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Questions tagged [principal-ideal-domains]

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

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Extension of basis over PID [closed]

Let $k[x] = R$ be ring and $L$ be free $k[x]$-module; let $v \in L$ be vector in $L$. Then how one can extend it to an $R$-basis for $L$?
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Non-zero ideals in ${\mathbb{Q}}_p$ are $p^n{\mathbb{Q}}_p$, $n\in\mathbb N_0$

How do I show that every non-zero ideal in ${\mathbb{Q}}_p$ is of the form $p^n{\mathbb{Q}}_p$ for some $n \in \mathbb{N}_0$, and investigate if ${\mathbb{Q}}_p$ is a principal ideal domain? If it ...
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Is there a way to find propotional gain schedule by iteration?

I have been reading two papers about controling a mechanical nonlinear model where the model structure is the same, but its parameters varying over time. The first controller is gain scheduling. ...
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0answers
28 views

Prove that a finitely generated $\mathbb F_p[t]$-module is a free $\mathbb F_p[t]$-module.

Specifically, I am asked to show that if $(G,\omega)$ is a $p$-valued group of finite rank, (meaning that the associated graded group $grG$ is finitely generated as an $\mathbb F_p[t]$-module), then $...
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2answers
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$R/I$ where if $I\ne 0$ contains $r^2$ for $r\in R$ then I contains $r$

Let $R$ be a PID and let $A \ne 0$ be an ideal of $R$ that satisfies the condition that $r^2\in A$, $r\in R$ implies that $r\in A$. Show that $R/A$ is isomorphic to a finite direct product of fields. ...
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45 views

Marcus Number Fields chapter 5 exercise 8

The exercise is the following: one should show that $R=\mathbb{Z}[\sqrt{223}]$ has ideal class group which is a cyclic group of order $3$. I tried to follow the standard path which Marcus uses to ...
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1answer
29 views

Can every maximal ideal of Dedekind domain be principal after restricting to a small enough distinguished open subset?

Let $S=\textrm{Spec }R$ where $R$ is a Dedekind domain, let $\mathfrak{p}$ be a maximal ideal of $R$, which is a closed point of $S$, can we find an distinguished open affine subset of $S$, say $\...
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1answer
28 views

ring of integers with golden ratio

How to show that the ring of integers of $\mathbb{Q}(\sqrt{5})$, i.e., $\mathbb{Z}[\phi]$ is a principal ideal domain (where $\phi$ is the golden ratio)? I want to prove that using as elementary ...
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1answer
55 views

In an integrally closed, Noetherian, local, integral domain of dimension $1$, the maximal ideal $P$ is eventually principal

Let $R$ be an integrally closed, Noetherian, local, integral domain of dimension 1 with unique maximal ideal $P$. Take an element $a \in P$ that is non zero. Show that for some $n$, $P^n$ is ...
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Need help understanding a step in a proof about modules over PIDs

This is Chapter 3, Theorem 7.3 in Algebra by W. Adkins & S. Weintraub (GTM). It's about the uniqueness of the cyclic decomposition for finitely generated modules over a PID. I highlighted the ...
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Help understanding a lemma on modules over a PID.

Lemma 1.84 Let $R$ be a PID. Suppose that $M\simeq R/(a)$ where $a$ is a nonzero element of $R$. Moreover, suppose $p$ is an irreducible element of $R$. Then (a) If $p\nmid a \implies M_p=\{...
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1answer
80 views

Prove that the quotient ring is not a PID

Show that $R = \mathbb{C}[x,y]/(y^2-x^3-1)$ is not a PID. My idea is to find an ideal of $\mathbb{C}[x,y]$ containing $(y^2-x^3-1)$ and show that its image is not principal. So I have $J = (x,y+1)$, ...
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3answers
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$Rx/Rax \simeq R/Ra$ whenever $\mathrm{Ann}(x)=\{0\}$

Let $R$ be a PID and $M$ an $R$-module. Let $x\in M$ be such that $\mathrm{Ann}(x)=\{0\}$. Prove that for any $a\in R$ we have $Rx/Rax \simeq R/Ra$. Indicate in your argument where you use the fact ...
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0answers
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Irreducible Subring in Q

Consider the subring R of Q[x,y] consisting of all $h(x,y)=a_0 + a_1x + a_2y + a_3xy + a_4x^2 + a_5y^2 + a_6x^2y + a_7xy^2 +’a_8x^3 + ...$ where all $a_i$ are in Q and a_1 = a_2 = 0. I’m suppose to ...
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1answer
32 views

Showing if $R$ is a PID and $0\neq r\in R$ is irreducible, then $R[X]/(r)\cong (R/(r))[X]$

I've been looking at the statements I found on these two Stack Exchange answers and I've been trying to prove them. The first claim is: If $R$ is a PID and $0\neq r\in R$ is irreducible, then $R[X]/...
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1answer
30 views

Showing that ideals are principal in ring of integers of cubic field

Let $K = \mathbb{Q}(\sqrt[3]{6})$. Factorise $\langle p \rangle$ into prime ideals in $\mathcal{O}_K$ for $p = 2, 5, 13$, checking that the factors are principal. I used the Dedekind-Kummer Theorem ...
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1answer
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Is this necessary for the first structure theorem for modules?

It seems to me that the proof of the first structure theorem for finitely generated torsion modules (over principal ideal domains) assumes that if there exists some decomposition then it must be an ...
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25 views

Is still true that every nonzero Prime Ideal is Maximal in a noncommutative PID?

Basically what I said in the title: I can't find a proof that doesn't make use of commutativity. I know that Commutativity is generally taken for granted in a domain, but I wanted to know in general, ...
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Set of all R-homomorphism is finitely generated

I have the following proposition: Proposition: If $R$ is a principal ideal domain (PID) and $M$, $N$ two finitely generated (fg) $R-$modules, then $\text{Hom}_R(M,N)$ is finitely generated. My idea: ...
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Prove polynomial ring over a discrete valuation ring quotient by powers of maximal ideal is regular?

Let $(R,\mathfrak{m},k)$ be a discrete valuation ring, (of characteristic $p$ if you need). Let $n\geq 1$ be an integer. Is the ring $\frac{R}{\mathfrak{m}^n}[x]$ regular? Note that: Regularity can ...
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Axes of a free module over a PID (2)

Let $R^n$ be a finitely generated free module of rank $n > 0$ over a principal ideal domain. I am trying to prove that for every non-zero element $a$ of $R^n$ there is a basis such that $a$ ...
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1answer
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$\mathbb{Z}$-Module exercise

I am trying to solve the following exercise on basic module theory and I am stuck. Any help would be more than welcome! So let $M\subseteq \mathbb{Z}^3$ the solutions to the following problem: $-3x+...
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1answer
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Axes of a free module over a PID

Let $R^2 = R \times R$ be a free module of rank $2$ over a principal ideal domain. I am trying to prove that for every non-zero element $a$ of $R^2$ there is a basis such that $a$ belongs to one of ...
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1answer
23 views

Given $k$, for every $n>1$, constructing a set of size $n$ of non-zero integers having gcd $k$ so that no proper subset has gcd $k$

For a finite subset $S \subseteq \mathbb Z \setminus \{0\}$, let us say $d=\gcd S$ iff $d>0 $ , $ d|a,\forall a \in S$ and $m|a,\forall a\in S \implies m|d$. My question is: Does there exist a ...
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1answer
25 views

Defining an evaluation map between $\mathbb{Q}\lbrack x \rbrack$ and $\mathbb{Q}\lbrack x \rbrack/(x^2-5)$

I need to define an evaluation mapping between $\mathbb{Q}\lbrack x \rbrack$ and $\mathbb{Q}\lbrack x \rbrack/(x^2-5)$. I know I want the identity to map to the identity, but I'm not sure what the ...
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0answers
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$I$ an irreducible ideal in a PID is a prime ideal

$I$ is irreducible ideal of $R$ when $R$ is a PID , I want to show it's a prime ideal. The definition I am familiar with: $I$ a proper ideal of $R$ is an irreducible ideal if for any two ideals $J,K$ ...
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2answers
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Is every ideal of Quotient Ring a PID?

We are given a principal ideal domain $R$ and an ideal of this domain, $I$. Is it true that every proper ideal of $\displaystyle\frac{R}{I}$ a principal ideal domain ? I have proven that every ...
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1answer
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Let $ R $ be a p.i.d. and $ A\in M_n(R) $. If $ \det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices.

Let $ R $ be a p.i.d. and $ A\in M_n(R) $. If $ \det(A)=1 $, prove or disprove that $ A $ can be expressed as products of elementary matrices. I know that we can express $ A $ as products of ...
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Determine the quotient and the remainder of the division:

Determine the quotient and the remainder of the division: ($1$).of $f\in \mathbb K[x]$ by $x^2-a$ in $\mathbb K[x],$Where $\mathbb K$ is a field. ($2$).of $x^m-1$ by $x^n-1$ in $\mathbb Z[...
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1answer
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Prove that for $\pi_g(\alpha)=\alpha g$, $\pi$ is a group action

Let $R$ a principal ideal domain and let $G$ the group of all the invertible members of $M_2(R)$. Let $\Omega :=R^2$. For all $g\in G$ let $\pi_g:\Omega\to\Omega$ defined by $\alpha \pi_g=\alpha g$ ...
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1answer
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Is the ideal generated by ${4,x}$ a principal ideal in $Z[x]$? [duplicate]

I've : $I=<p,x>$ is not a principal ideal in $Z[x]$ where p is prime. My question : Is $I=<p,x>$ a principal ideal in $Z[x]$ where p is not a prime? More particularly, is the ideal ...
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1answer
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Integral Domain, PID and gcd

Let $R$ be a PID and $R'$ a Integral Domain and $R\subseteq R'$. Let $a,b,d \in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$. Proof: $d$ is a gcd of $...
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2answers
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When is a Unique Factorization Domain a Principal Ideal domain [duplicate]

"Let $R$ be a Unique Factorization Domain and $(a,b)=(c)$ for $a,b,c \in R$. Show that $R$ is a Principal Ideal domain." To be honest I found this very hard, here is my naive try: Lets assume $R$ is ...
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1answer
70 views

Showing that an ideal in $\mathbb{Z}[\sqrt{-21}]$ is principal

I have $\mathfrak{a}=(5,\sqrt{-21}-2).$ Can anyone tell me why $\mathfrak{a}^2$ is principal? I have multiplied the ideals out to obtain $$(5, 5\sqrt{-21}-10,-17-4\sqrt{-21}) $$ How does this reduce ...
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1answer
68 views

Showing that an ideal is not principal in $\mathbb{Z}[\sqrt{-21}]$

I am trying to show that the ideal $$(2,\sqrt{-21}-1)(3, \sqrt{-21}) $$ is not principal in $\mathbb{Z}[\sqrt{-21}]$. Can anyone help with this?
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Gaussian Integers form an Euclidean Ring

A Ring $R$ is called euclidean if a map $f:R\backslash {0} \rightarrow \mathbb{N}$ exists with the following properties: For two elements $a,b \in R$ with $b\neq 0$ there exist $q,r\in R$ with: (i) $...
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Step in proof of Structure Theorem for finitely generated modules over a PID

I'm looking for help with problem #6 on page 189 of Jacobson's "Basic Algebra I". In particular, we suppose $D$ is a PID and $M$ is a finitely generated module over $D$ with generators $x_1, ... x_n$....
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Invariant factors of a module over a $PID$

so i need some help. Imagine D is Principal Ideal Domain, and $M$ and $N$ are cyclic modules with order $a$ and $b$ respectevly, and $mdc(a,b)\neq 1$. So im asked to prove that the invariant factors ...
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Counterexample PID [duplicate]

We know that if F is a field, then the polynomial ring over F is a PID. Do you have a counterexample that shows that if F isn’t a field than the polynomial ring over F isn’t a PID?
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If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $\frac{R}{s}$ has finitely many prime ideals .

If $R$ is a principal ideal domain and $s$ is a non-zero prime ideal, then $\frac{R}{s}$ has finitely many prime ideals . How can I prove it? My attempt: As $R$ is a PID then $s$ will be ...
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2answers
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$R=\{f(x)\in \mathbb{Q}[x] | f'(0)=0 \}$ is not PID

Suppose $R=\{f(x)\in \mathbb{Q}[x] | f'(0)=0 \}$ subring of the principal domain $\mathbb{Q}[x]$. I have already proven that $x^2$ is irreducible in $R$. Since elements in $R$ have the form $f(x)=a_0 ...
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Examples of PIDs that are not Fields

If $R=\mathbb{Z}[\frac{1}{2}(1+\sqrt{-19})]$ is an example of a PID which is not a Euclidean domain. But since Euclidean domains may not necessarily be fields, this doesn't say much unless $R$ in this ...
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Some counterexamples in basic ring theory

Give an example if possible, and if not possible explain why not. a) A subring of a PID that is not PID. b) A PID that is a subring of a non-PID. c) A subring of a PID that is not UFD. My approach:...
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1answer
24 views

Is it possible for an integral domain to have an ideal that cannot be generated by a countable set?

This question came up when I was working on the following problem Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain: i) any two ...
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Is the product of two Smith Normal Forms a the Smith normal form of the product?

Suppose A and B are square matrices of the same size over a PID R. Does the Smith Normal form of AB equal the product of the Smith normal form of A and B? I think this should be false. However, I can'...
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1answer
198 views

When does $(a,b)=(\gcd(a,b))$ hold?

I had a look here to understand why $K[X,Y]$ is not a PID. So one of the conclusions was that $(x,y) \neq (1) = \gcd(x,y)$, but I thought that $(a,b)=\gcd(a,b)$ was always true so obviously I was ...
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1answer
51 views

Existence of Principal Ideal

My first question is DOES a Principal Ideal always exist in a ring? (My thoughts: By it's very definition any element can generate a PI and hence it exists) Follow-up question: (If answer to 1st ...
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Position of $\mathbb{Z}[\sqrt{-d}]$ in class hierarchy for all integers $d$

For any real number $d,$ define $$\mathbb{Z}[\sqrt{-d}]: = \{a+b\sqrt{-d}: a,b\in \mathbb{Z}\}.$$ It is well-known that $\mathbb{Z}[\sqrt{-2}]$ and $\mathbb{Z}[\sqrt{-1}]$ are ED but not fields, $$\...
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1answer
73 views

Why $K[[X]]$ is PID and what's the form of the ring's ideals?

Let $K$ be a field. We write $K[[X]]$ for the ring of all formal power series with coefficients from the field $K$. Then, we will try to prove the next theorem. Theorem. If $K$ is a field then: ...
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33 views

Factorization Domain in which every domain generated by two elements is a principal domain, is a PID

Let $R$ be a factorization domain in which every ideal generated by two elements is a principal domain. Show that $R$ is a PID. So let $I \subset R$ be an Ideal, then I have to show that there exists ...