For questions about principle ideal domains: rings without zero divisors where every ideal is principle.

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Localization of a PID is a PID

I would like a verification of a proof for the following statement. Let $S$ be a multiplicatively closed subset of a ring $R$. If $R$ is a PID, then $S^{-1}R$ is a PID. Let $I = \left<r_1/s_1, ...
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1answer
54 views

Prove that $a$ is a prime element of $R$

Let $R$ be a PID and $P = (a)$ is a prime ideal of $R$. Prove that $a$ is a prime element of $R$. Since $P$ is a prime ideal of $R$, let $x,y \in R$ s.t. $xy \in P = (a).$ (WTS $a \mid x$ or $a\mid ...
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73 views

The ring is a principal ideal domain, especially an integral domain.

The following holds for the ring $ \mathbb{Z}_p, p \in \mathbb{P}$: The ring $ \mathbb{Z}_p $ is a principal ideal domain, especially an integral domain. I try to understand the following proof: ...
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65 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 ...
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1answer
37 views

What's the importance of invariant factors?

I understand why the following theorem holds: If $R$ is a PID and $M$ is a finitely-generated torsion $R$-module, then there exist $q_1,...,q_s$, non-invertible elements of $R$, such that $q_i ...
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1answer
50 views

Domain of $\ln\left(\frac{6}{6+x-x^2}-1\right)+\arcsin\left(\frac{x+1}{3}\right)$

blob:https%3A//mail.google.com/ea67134d-45a0-4cc0-9ec7-abf6d5a50852 I believe that my first condition is wrong but I don't understand why. Can somebody please help?
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1answer
68 views

Decomposition of Free Module over PID

Let $M$ be a free module with rank $2$ over the PID $R$ having basis $B=\{b_1,b_2\}$. I have a few questions regarding the submodule $Rm$, where $m$ is some element of our module $M$. Suppose $m$ is ...
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1answer
32 views

Torsion elements with relatively prime orders

Let $x$ and $y$ be torsion elements in an R-module $M$ having orders $a$ and $b$. With $a,b \in R$, $a$ and $b$ are relatively prime, and $R$ is a PID. I want to show that $x + y$ has order $ab$. So ...
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81 views

Prime ideal being maximal ideal and PID

Q1. Does there exist an ID R in which every non zero prime ideal of type pR is maximal ideal but R is not PID? Q2. Does there exist an ID R in which every non zero prime ideal is maximal ideal but R ...
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35 views

Problem of Module on a PID

Suppose that $R$ is a $\mathrm{PID}$. Suppose that $a$ and $b$ are nonzero elements of $R$ which are relatively prime. Let $M$ be an $R$ module so that $abM=\{0\}.$ Show that $aM=M_b$ and $bM=M_a$ ...
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1answer
117 views

Trapezoidal Motion Profile Using Discrete Method

I'm trying to program an arduino to generate a Trapezoidal Motion Profile to control a DC motor with a quadrature encoder. Essentially, the user will input the desired Target Position, Max Velocity ...
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1answer
88 views

Suppose that $K$ is a field and that $f$ and $g$ are relatively prime in $K[x]$. Show that $f - Yg$ is irreducible in $K(y)[x]$.

I'm a bit confused of the notation $K(y)[x]$, is that simply $K[y][x]$ so... $K[y,x]?$ Anyways, here's my attempt at trying this before I get stuck. Since $f$ and $g$ are relatively prime, that ...
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1answer
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algebraic poset

I learn domain theory and stack in definition of algebraic poset. Recall $P$ is algebraic if for every $x\in P$,the set of compact element $y$ below $x$ is directed and has $x$ as least upper bound. ...
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213 views

Reducing multivariate rational fractions to lowest terms

I wish to simplify multivariate rational fractions to a canonical form. Thanks to some very helpful mathematically inclined people who verified that my understanding of Wikipedia was correct, I'm now ...
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133 views

Applications of the Dedekind-Hasse criterion

It is a fact that an integral domain $R$ is a principal ideal domain if and only if there is a Dedekind-Hasse function $|R|\setminus\{0\}\xrightarrow{\ \ \delta\ \ }\mathbb{N}$ on $R$, i.e. a function ...
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Elementary divisors for chains of submodules

Given free modules $N \le M$ of finite rank over a PID $R$, it's well-known that there is a basis $\{x_1,\ldots,x_n\}$ of $M$ and there are $e_1,\ldots,e_n \in R$ such that $\{e_ix_i\mid e_i \neq ...
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53 views

Shortest proof for showing $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID.

I'm looking for an easy proof for that $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID. One proof I know is to show that the field norm is a Dedekind-Hasse norm, but this proof is quite dirty( that it ...
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40 views

Are the ring of integers of the constructible numbers a Euclidean domain?

I suspect that since Euclid uses the Euclidean Algorithm to perform division on constructible numbers in Elements, the ring of integers of the constructible numbers are a Euclidean Domain, but I have ...
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Property of free submodules for a module over a PID

It's possible to produce an example of an integral domain $R$ and a free $R$-module $M$ with free submodules $L, L'$ such that $L+L'$ is not free. We can take $R=M=K[x,y]$ , $L=<x>$ , ...
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Existence of alternative basis element in free module over a PID

first question on stack exchange, please let me know if I have made any errors with formatting or in general! :) Let $f_1,f_2, ...,f_s$ be a basis of a free module $V$ over a PID $R$. Suppose that ...
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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 finite generated case. Let $M$ be an free abelian group, $N$ a subgroup ...
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62 views

$\mathbb{Z}[i]$ is principal. And what are the units

I have elements of the form $a+bi$. I have attempted to consider arbitrary ideals in $\mathbb{Z}[i]$. If $N$ is ideal and $N=\{0\}$ then it is generated by $0$. If $N$ is not trivial, then exists ...
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61 views

Is quantum torus a principal ideal domain?

For a quantum torus $C_q[x_1^{\pm1}, ...,x_n^{\pm1}]$ satisfying $x_ix_j=q_{ij}x_jx_i$. Question: Is this quantum torus a principal ideal domain?
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458 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 ...
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Showing that the integers localized at a prime, p, is a Euclidean Domain

I want to show that the integers localized at some prime natural number $p$: $$R=\Biggl\{\frac mn \in \Bbb Q ~\Bigg\vert~ m,n \in\Bbb Z,\ n\notin p\Bbb Z\Biggr\}$$ is a Euclidean Domain, but I can't ...
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39 views

For finitely generated $B$ all modules in exact sequence are finitely generated in PID

Let there be an exact sequence of $R$-modules: $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ where $R$ is a principal ideal domain and $B$ is finitely generated. Are $A$ and $C$ ...
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64 views

Rank of a matrix over a principal ideal domain

I apologize if my question is stupid but I'm not very familiar with matrices over a principal ideal domain $R$ (For example, $R=\mathbb{Z}$ or $R=\mathbb{R}[X]$). I was wondering how to define the ...
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torsion free RG-module

Let $R$ be a PID, $G$ be a cyclic group, $M$ be an $RG$-module and $N$ be a submodule of $M$. How can we test whether $M/N$ is torsion free as an $RG$-module or not? (I know how if we consider it as ...
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412 views

Proof for maximal ideals in $\mathbb{Z}[x]$

I have been trying to prove the following theorem: Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where $p$ is prime integer and $f$ is primitive integer polynomial that is ...
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0answers
33 views

Computing the invariant factors and elementary divisors of finitely generated module over a PID

I have just encountered this question in my abstract algebra class dealing with finitely generated modules over PIDs stating the following: Let $ D = \mathbb{R}[x] $ be the ring of polynomials ...
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0answers
40 views

Ideal class group of $ \mathbb{Z}[ \sqrt{2} ] $

How does one compute the ideal class group for $ \mathbb{Z}[\sqrt{2}]$? Motivation: I wish to prove that $ \mathbb{Z}[\sqrt{2}]$ is a PID. I have seen proofs which use norm and go on to show that it ...
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0answers
26 views

Under what conditions is a tower of quadratic extensions a UFD, GCD domain, or just an Integral Domain?

I have been studying towers of quadratic extensions to $\mathbb Q$ and have noticed the following: $\mathbb Q[\sqrt 2]$ and $\mathbb Q[\sqrt 2][\sqrt 3]$ are unique factorization domains(UFDs), but ...
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168 views

Is $\mathbb Z\left[\frac{1+\sqrt{-15}}{2}\right]$ a PID?

$1)$ Let $R:=\mathbb Z[w]$, where $w=\frac{1+\sqrt{-15}}{2}$. What is the norm $N_{R/\mathbb Z}(x+yw)$ in terms of $x,y\in\mathbb Z$? Which of the integers $1,\dots,10$ occur as the norm of some ...
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$\mathbb{Z}[\zeta_n]$ is a PID for $n=3,4,5$ using Minkowski theory

I want to show that $\mathbb{Z}[\zeta_n]$ is a PID for $n=3,4,5$ using Minkowski theory. I know that if the class group is trivial, then it is a PID. Is this helpful to show the claim or how else can ...
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Examples of PIDs and prime ideals

(a) Give a specific example of a PID with exactly two prime ideals. Give a brief proof of your answer. (b) Give an specific example of a PID with infinitely many prime ideals. Give a brief proof of ...
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181 views

Subring of the field of rational numbers

Let $R=\{a\cdot2^n\mid a,n \in \mathbb{Z}\}$ be a subring or the field of rational numbers $\mathbb Q$. i) What kind of elements are invertible in $R$? ii) Prove that $R$ is a principal ...
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Inclusion-minimality of a lattice basis

An integer lattice is a subgroup of $\mathbb{Z}^n$. Since $\mathbb{Z}$ is PID, each lattice has a well-defined rank and a generating set of rank many elements is a basis. I wonder if there is a way ...
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39 views

Proving the Gaussian Integers are a Principal Ideal Domain

Is there a good way to show that the Gaussian integers are a Principal Ideal Domain without using the fact that they are a Euclidean Domain? It seems like a lot of extra structure to need to prove ...
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75 views

Finitely generated modules over principal ideal domain

Let $A$ be principal ideal domain with field of fractions $K$. $L$ is finite separable extension of $K$ and $B$ is the integral closure of $A$ in $L$. It is obvious that there exists a constant $d$ in ...
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Is any Direct Summand of a Free Module over a PID also Free?

Let $R$ be a PID and $F$ be a free module over $R$. Suppose we have $F=A\oplus B$ for some $R$-modules $A$ and $B$. Then are $A$ and $B$ necessarily free? If $F$ is finitely generated, then I ...
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Let $R$ be a PID. Prove that $\exists c \in R$ such that $c\mid a, c\mid b$ and $c = ax + by$.

Let $R$ be a PID and $a,b \in R$. Prove that $\exists c \in R$ such that $c\mid a, c\mid b$ and $c = ax + by$ for some $x,y \in R$.
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Another question about free modules of finite rank over a PID

I am now trying to prove the following: Let $R$ be a PID and let $M$ be a free $R$-module of finite rank. If $N$ is a submodule of $M$ and $M/N$ is finite, then rank$(N)$ = rank$(M).$ Attempt at a ...
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Bases for submodules of free modules over a PID

I have proved the following: If $G$ is a free abelian group of rank $n$ and $H$ is a subgroup of $G$, then $H$ is free of rank $m\leq n$. Moreover, there exists a $\mathbb{Z}$-basis $x_1,\ldots,x_n$ ...
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130 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 ...