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

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78 views

Prove that $(2)$ is a prime ideal in $\mathbb Z[w]$

Let $w\in\mathbb C$ be such that $w^3=1$ and $w\neq1$. Prove that $(2)$ is a prime ideal in $\mathbb Z[w]$, and describe $\mathbb Z[w]/(2)$. What I wanted to do is to show that $\mathbb Z[w]$ is a ...
2
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1answer
77 views

do you need $P$ prime to show that $R/P$ is a PID if $R$ is a PID?

My question relates to this question, which is exercise 3 in Section 8.2 of Dummit and Foote. They ask to prove that a quotient of a PID by a prime ideal is again a PID. The answers to the previous ...
0
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0answers
32 views

Elementary Divisors on a PID

Let $N$ be a submodule of $\mathbb{Z}^3$ generated by $\{e_1-e_3,2e_1+3e_2+e_3,3e_1+e2+5e_3\}$, with $\{e_1,e_2,e_3\}$ the canonical basis. I am asked to compute a base for $N$ by the structure ...
3
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2answers
68 views

In $\Bbb Z$, what element generates the ideal $(4,7)$?

I have a really silly question. $\mathbb{Z},+,\cdot$ is a HID, so all ideals are principal ideals. Now, $(4,7)$ is an ideal in $\mathbb{Z}$, so it must be a principal ideal, but which element is its ...
4
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1answer
156 views

Ring of formal power series over a principal ideal domain is a unique factorisation domain

An exercise in my algebra course book asks to prove that if $R$ is a PID, then $R[[x]]$ is a UFD, where $R[[x]]$ is the ring of formal power series over $R$. After some failed attempts at proving the ...
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0answers
109 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 ...
1
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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
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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
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1answer
58 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
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1answer
43 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
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1answer
232 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
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0answers
62 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
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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
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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
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1answer
43 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
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2answers
67 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
85 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 ...
3
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1answer
101 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
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1answer
49 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 ...
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2answers
133 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
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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 $a \in \mathbb{Z}[\sqrt{-2}]$ ...
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0answers
14 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
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2answers
62 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
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1answer
74 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
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1answer
37 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
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1answer
109 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 ...
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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
112 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
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3answers
44 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
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1answer
69 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/ ...
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1answer
96 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
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1answer
81 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
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2answers
92 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
214 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.
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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
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1answer
68 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 ...
4
votes
1answer
59 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
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1answer
72 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
162 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
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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
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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 ...
2
votes
1answer
47 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
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1answer
114 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
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1answer
67 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. ...
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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
198 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
136 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 ...
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2answers
40 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 $g\circ i=f$.