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

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2
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1answer
34 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 ...
3
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1answer
47 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 ...
13
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3answers
443 views

Prove that $n^2+n+41$ is prime for $n<40$

Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it. Let $n\in\{0,1,...,39\}$. Prove that $n^2+n+41$ is prime. I shall provide my own ...
4
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0answers
59 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 ...
4
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1answer
60 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 ...
0
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0answers
29 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
58 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 ...
1
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1answer
23 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
36 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
48 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
32 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
124 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
54 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 ...
1
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2answers
85 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 ...
0
votes
1answer
40 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
57 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
31 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 $\mathcal{O}_v$ ...
2
votes
1answer
203 views

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR.

Let $f:R\longrightarrow S$ be a surjective ring homomorphism. If $R$ is PID, then $S$ is PIR. I think I have proved this: Let $J$ be an ideal of $S$. Then $f^{-1}(J)=(a)$ is a principal ideal of ...
1
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1answer
70 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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0answers
100 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 ...
2
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1answer
63 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 ...
5
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5answers
2k views

Is $\mathbb{Z}[x]$ a principal ideal domain?

Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a ...
2
votes
2answers
46 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 ...
0
votes
2answers
62 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?
8
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1answer
808 views

Ring of Polynomials is a Principal Ideal Ring implies Coefficient Ring is a Field?

I read this proof that if $D$ is an integral domain and $D[X]$ is a principal ideal domain, then $D$ is a field. My question is if the requirements can be relaxed a bit, namely: Is it true that ...
1
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1answer
36 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
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1answer
27 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
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1answer
65 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 ...
0
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0answers
11 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
52 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 ...
3
votes
1answer
99 views

Proof for Unique Factorization Domain

Prove that the quotient ring $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a unique factorization domain. I am trying to prove first it is a principal ideal domain. However I am really stuck on this problem
2
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1answer
31 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 ...
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0answers
50 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
97 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
29 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
58 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
64 views
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1answer
74 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
69 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 ...
3
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1answer
787 views

For which $d$ is $\mathbb Z[\sqrt d]$ a principal ideal domain?

Is there any general idea about for which $d$, $\mathbb Z[\sqrt d]$ a principal ideal domain (PID)? As for example $\mathbb Z[\sqrt{-1}]$ and $\mathbb Z[\sqrt 2] $ are PIDs, but $\mathbb Z[\sqrt{-5}] ...
5
votes
1answer
85 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.
0
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0answers
151 views

Homology out of Smith normal form: simultaneous or independent diagonalization?

Let $R$ be a PID and $R^m\overset{A}{\longrightarrow} R^n\overset{B}{\longrightarrow} R^o$ matrices with $BA=0$ and Smith normal forms ...
1
vote
1answer
21 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
50 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 ...
0
votes
0answers
30 views

Domain of reciprocal of log in complex plane

Ok so what i already know is that the function f(z) = log(z) is undefined when z = negative or zero which is quite a basic concept. Can someone explain the mystery ...
2
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1answer
56 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
58 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 ...
3
votes
1answer
129 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
28 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
31 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 ...