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

learn more… | top users | synonyms (1)

1
vote
1answer
29 views

Divisibility lemma: $\exists n_0\mid n,\,\, m_0\mid m,\,(n_0,m_0) = 1,\text{ and }\,[n_0,m_0] = [n,m]$

I want to prove that, in a commutative group, there always exists an element whose order is $\mathrm{lcm}$ of the orders of two other elements. The exercise indicates that it follows easily from the ...
3
votes
2answers
42 views

Does $\text{rank}(M/M')=\text{rank}(M)-\text{rank}(M')$ hold for a free module over a PID?

Let $M$ be a free module over a PID, $\text{rank}(M)<\infty$, $M'$ submodule of $M$, then $\text{rank}(M/M')=\text{rank}(M)-\text{rank}(M')$?
0
votes
1answer
54 views

Over a PID, $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$?

Let $D$ a PID, $F$ a free module rank $n$, $N$ a submodule of $F$. I want to prove (or find a counterexample) of: $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$ ...
1
vote
1answer
40 views

local PID that is not a field is a DVR

I would be very happy if someone would check my proof of the fact that a local PID that is not a field is a DVR: Let $A$ be a local PID that is not a field. Since irreducibles generate maximal ideals ...
5
votes
1answer
222 views

Proving a subring of $\mathbb{Q}$ containing $\mathbb{Z}$ is a PID

Let $S$ be a subring of $\mathbb{Q}$ containing $\mathbb{Z}$. Prove that it is a principal ideal domain. So here is what I tried. Take any ideal $I\subset S$. Take any two elements, say $a=p/q, ...
2
votes
0answers
61 views

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 ...
0
votes
1answer
44 views

Tensor product of quotient and kernel

In my problem I have a PID $R$, elements $0\neq a,b\in R$ and a map $\phi_a:R\rightarrow R$ where $r\mapsto ar$. Assuming I have done all my previous calculations right I need to prove that ...
1
vote
1answer
68 views

Integral closure of a PID is torsion free

Can anyone explain me why the integral closure of a PID $A$ in a separable finite extension of its fraction field is a torsion free $A$-module? I know that it is a finitely generated A-module ...
2
votes
1answer
38 views

Negative degree valuation: valuation ring and its maximal ideal

I know that the $v: f \mapsto -\deg(f)$ is a discrete valuation on the field of complex rational functions $\mathbb{C}(X)$ (the quotient field of $\mathbb{C}[X]$). The valuation ring $\mathcal{O}_v$ ...
1
vote
1answer
72 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: ...
0
votes
2answers
65 views

Showing that $1 + \sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$

Consider the ring $\mathbb{Z}[\sqrt{5}]$. How can we show that the element $1 + \sqrt{5}$ is irreducible in this ring?
2
votes
2answers
80 views

Integral domain, UFD and PID related problem

(i) Let $R$ be an integral domain that has irreducible elements. Prove that $R[X]$ is not A PID. (ii) Let $R$ be a UFD and $K$ its field of fractions. Let $f \in R[X]$ be a monic polynomial ...
2
votes
1answer
87 views

Quotient of polynomial ring in two variables is a PID

With $K$ a field and $K[x,y]$ the polynomial ring over it in two variables, the quotient ring of it over the ideal generated by $1-xy$ is a PID. I've tried using Noetherianess but haven't gotten ...
1
vote
1answer
46 views

Finitely generated module with free submodule $S$ and $M/S$ torsion free implies $M$ is free

Let $R$ be a PID and $M$ a module. Show the following: (i) If $M$ is finitely generated and $S$ is a free submodule with $M/S$ torsion free, then $M$ is free. (ii) If $M$ is torsion free ...
1
vote
2answers
129 views

Finitely generated module over PID; Dummit and Foote, Exercise 12.1.12

Let $R$ be a PID and let $p$ be a prime in $R$. (a) Let $M$ be a finitely generated torsion $R$-module. Use the previous exercise to prove that $p^{k-1}M/p^kM \cong F^{n_k} $ where $F$ is the field ...
1
vote
1answer
31 views

Factor $55 - 88 \sqrt{-2}$ as a product of primes in $\mathbb{Z}[\sqrt{-2}]$

To solve this problem, I let $K = \mathbb{Q}(\sqrt{-2})$, and I thought to take the norm $$N(55 - 88 \sqrt{-2}) = 55^2 + 2 \cdot 88^2 = 18513 = 3^2\cdot11^2 \cdot 17$$ If $a \in \mathbb{Z}[\sqrt{-2}]$ ...
0
votes
0answers
12 views

What can we say about the function $f(x)$ in this case?

Alright, I'm little bit confused about what's happening here to the function $f(x)$, i thought that the formula of $f(x)$, have nothing to do with its behavior or domain. there are two or many ...
0
votes
2answers
61 views

Looking for help, Aluffi Exercise 5.13, Chapter 6: characterization of PIDs

I quote: "Let $M$ be a finitely generated module over an integral domain $R$. Prove that if $R$ is a PID, then $M$ is torsion-free if and only if it is free. Prove that this property characterizes ...
0
votes
1answer
73 views

continuous poset w.r.t. Scott topology

I am learning continuous poset by myself. I have conclusion as follows: If $P$ is a continuous poset w.r.t. Scott topology then there is $x\in P$ s.t. for any $y\in P$ and for any open sets $U_x$ and ...
2
votes
1answer
36 views

Questions about ring of smooth functions

First of all, this is a homework problem. Let $C^{\infty}(\mathbb{R})$ denote the ring of smooth functions. Let $I_n$ denote the set of $f\in C^{\infty}(\mathbb{R})$ such that $$f^{(k)}(0)=0, \ 0 ...
3
votes
1answer
101 views

When a holomorphy ring is a PID?

I will use the notation and language of Stichtenoth, Algebraic Function Fields and Codes. Let $F$ be a function field over a finite field $\mathbb F_q$, $S$ a non empty set of places (possibly ...
1
vote
0answers
52 views

When is a holomorphy ring a PID? [duplicate]

I will use the notation and language of Stichtenoth, Algebraic Function Fields and Codes. Let $F$ be a function field over a finite field $\mathbb F_q$, $S$ a non empty set of places (possibly ...
3
votes
2answers
109 views

PID and finitely generated module

I am trying to prove the following statements: Let $R$ be a PID and $M$ a finitely generated $R$-module. Prove: (a) $M$ is torsion module iff $\operatorname{Hom}_R(M,R)=0$ (b) $M$ is an ...
2
votes
3answers
42 views

Torsion-free module morphism

I am trying to prove the statement: Let $R$ be a PID but not a field and let $M$ be an $R$-module. Then $$ M \space \text{is torsion-free $R$-module} \space \text{iff} \space ...
2
votes
1answer
67 views

PID and module problem

Let $R$ be a principal ideal domain but not a field, and let $M$ be an $R$-module. Show the following: (i) Let $p \in R$ be an irreducible element and $r \in R \setminus \{0\}$. Then $(R/ ...
-2
votes
1answer
93 views

how can i prove that every subring of $\mathbb{Q}$ is PID? [duplicate]

How can I prove that every subring of $\mathbb{Q}$ is PID?
1
vote
1answer
80 views

Challenging problem in Hungerford's $\textit{Algebra}$

Here is a problem from the text $\textit{Algebra}$ by Hungerford, which I seem to be stuck on for quite some time now: Let $A$ be a cyclic $R$-module ($R$ is assumed to be a principal ideal domain) ...
2
votes
2answers
82 views

Does the statement ''PID of dimension $0$ $\Longrightarrow$ field'' actually use Zorn's Lemma?

Everywhere I look seems to blow by the statement that PIDs which are not fields have Krull dimension $1$. This relies on the fact: A PID with Krull dimension $0$ is a field. (*) It seems that the ...
5
votes
1answer
182 views

Proof that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a PID

How would one prove that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a principal ideal domain (PID)? It isn't a Euclidean domain according to the Wikipedia article on PIDs.
1
vote
1answer
25 views

What exactly does it mean for a maximal ideal to be unique in a principal ideal domain?

I'm currently reading about PIDs and have come across a question involving maximal ideals which at one point reads "Suppose that a Euclidean domain $R$ had a unique maxima ideal $P$". Does this mean ...
0
votes
1answer
67 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 ...
3
votes
1answer
57 views

Principal Ideal Domains with Many Primes of Index 2

By the index of a prime $p$ in a principal ideal domain $R$, I mean the number of cosets of the ideal $(p)$ in $R$ (i.e. $\vert R/(p)\vert$). If I am given a positive integer $m$, I am wondering ...
0
votes
1answer
68 views

$\Bbb Z[x]$ is not a principal domain

I know this is already answered here but I am wondering that if the following way to prove is also correct - let $$f(x) = 4x^2+4x+1$$ $$g(x)=4x^2-1$$ Since this is a UFD so a unique gcd will ...
5
votes
1answer
158 views

Problem with Smith normal form over a PID that is not an Euclidean domain

This is an homework exercise of the Algebra lecture. I need to evaluate the Smith normal form of the following matrix $$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 ...
0
votes
1answer
31 views

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. ...
0
votes
1answer
35 views

Principal Ideal Domain Modulo a Power of a Prime Ideal

If $R$ is a PID, $p$ is a prime element of $R$, $R/(p)$ is finite, and $\alpha$ is a positive integer, is it true that $\vert R/(p^{\alpha})\vert=\vert R/(p)\vert^{\alpha}$? I seem to recall seeing ...
1
vote
1answer
45 views

The number 2 in a PID

Let $R$ be a PID. Then $R$ is a commutative ring with multiplicative identity $1$. We can then define $2=1+1$. From here, what is known about $2$ and its prime factors? I suppose this breaks into two ...
2
votes
1answer
101 views

$R$ is a commutative integral ring, $R[X]$ is a principal ideal domain imply $R$ is a field

I've just read a proof of the statement: Let $R$ be a commutative integral ring. If $R[X]$ is a principal ideal domain, then $R$ is a field. In one part of the proof there is a step which I ...
1
vote
1answer
64 views

Difference between PID and principal ideal ring

All rings are commutative, associative and with 1. Wikipedia states that the difference between PID and Principal Ideal Ring is that the former has to be integral domain while the latter does not. ...
1
vote
1answer
30 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 ...
2
votes
2answers
50 views

Is the ideal $(2,x^4+x^2+1)<\mathbb{Z}[x]$ maximal?, principal?

I'm trying to solve the following problem: Let I=$(2,x^4+x^2+1)<\mathbb{Z}[x]$ be an ideal. Is $I$ maximal? Is $I$ principal? Any help would be appreciated.
2
votes
2answers
193 views

Is $\mathbb{C}[x,y] / (y^2-x^3)$ a PID?

First, I'd like to show $\mathbb{C}[x,y] / (y^2-x^3)$ is an integral domain. Then I need to find out whether or not it is a PID. For the first part, I want to show $y^2-x^3 \: | \: fg \implies ...
3
votes
2answers
128 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 ...
-2
votes
2answers
39 views

R is a PID, and a is a nonzero nonunit in R. How can we show R/Ra is an injective module over R?

If we use Baer's criterion then it suffices to show that if there exist a map from an ideal $I$ to $R/Ra$ we must find a map $g$ such that $g\circ i=f$.
4
votes
3answers
564 views

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal. What could be wrong in this approach?

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which ...
19
votes
2answers
487 views

Quotient of polynomials, PID but not Euclidean domain?

While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$ is ...
15
votes
2answers
735 views

Ring of integers is a PID but not a Euclidean domain

I have noticed that to prove fields like $\mathbb{Q}(i)$ and $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ have class number one, we show they are Euclidean domains by tessalating the complex plane with the ...
0
votes
0answers
75 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
2answers
68 views

Intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ is not finitely generated.

Consider the subring $\mathbb{Z}[2x,2x^2,2x^3,\dots]\subset \mathbb{Z}[x]$. Then show that the intersection of ideals $I=(2x)$ and $J=(2x^2)$ of $\mathbb{Z}[2x,2x^2,2x^3,\dots]$ i.e., $I\cap ...
2
votes
1answer
22 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 $R$ ...