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

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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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34 views

$R/Ra$ is an injective module over itself

Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself. If $R$ is a PID, every $R$- divisible module is injective, but the question concerns ...
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94 views

Is this module injective? [duplicate]

Here's the problem: (This problem is from Hungerford's Algebra Chapter 5 exercise 6.7 ) Let R be a principal ideal domain and $p$ a prime in $R$ and $n$ a positive integer. Then $R/(p^{n})$ is an ...
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33 views

Show that $R[x,y]$ is not a Principal Ideal Domain [closed]

Let $R$ be a commutative ring with $1$. Prove that a polynomial ring in more than one variable over $R$ is not a P.I.D.. In order to show this is not a P.I.D. what do i need to show.
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A principal ideal domain if and only if proof

Show an ideal $(p)$ in a principal ideal domain in a maximal ideal if and only if $p$ is irreducible. This is a new concept i do not know how to go about this.
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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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2answers
50 views

Why is F[x] a UFD? [duplicate]

When reading the proof for if $R$ is a UFD, then $R[x]$ is a UFD, the author uses a fact that $F[x]$ is a UFD. I don't quite understand this. Why $F[x]$ is a UFD? ($F$ is the fraction field of ...
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42 views

$F$ is a field iff $F[x]$ is a Principal Ideal Domain

A commutative ring $F$ is a field iff $F[x]$ is a Principal Ideal Domain. I have done the part that if $F$ is a field then $F[x]$ is a PID using the division algorithm and contradicting the ...
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64 views

A question regarding Kummer [closed]

As you know, Ernst Kummer noticed that examples such as $$6 = 2\cdot 3 \text{ or } 6 = 3 \cdot 2 \text{ and, crucially } 6 = (1 + \sqrt{-5}) (1 -\sqrt{-5}) $$ proved the failure of unique ...
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For which values of $d<0$ , is the subring of quadratic integers of $\mathbb Q[\sqrt{d}]$ is a PID?

The "integers" of quadratic field $\mathbb Q[\sqrt{d}]$ , for a squarefree integer $d$ , forms an integral domain . I know that for $d<0$ , the quadratic integers of the quadratic number fields ...
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69 views

If R is a PID, is it true that $R/\ker \phi$ is also a PID?

I came across this solution that seeks to prove that any submodule of a cyclic module is cyclic. Proof: Let $M$ be a cyclic module, so that $\phi:R \rightarrow M$ is a surjection under $\phi(r)=r ...
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1answer
45 views

Does validity of Bezout identity in integral domain implies the domain is PID ?

Let $D$ be an integral domain such that for any $a,b \in D$ , $Da+Db$ is a principal ideal , then must $D$ necessarily be a principal ideal domain i.e. should all the ideals of $D$ be principal ? ...
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78 views

Quotient of direct sum of $l$ copies of a PID by an ideal.

Let $R$ be a PID, and $\psi:R^k\to R^l$ an homomorphism. I would like to know under what circumstances it's true that: $$ R^l/\operatorname{Im}\psi \cong R/I_1\times \cdots \times R/I_l, $$ for some ...
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20 views

prove that the quotient ring S3/T3 is isomorphic to D3

Could you please help with this question? I've already shown that T_3 is an ideal of S_3. Thanks,
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1answer
22 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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14 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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1answer
32 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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40 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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22 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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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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98 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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151 views

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

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$? 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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39 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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14 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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29 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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65 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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68 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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30 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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66 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
103 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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80 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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26 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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41 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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53 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
37 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
170 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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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
43 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
64 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 ...
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1answer
32 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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64 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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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 ...
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1answer
76 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
42 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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120 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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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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58 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
70 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 ...