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

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29 views

What is the dimension of $\mathbb R[x] / \langle x^3-x\rangle$ as a vector space over $\mathbb R$ ?

What is the dimension of $\mathbb R[x] / \langle x^3-x\rangle$ as a vector space over $\mathbb R$ ? Can someone please give some links , articles where I can study about polynomila rings and its ...
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2answers
35 views

The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$)

The rings Z[$\sqrt{6}$] and Z[$\sqrt{7}$] are PIDs. Exhibit generators for their ideals (3,$\sqrt{6}$), (5, 4 + $\sqrt{6}$), (2, 1 + $\sqrt{7}$) Can I get walked through one of them so that I ...
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0answers
15 views

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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11 views

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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79 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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4answers
128 views

Why is $(2, 1+\sqrt{-5})$ not principal?

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbf Z[\sqrt{-5}]$? Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb ...
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31 views

Smith Normal Forms for Fields and PID

I need to reduce the following matrices, $M_{1}, M_{2}$, into Smith Normal form, in the field $\mathbb{Z}/2\mathbb{Z}[x]$ $M_{1} = \left ( \begin{array}{ccc} x & 1 & 0 \\ 0 & x & 1 ...
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13 views

Inclusion with some principal ideals

Let $A$ be a principal ideal domain, $M$ a free $A$-module of rank $n$, $M'$ a submodule of $M$ with $M' \ne (0)$, and $L(M,A)$ the set of linear forms on $M$. For $v \in L(M,A)$, we can write ...
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0answers
28 views

$\mathbb{Z}[\zeta_n]$ 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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1answer
50 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 ...
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1answer
59 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 ...
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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 ...
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2answers
63 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 ...
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1answer
76 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
72 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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1answer
25 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 ...
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2answers
40 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 $\text{rank}(M/M')=\text{rank}(M)-\text{rank}(M')$?
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1answer
51 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: $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$ ...
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1answer
36 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 ...
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1answer
136 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, ...
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0answers
60 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 ...
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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 ...
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1answer
63 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
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 $\mathcal{O}_v$ ...
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1answer
71 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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2answers
63 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?
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2answers
52 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
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1answer
70 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 ...
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1answer
38 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
102 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 ...
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1answer
29 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
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 ...
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2answers
55 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 ...
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1answer
69 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 ...
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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 ...
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2answers
100 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 ...
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3answers
34 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 ...
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1answer
59 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
67 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?
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1answer
78 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) ...
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2answers
76 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 ...
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1answer
111 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
23 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 ...
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1answer
57 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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0answers
31 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 ...
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1answer
63 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
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1answer
137 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
30 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. ...