion, they say that one should ...
3
votes
0answers
46 views

Inverse limit of flat algebras is flat?

Let $k$ be a field, let $R$ be a $k$-algebra, let $\{ S_i \}_{i \in I}$ be an inverse system of $k$-algebras, and let $R \to S_i$ be a $k$-algebra homomorphism making $S_i$ into a flat $R$-module. Is ...
0
votes
0answers
26 views

Singular ideal of an idealization

Let $S$ be a commutative ring, and let $A$ be a faithful $S$-module. Through idealization, we can make the abelian group $R=S⊕A$ into a commutative ring using the multiplication $(s,a)(s',a')=(ss',sa'+...
2
votes
1answer
34 views

Where am I going wrong in computing the cokernel of this map?

Let $A:\mathbb {Z^3\to Z^3}$ be the map given by $$A=\begin{pmatrix} 1&0&0\\ 1&2&0\\ 1&0&2 \end{pmatrix}$$ Then the cokernel of $A$ is $\mathbb Z^3/\text{ Im }A$ which can ...
0
votes
1answer
38 views

Proof of the exactness of the tensor product

In Atiyah and MacDonald, Prop 2.18 establishes that for any exact sequence $$M'\xrightarrow{f}M\xrightarrow{g}M''\xrightarrow{}0\tag{1}$$ of $A$-modules and homomorphisms, and for any $A$-module $N$, $...
2
votes
2answers
63 views

Proving/Disproving $M$ has the structure of an $R$-module

Given an abelian group $M$ and a ring $R$, how can one prove or disprove that $M$ has the structure of an $R$-module? When proving $M$ is an $R$-module, if it is not obvious how to define an action $R\...
1
vote
1answer
44 views

On boolean algebras as rings, modules and/or R-algebras

After trying to make sense of first order logic from an algebraic point of view I started to read about boolean algebras (similar to the explanations given here: wikipedia on boolean algebras. I also ...
2
votes
0answers
53 views

Finding the structure of an $F_p[X]$module

$p$ is a prime and $M$ is an $F_p[X]-$ module. Given $(X-1)^3M=0, \vert (X-1)^2M\vert=p, \vert (X-1)M\vert=p^3$ and $\vert M\vert =p^7$, determine $M$ as an $F_p[X]-$module, up to isomorphism. Using ...
1
vote
2answers
21 views

Element separated from a submodule by a homomorphism

Let $M$ be a module over a commutative ring $A$, $M'$ a submodule, and $y\in M\setminus M'$. Then $y$ can be separated from $M'$ by homomorphism. What I wish to prove is that there exists an $A$-...
2
votes
1answer
37 views

Socles and factors

Let $A$ be a finite dimensional algebra over a field $K$ and let $M$, $M'$ and $N$ be $A$-modules. Suppose that $M'\subseteq M$ and assume that $N$ has simple socle. Let $f: M \longrightarrow N$ be ...