1
vote
1answer
23 views

Help understanding statement relating to structure of modules over PIDs

Lemma IV.6.11 of Hungerford's Algebra is Lemma 6.11. Let $R$ be a principal ideal domain. If $r \in R$ factors as $r = p_1^{n_1} \cdots p_k^{n_k}$ with $p_1,\ldots,p_k \in R$ distinct primes and ...
3
votes
2answers
57 views

module over a quotient of a principal ideal domain

The Statement I suspect the following proposition is well known, but I found no reference. Proposition If $A$ is a principal ideal domain, if $I$ is a nonzero ideal of $A$, and if $M$ is an ...
0
votes
0answers
24 views

Smith normal forms and a math program

I am interested to know the Smith normal form of $4 \times 2$ matrices $M$: The two cases of my interests are: (1). $$M_1= \begin{pmatrix} 3 & 0\\ -5 & 4\\ 4 & -5\\ 0 & 3 ...
2
votes
1answer
21 views

Show that if $R$ is principal, $N $ is pure and $Ann(x+N)= Rd$ then there exists $y \in M$ such that $x+N=y+N$ and $Ann(y)=Rd$

Let $R$ a integral domain and $M$ a $R$-module. A submodule $N$ of $M$ is pure if for all $x \in M$ and $a \in R$ such that $ax \in N$, there exists $y \in N$ such that $ax=ay$. Show that if $R$ ...
7
votes
2answers
146 views

$M \oplus M \simeq N \oplus N$ then $M \simeq N.$

Let $M$ and $N$ be finitely generated $R$-modules where $R$ principal domain. Show that if $M \oplus M \simeq N \oplus N$ then $M \simeq N.$
0
votes
2answers
51 views

Show that an integral domain $R$ is principal if and only if every submodule of a cyclic $R$-module is cyclic.

Good morning, I have difficulty with this problem: Show that an integral domain $R$ is principal if and only if every submodule a cyclic $R$-module is also cyclic.
-1
votes
2answers
24 views

$Hom(F,T)$, where $F$ is torsion-free and $T$ is torsion module [closed]

If the underlying ring is a PID, then is it true that $Hom(F,T)=0$?
2
votes
1answer
20 views

Quick question: SES where base ring is a PID

Let $M$ be an $R$-module, where $R$ is a PID. Assume there is a free $R$-module $F$ and a surjective map $\phi :F\rightarrow M$. Then why is $\ker(\phi)$ also free? Thank you.
0
votes
1answer
45 views

A submodule of a free module over a PID

$R$ is a principal ideal domain and $F$ is a free module over $R$ of infinite rank with basis $\{e_1,...,e_n,...\}$. Is it true that the $R$-submodule of $F$ spanned by $\{e_1,...,e_n\}$ has a ...
3
votes
1answer
73 views

Structure Theorem For PIDs

So, I'm a biologist at KCL, but I quite like mathematics and so am going through a book of exercises in algebra. Unfortunately, I've run into a problem in trying to answer some of the questions. I've ...
1
vote
1answer
45 views

Projective Modules on PIDs

We are given a PID. Then we have to show that if F is a finitely generated free module of rank n, then a submodule M of F is free. This is fine. However, it then says to deduce that every finitely ...
0
votes
1answer
24 views

Show $M(\mathfrak{p})$ is an $R$-submodule of $M$

Let $R$ be a PID, $\mathfrak{p}$ be a prime ideal in $R$, and $M$ an $R$-module. Then $$M(\mathfrak{p})=\{m\in M \mid \operatorname{Ann}_Rm=\mathfrak{p}^j, \, \text{ for some } j\geq 0 \}$$ is ...
-1
votes
1answer
123 views

An ideal in a domain is free if and only if it is principal [closed]

Let $R$ be a domain. Show that every ideal in $R$ is $R$-torsion-free, but is free if and only if it is principal. In particular, show that $R$ is a PID if every submodule of a free module is free. ...
3
votes
2answers
53 views

If $N\cap rM=rN$ for all $r\in R$, then is $M=N\oplus K$ for some $K$?

Suppose $M$ is a finitely generated free module over a principal ideal domain $R$, and $N$ a submodule. Why does the condition $N\cap rM=rN$ for all $r\in R$ implies that $M=N\oplus K$ for some ...
1
vote
1answer
100 views

Infinite direct product of rings free.

Let $A$ be a commutative ring (viewed as an $A$-module over itself) that is not a field. Are there some conditions that guarantee that $\prod_{k=0}^\infty A$ is free? What if $A=\mathbf{Z}$ or more ...
1
vote
1answer
38 views

irreducible elements of polynomial rings

Let $p$ be a prime integer. For $x\in\mathbb{Z}$, let $x'$ be the remainder of $x$ when divided by $p$. Let $\sum_{i=0}^{n}a_iX^i\in \mathbb{Z}[X]$ with $p$ does not divide $a_n$ in $\mathbb{Z}$. Then ...
0
votes
1answer
127 views

Prove that $M$ is a free module if and only if $M$ is a projective module over $PID$.

Let $R$ be a principal ideal domain and $M$ a finitely generated $R$ module. Prove that $M$ is a free $R$-module if and only if $M$ is a projective $R$-module. I am quite confused and totally not ...
0
votes
0answers
48 views

Invariant factors of torsion module under homomorphism

Let M be a torsion module for a PID D with invariant factors $(d_1) \supseteq (d_2) \supseteq...\supseteq (d_r) $ (which means $M = Dz_1 \bigoplus Dz_2 \bigoplus ... \bigoplus Dz_r $ s.t $ann(z_i) ...
0
votes
1answer
141 views

Find a non free submodule of a free module over R which is not PID

I try to solve following question. Show that $R=\mathbf{Q}[x,x^{-1}]$ is not a PID. Construct a free module over $R$ having a non free submodule. One may give some examples for free modules but not ...
0
votes
1answer
52 views

“Two ways” to reduce a module

Let $M$ be a module over a principal ideal domain $R$ and $\mathfrak{m}$ a maximal ideal of $R$ with residue field $R/\mathfrak{m}=k$ of characteristic $p$. Under what circumstances are the modules ...
2
votes
1answer
85 views

Reference on Operators on Modules over PID

The operators on finite dimensional vector spaces over any field can be seen in nice form w.r.t. some choice of basis such as "diagonal, triangular, Jordan form, rational form etc." But, for ...
3
votes
0answers
345 views

Exercise on modules over PID involving injective modules, Baer's criterion.

I'm interested if I solved this somewhat correctly, and would like to be set straight if it is wrong. This is an exercise from an introductory text. Let $A$ be a module over a principal ideal ...
4
votes
1answer
189 views

Question about torsion submodules and Decomposition theorem

Let $A$ be an principal ideal domain, and $M$ an $A$-module. If $p$ is irreducible in $A$, let's define $$\mathrm{Tor}_p(M):=\{m\in M\mid p^km=0\text{ for some }k\in\mathbb{N}\}.$$ I need to show ...
3
votes
1answer
1k views

Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
2
votes
1answer
156 views

How can we write this $\mathbb{Q}[x]$-module as a direct sum of cyclic $\mathbb{Q}[x]$-modules?

If $L$ is the submodule of $\mathbb{Q}[x]^{(3)}$ generated by $(2x-1,x,x),(x,x,x),(x+1,2x,x)$. How do we write $\mathbb{Q}[x]^{(3)}/L$ as a direct sum of cyclic modules?
3
votes
1answer
134 views

Do a matrix and its transpose have the same invariant factors over a PID?

I suspect this is true since it holds in the case over a field. But suppose $A\in M_{m\times n}(R)$ where $R$ is a PID. Does it still hold that $A$ and $A^{T}$ have the same invariant factors? ...
2
votes
2answers
127 views

Sub-module over the Fraction field of a PID

R is a PID with field of fractions K. $M \subseteq K$ is a fin. generated R-Submodule. I am trying to show M is in fact generated by one element.
1
vote
1answer
290 views

Isomorphism between set of all R-modules homomorphisms

Let $S$ be an $R$-module where $R$ is an integral domain and let $P = \langle p \rangle $ be a prime ideal. Define: $S_{P}=\{s \in S: p^{n} s=0 \ \textrm{for some natural }n\}$. As usual, denote ...
10
votes
1answer
347 views

Finitely generated modules over PID

Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus $ $B$ $\cong$ $C\oplus $ $D$ and $A\oplus $ $D$ $\cong$ $C\oplus $ $B$ . Prove that $A$ $\cong$ $C$ and $B$ ...
8
votes
1answer
565 views

Proofs of the structure theorem for finitely generated modules over a PID

I'm looking for different proofs (references or sketch of main ideas) of the structure theorem for finitely generated modules over a PID. If possible, a comparison in terms of clarity, elegance or ...