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

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4
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2answers
62 views

Proof of the properties of tensor product

On page 25 of Atiyah-Macdonald "Introduction to commutative algebra", the author says that "We shall never again need to use the construction of the tensor product given above and the reader may ...
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2answers
42 views

$\Bbb{Z}[X]/\langle f\rangle$ is a finitely generated $\Bbb{Z}$-module

Let $f$ be a monic polynomial in $\Bbb{Z}[X]$. Show that $\Bbb{Z}[X]/〈f〉$ is a finitely generated $\Bbb{Z}$-module. I don't even know how to start. If $g\in\Bbb{Z}[X]/〈f〉$, we are trying to find ...
3
votes
2answers
44 views

$\Bbb{Q}$ is not a finitely generated $\Bbb{Z}$-module

I'm trying to show that $\Bbb{Q}$ is not a finitely generated $\Bbb{Z}$-module. Assume to the contrary that $$\Bbb{Q}=\Bbb{Z}\dfrac{a_1}{b_1}+...+\Bbb{Z}\dfrac{a_n}{b_n}$$ where $a_i,b_i\in\Bbb{Z}$. ...
0
votes
1answer
26 views

Isomorphism of quotient of direct sum modules

Let $M, N, M'$ and $N'$ be R-modules. If $M'$ and $N'$ are submodules of both $M$ and $N$ then is it true that \begin{equation} \frac{M}{M'} \oplus \frac{N}{N'} \cong \frac{M \oplus N}{M' \oplus N'} ...
3
votes
1answer
33 views

Homomorphism from a finitely generated module to a direct sum of modules

Let $R$ be a commutative ring with unit. If $M$ and $N_i$ are arbitrary $R$-modules, the module $\operatorname{Hom}_R(M,\bigoplus_{i\in I}N_i)$ is not isomorphic to $\bigoplus_{i\in ...
1
vote
1answer
44 views

Example of non noetherian ring and noetherian $\Bbb Z$-module

a non Noetherian ring that is a Noetherian $\Bbb Z$-module a Noetherian ring that is a non Noetherian $\Bbb Z$-module I have no idea in 1, and I'm not sure if $\mathbf{Q}$ is right for 2? ...
4
votes
2answers
66 views

Free finitely generated modules

Let $A$ be a ring and consider the free modules $A^{\oplus n}$, $A^{\oplus k}$, with $n,k\in \mathbb{N}$. Can $A^{\oplus n}$ be isomorphic to $A^{\oplus k}$ if $k\neq n$? Thanks in advance for the ...
2
votes
1answer
41 views

Are the two ways of creating an $S^{-1}A$ algebra equivalent?

Let $f:A\to B$ be a ring homomorphism and $S$ be a multiplicative set, define $S^{-1}B$ to be $B\times S$ with equivalence relation $(b,s)\sim(b',s')$ iff $\exists t\in S$ such that $t(sb'-s'b) = 0$. ...
3
votes
4answers
160 views

How to show $\mathbf{Q} $ is not free

We know that torsion free plus finitely generated $\rightarrow$ free and that $\mathbf{Q}$ is torsion free is easy. But how to show $\mathbf{Q}$ is not finitely generated and not free?
0
votes
1answer
23 views

When is a homomorphism an epimorphism?!

I want to prove the following characterization of an $R$-module homomorphism $g$ to be surjective: "whenever the composition of $g:M→N$ and $k:N→Y$ is zero, then $k=0$". It is easy to go one side: ...
4
votes
1answer
19 views

$M$ noetherian, $f$ endomorphism of $M$, $\operatorname{coker}f$ has finite length, then $\operatorname{coker}f^n$ and $\ker f^n$ have finite length.

Let $M$ be noetherian and let $f$ be an endomorphism of $M$. Suppose that $\operatorname{coker}f$ has finite length. Prove that both $\operatorname{coker}f^n$ and $\ker f^n$ have finite length ...
0
votes
1answer
22 views

Free module over a ring with identity with a basis of size m, ∀m≥n

Please, help on this Exercise [Hungerford's Algebra, IV.2.12] If $F$ is a free module over a ring with identity such that $F$ has a basis of finite cardinality $n\geq 1$ and another basis of ...
5
votes
2answers
139 views

Is this particular module flat?

Let $A=k[x^2,xy,y^2]\hookrightarrow B=k[x,y]$, where $k$ is a field. Is $B$ flat over $A$? I am guessing the answer is no. My first thought is, since $B$ is integral over $A$, so it's finitely ...
1
vote
2answers
65 views

On a proof that left artinian implies left noetherian

Questions: [Refer to below] Could one elaborate on $\rm\color{#c00}{(a)}$, $\rm\color{#c00}{(b)}$ and $\rm\color{#c00}{(c)}$ ? My thoughts : $\rm\color{#c00}{(a)}$ For $r+J\in R/J$ and ...
3
votes
1answer
23 views

bilinear maps with respect to noncommutative rings

Consider a noncommutative ring with unity $R$, three left $R$-modules $M,N,P$ and a map $f\colon\;M\times N\to P$ such that: $ f(m+m',n)=f(m,n)+f(m',n)\\ f(m,n+n')=f(m,n)+f(m,n')\\ ...
0
votes
2answers
58 views

not free modules

I'm not sure if this result is true or not. Let $R$ be a commutative ring and M a left $R$-module, if P is not free as a sub-module of M, can we say that M is not free either? thank you for your time
4
votes
1answer
30 views

Picard group of $\mathbb Z[\sqrt{-5}]$

I search for a simple proof for the fact that $\operatorname{Pic}(\mathbb Z[\sqrt{-5}])=\mathbb Z/2\mathbb Z$, where $\operatorname{Pic}(R)$ is the Picard group of the ring $R$ - the set of ...
2
votes
1answer
22 views

Isomorphism between modules over a semisimple ring

If $P$ is a module over the semisimple ring $R/J$, where $R$ is a semilocal ring having $1$, and $J$ is its Jacobson radical, does any isomorphism $P⊕...⊕P≅P'⊕...⊕P'$ with the same (finite) number of ...
0
votes
0answers
14 views

From progenerators to progenerators

I know that if $R$ is a ring (with identity) and $P$ is a progenarator right $R$-module (a f.g. projective generator) then $P/PJ$ is clearly f.g. when $J$ is the Jacobson radical of $R$; but, how ...
3
votes
1answer
40 views

Are finitely presentable modules closed under extensions?

If $0 \to A \to B \to C \to 0$ is an exact sequence of modules, and $A$ and $C$ are finitely presentable, then is $B$ finitely presentable? The answer is "yes" if we replace modules with groups, ...
1
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1answer
60 views

Given a, b How many solutions exists for x, such that: $a \bmod{x}=b $

Given $a, b$. How many solutions exists for $x$, such that: $$a \bmod{x}=b $$ By example: $a = 21$ and $b = 5$ $21 \bmod{8} = 21 \bmod{16} = 5$ Then $x$ has 2 solutions
7
votes
2answers
56 views

Is Orzech's generalization of the surjective-endomorphism-is-injective theorem correct?

In math.stackexchange answer #239445, Makoto Kato quoted a statement from the paper Morris Orzech, Onto Endomorphisms are Isomorphisms, Amer. Math. Monthly 78 (1971), 357--362. The statement ...
1
vote
1answer
20 views

Prove that if $M$ is an $R-$ projective left module then $M/IM$ is an $R/I-$ projective left module. [duplicate]

Let $I$ ba a two-sided ideal of a ring $R$ and $M$ be an $R-$ left module. Prove that if $M$ is an $R-$ projective left module then $M/IM$ is an $R/I-$ projective left module. It is easy to see that ...
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vote
1answer
14 views

Let $M$ be an $R-$module and $x\in M\setminus\left\{ 0\right\} $. Prove that there exists a left ideal of $R$, say $I$ such that $Rx\cong R/I $.

Let $M$ be an $R-$module and $x\in M\setminus\left\{ 0\right\} $. Prove that there exists a left ideal of $R$, say $I$ such that $Rx\cong R/I $. Help me some hints. Thank you in advance.
0
votes
0answers
42 views

Two properties related to semisimple rings

Let $R$ be a semisimple ring Show the following (i) If $xy=1 \in R$, then $yx=1$. (ii) If $x \in R$ is such that $xR$ is a left ideal of $R$, then $xR=Rx$. I am pretty lost with the two items. I ...
1
vote
2answers
28 views

IM is a submodule of M

In the fisrt answer of this question where $R$ is a ring, $I$ is a left ideal of $R$ and $M$ an $R$-module; I don't know why can't I see that $IM$ is closed under addition. If we take two elements $x$ ...
1
vote
1answer
23 views

Equivalence of categorical coproduct proof

quiLet $C$ be an abelian category and {$X_1$,...,$X_n$} a finite family of objects in that category. ( $X$,($M_i$: $X_i$$\to$ $X$) where $i_1$=1,....n a coproduct of the finite family if and only if ...
0
votes
2answers
30 views

Prove that an exact sequence splits

Let $0 \to r\mathbb{Z}_n \to \mathbb{Z}_n \to s\mathbb{Z}_n \to 0$ where n =rs an exact sequence of $\mathbb Z$ modules the how can I prove the sequence splits if and only if $(r,s)=1$ The only thing ...
2
votes
1answer
20 views

Semisimplicity of the ring $\mathbb Z_n$

I am being asked to figure out when $\mathbb Z_n$ is a semisimple ring. It is clear to me that if $n$ is prime then $\mathbb Z_n$ is simple, which implies it is semisimple. If $n=p_1...p_n$ is a ...
0
votes
0answers
20 views

Simple problem about morphism in abelian categories

$f$ : $X\to$ $Y$ and $g$ : $Y\to$$Z$ a sequence in abelian categories. Show that if $gf$=$0$ if and only if exist a monomorphism $h$:$Im(f)$ $\to$ $Ker(g)$ such $kh$=$j$, where $j$:$Im(f)$$\to$ $Y$ ...
0
votes
1answer
24 views

Exact sequence of modules exercise

Show that if $$0 \rightarrow M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3$$ is an exact sequence of $R$-modules, then for all $R$-module $$0 \rightarrow \operatorname{Hom}_R(M,M_1) \xrightarrow{f_*} ...
3
votes
1answer
36 views

Finite abelian groups (application of structure theorem)

Problem Find all finite abelian groups that simultaneously have exactly $7$ elements of order $2$, exactly8 elements of order $3$, exactly $8$ elements of order $4$, at least an element of order ...
1
vote
1answer
52 views

Constructing pullback and pushout problem

i) Let $p$ be a prime and $f: \Bbb Z \rightarrow \Bbb Z_p$ and $g: \Bbb Z_{p^2}\rightarrow\Bbb Z_p$ be the canonical epimorphism. Show that the pullback of $f$ and $g$ is isomorphic to $\Bbb Z ...
3
votes
1answer
57 views

Are two bimodules isomomorphic as left and right modules also isomorphic as bimodules?

let R be a commutative ring, and M, N two bimodules over R, such that there exists f : M -> N an isomorphism of left R-modules, and g : M -> N an isomorphism of right R-modules. Then are M and N ...
1
vote
1answer
35 views

Exact sequence and Noetherian modules

Let $R$ be a ring, $X,Y,Z$ and $T$ four $R-$ modules such that there exists a short exact sequence $$0 \rightarrow X \xrightarrow{f_1} Y \xrightarrow{f_2} Z \xrightarrow{f_3} T \rightarrow 0 $$ Prove ...
3
votes
1answer
38 views

Let $R=M_n(D)$, $D$ is a division ring. Prove that every $R-$simple module is isomorphic to each other. [duplicate]

Let $R=M_n(D)$, $D$ is a division ring. Prove that every $R-$simple module is isomorphic to each other. I need some hints to prove it. Thank you very much.
3
votes
2answers
49 views

If $M/N$ is isomorphic to $M/N'$, can we conclude that $N$ is isomorphic to $N'$?

Let $N$ and $N'$ be submodules of $R-$module $M$. If $M/N$ is isomorphic to $M/N'$, can we conclude that $N$ is isomorphic to $N'$? I believe the answer is no but I can't find a counter example. Help ...
1
vote
1answer
21 views

Finitely generated modules over a PIR (structure theorem application)

Let $R:=\mathbb R[x]/\langle (x^2+1)^2 \rangle$ and $J:=\langle x^2+1 \rangle \lhd R$. Prove that every finitely generated $R$-module is isomorphic to $R^m \oplus (R/J)^n$ for a unique pair of non ...
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vote
0answers
29 views

Group presentation and Smith normal form

Problem Let $A=\langle a,b: a-3b=0,3a=3b \rangle$ and $B=\mathbb Z^3/S$ with $S=\{m \in \mathbb Z^3: m_1+2m_2+m_3=0, 5|m_3 \}$. Calculate $\operatorname{Hom}_{\mathbb Z}(A,B)$. My attempt at a ...
3
votes
1answer
92 views

Atiyah-MacDonald Ch. 4 exercise 20: what's the module analogue of $\sqrt{\mathfrak{a}+\mathfrak{b}} = \sqrt{\sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}}$?

Atiyah-MacDonald exercises 20-23 in chapter 4 develop a theory of primary decomposition for modules, in analogy with the theory developed in the chapter for rings. Exercise 20 begins with this ...
0
votes
1answer
20 views

Representatives of simple $\mathbb C[\mathbb D_3]$-modules (left modules)

Problem Let $\mathbb D_3$ be the symmetry group of the equilateral triangle. Give a complete list of the representatives of the simple left $\mathbb C[\mathbb D_3]$-modules. My attempt at a solution ...
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votes
0answers
27 views

Module of representation matrix

Can someone please show me why the module of any representation matrix in a one-dimensional representation of a finite group is equal to 1? and please define module of a representation as well. ...
2
votes
0answers
27 views

Characterize all $\mathbb Z[i]$ modules of order $21$ and $65$

I am trying to figure out how many $\mathbb Z[i]$ modules (up to isomorphisms) are of $21$ and $65$ elements. I've done a few similar exercises for finite abelian groups ($\mathbb Z$ modules) but I am ...
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vote
0answers
25 views

Let $p,q$ and $r$ be positive prime numbers. Determine the number of abelian groups of order $p^6q^3r$

Let $p,q$ and $r$ be positive prime numbers. Calculate the number of non isomorphic abelian groups of order $p^6q^3r$. I've tried to use the structure theorem. So we have $$G \cong \mathbb Z/\langle ...
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vote
0answers
16 views

Invariant factors of finite abelian group

Calculate the invariant factors of the group $G=\mathbb Z_{12} \oplus \mathbb Z_{21} \oplus \mathbb Z \oplus \mathbb Z \oplus \mathbb Z_{20} \oplus \mathbb Z_{9} \oplus \mathbb Z_7$. Applying the ...
0
votes
0answers
39 views

$G$ finite abelian group, $p$ prime that divides order of $G$

Problem Let $G$ be a finite abelian group and $p$ a positive prime that divides $|G|$. Show that the number of elements of order $p$ in $G$ is coprime with $p$. Let $|G|=p^nm$ with $n \geq 1$ and ...
2
votes
1answer
33 views

How can I visualize or understand a module in concrete terms?

As I understand it, a module is a ring action on an abelian group, usually a group that carries the operation of addition. I get this because I think of it as a set of elements where these abstract ...
2
votes
1answer
27 views

Is {0} a free module?

Is $\{0\}$ a free module (over any ring $R$) ? A free module is isomorphic to $R^n$, but is $n=0$ allowed? Alternatively, a free module is defined to have a set of linearly independent generating ...
2
votes
1answer
44 views

Is there a non-projective submodule of a free module?

Is there an example of a commutative ring $R$ with identity such that there exists a free $R$ module $M$ that has a non-projective submodule? I tried experimenting with modules over $\mathbb Z$ but ...
3
votes
1answer
30 views

Abelian group of order $p^2q^2$ ($p$,$q$ distinct primes) determine number of elements of order $pq$ and $pq^2$

Problem For each abelian group of order $p^2q^2$ determine the number of elements of order $pq$ and the number of elements of order $pq^2$ in $G$. By the structure theorem we have that $$(1) ...