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

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3
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1answer
51 views

Suppose A is a principal ideal domain with every ideal of finite index. Must A be a Euclidean domain?

Suppose $A$ is a principal ideal domain with every ideal of finite index (except the zero ideal). Must $A$ be a Euclidean domain? If it's not known, are there any relevant partial results?
-1
votes
1answer
35 views

Is $\Bbb Z / p$ where $p$ is not prime a PID? [closed]

Is it possible for a finite ring with unity in the form of $\Bbb Z / p$ where $p$ is not prime to be a PID?
6
votes
1answer
118 views

$\mathbb{Q}(\sqrt[3]{17})$ has class number $1$

Let $\alpha:=\mathbb{Q}(\sqrt[3]{17})$ and $K:=\mathbb{Q}(\alpha)$. We know that $$\mathcal{O}_K=\left\{\frac{a+b\alpha+c\alpha^2}{3}:a\equiv c\equiv -b\pmod{3}\right\}.$$ I have to show that $K$ has ...
1
vote
1answer
14 views

A problem involving proving cyclic module defined via invertible linear operator has cyclic inverse module

I have met this recently in my abstract algebra course dealing with modules over PIDs and we are dealing with cyclic modules at the moment, the problem I am having difficulty with is as follows: ...
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0answers
12 views

Computing the invariant factors and elementary divisors of finitely generated module over a PID

I have just encountered this question in my abstract algebra class dealing with finitely generated modules over PIDs stating the following: Let $ D = \mathbb{R}[x] $ be the ring of polynomials ...
1
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2answers
28 views

Coprime elements in a PID satisfy that any of their powers are coprime

I have recently met this problem in my abstract algebra dealing with PID rings and coprimes stating: Let D be a PID ring $ a,b \in D $ two coprime elements. We are to show that for all $ m,n \in ...
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4answers
175 views

Description of ideals of ring $F[x]/(x^n)$?

What is a description of the ideals of the ring $F[x]/(x^n)$, where $F$ is a field?
4
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3answers
70 views

Ideal of $\mathbb Q[x]$ which contains two polynomials

Suppose $I$ is an ideal of $\mathbb Q[x]$ which contains $x^2 + 2x +4$ and $x^3 - 3$. Prove $I =\mathbb Q[x]$. This is an exercise in my abstract algebra text book. I know the definition of an ...
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2answers
38 views

Localization of a PID is a PID

I would like a verification of a proof for the following statement. Let $S$ be a multiplicatively closed subset of a ring $R$. If $R$ is a PID, then $S^{-1}R$ is a PID. Let $I = \left<r_1/s_1, ...
0
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2answers
19 views

Characteristic and Principal Ideal.

This might be a simple question for some of you, but I am quite confused on the whole concept of principal ideals. Question 1: What is the characteristic of $\mathbb{Z}_2[X,Y]$ where it is the ring ...
1
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1answer
33 views

A commutative ring with unity which is not a PIR has a non-trivial ideal generated by two elements which is not a principal ideal?

Let $R$ be a commutative ring with unity which is not a principal ideal ring . Then is it true that $\exists 0\ne x,y \in R$ such that the ideal $\langle x, y \rangle$ is not a principal ideal ?
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2answers
60 views

Ring of polynomial functions on unit hyperbola is PID

Let $R=\mathbb{R}[X,Y]/(XY-1)$ be the ring of polynomial functions pn the unit hyperbola. How do I prove that $R$ is a principal ideal domain with unit group ...
2
votes
1answer
48 views

Is the polynomial ring over a PID also a PID?

As stated in the title, given a principal ideal domain $R$, is the polynomial ring $R[x]$ necessarily a principal ideal domain? In particular, is the polynomial ring $(\mathbb{Z}[i])[x]$ over the ...
1
vote
3answers
57 views

Is $\mathbb Z_3[x]$ isomorphic with $\mathbb Z$?

Is $\mathbb Z_3[x]$ isomorphic with $\mathbb Z$ ? (This question arose in trying to determine whether there is a commutative ring $R$ with unity such that $R[x]\cong\mathbb Z$ . It is easy to see ...
0
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2answers
93 views

Higher Ext's vanish over a PID

Let $R$ be a PID and $M$, $N$ be $R$-modules. I am trying to show that $$\forall n\ge 2~: \operatorname{Ext}_{R}^{n}(M,N)=0.$$ For example $\forall n\ge 2~: \operatorname{Ext}_{\mathbb ...
2
votes
0answers
38 views

For finitely generated $B$ all modules in exact sequence are finitely generated in PID

Let there be an exact sequence of $R$-modules: $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ where $R$ is a principal ideal domain and $B$ is finitely generated. Are $A$ and $C$ ...
0
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1answer
40 views

Is $\mathbb{Q}$ a principal ideal domain?

Is $\mathbb{Q}$ a principal ideal domain? I know that $\mathbb{Z}$ is, but I'm not sure about $\mathbb{Q}$.
2
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3answers
21 views

About the rank of submodules over PID

If $M$ is a free $R$-module of finite rank $n$ and $R$ is a PID, do proper submodules of $M$ have strictly less rank than $M$? I know that in this case, every submodule of $M$ is free and has ...
2
votes
1answer
22 views

Does torsion-free implies flat over a “locally principal” domain?

Let $R$ be a commutative ring. An $R$-module $M$ over $R$ is said to be torsion free if for every $r\in R$ which is not zero divisor and for every $0\neq m\in M$, we have $r\cdot m\neq 0$. I know ...
4
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1answer
58 views

Abstract algebra - UFDs

I'm having trouble coming up with a counterexample for the following (I'm not interested in a full solution, rather I want ideas/hints): (*) For an injective homomorphism of rings $\phi:R \rightarrow ...
6
votes
2answers
52 views

Something wrong my proof that $\mathbb{Z}[x]$ is not a PID nor an Euclidean domain?

My proof goes as follows: Suppose for contradiction that $\mathbb{Z}[x]$ is a PID. Then the ideal generated by any irreducible element is maximal. We know that $x^2+1$ is irreducible in ...
2
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1answer
71 views

Goldbach property for polynomials with base ring a PID with infinitely many maximal ideals

Is it true that if $R$ is a PID with infinitely many maximal ideals, then every element of $R[x]$ of degree $n\ge1$ is a sum of two irreducible polynomials in $R[x]$? Even if this is not true in ...
1
vote
1answer
11 views

Decomposing a Module over a PID

Suppose that $f: M \to R$ is a $R$-morphism, where $R$ is a PID and $M$ is an $R$-module. Decompose $M=X \oplus \ker f$ for some $X \leq M$. I have been staring at this for a bit and have yet to ...
8
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2answers
417 views

What information do we gain from PIDs

I am self-learning some algebraic number theory and my question is regarding the advantages to studying PIDs. I have seen that Euclidean Domains $\subseteq$ Principal Idea Domains $\subseteq$ Unique ...
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0answers
36 views

prove that in PID every ideal has a unique factorization into prime ideals

this is my attempt: any ideal A in pid R is a principal ideal. so A=(a) for some element a in R. pid is ufd so a has a unique factorization into prime elements say a=p1*p2...*pn. then since pid is a ...
0
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1answer
18 views

Question in proving “Any principal ideal domain is a unique factorization domain”

In proving Any principal ideal domain is a unique factorization domain. Let $D$ be a principal ideal domain, and let $d$ be a nonzero element of $D$ that is not a unit. Suppose that $d$ cannot be ...
2
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1answer
79 views

Find an Ideal of $\mathbb{Z}+x \mathbb{Q}[ x ]$ that is NOT principal

The ring $\mathbb{Z}+x \mathbb{Q}[ x ]$ cannot be a principal ideal domain since it is not a unique factorization domain. Find an ideal of $\mathbb{Z}+x \mathbb{Q}[ x ]$ that is not principal. ...
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2answers
34 views

When is $\Bbb{Z[\zeta_n]}$ a PID?

When is $\Bbb{Z[\zeta_n]}$ a PID? I was just wondering if $\Bbb{Z[\zeta_n]}$ is PID or not where $\zeta_n$ is $n$th primitive root of unity for arbitrary positive $n$
1
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1answer
30 views

Finitely generated torsion free module over $A$ is locally free

$A$ is an integral domain whose local rings $A_p$ are principal ideal domains, then any finitely generated torsion free module over $A$ is locally free. I know that finitely generated torsion free ...
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2answers
58 views

$R=\{a/b: a,b \in \mathbb{Z}$ and $b$ is odd}. Show that the ring $R$ is a PID.

Let us have a ring $R$ defined as $R=\{a/b: a,b \in \mathbb{Z}$ and $b$ is odd}. I want to show that $R$ is a PID. I think I should start with that $I\cap Z = n \mathbb{Z}$ for some $n \in ...
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0answers
29 views

Ideal class group of $ \mathbb{Z}[ \sqrt{2} ] $

How does one compute the ideal class group for $ \mathbb{Z}[\sqrt{2}]$? Motivation: I wish to prove that $ \mathbb{Z}[\sqrt{2}]$ is a PID. I have seen proofs which use norm and go on to show that it ...
0
votes
1answer
90 views

Is a localization of an UFD at a prime ideal a PID?

I have this problem: Let $R$ be an UFD and $p$ a prime element. Prove $R_{(p)}$ is a PID. I think I'm just struggling to get started. I know that $R_{(p)}=\{\frac{a}{b}: a \in R, b \in ...
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2answers
63 views

How do I show that the unit group of $\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order 10? [closed]

How do I show that the unit group $R^*$ of $R=\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order 10? I am allowed to use the fact that $R$ is isomorphic with $\mathbb{Z}[X]/(5X,X^2)$. ...
0
votes
1answer
28 views

$x=(0,\overline{1})$ and $y=(0,\overline{2})$ generate the same ideal in $R=\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$

How do I show that $x=(0,\overline{1})$ and $y=(0,\overline{2})$ generate the same ideal in $R=\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}$, but that there is no $u\in R^*$ such that $y=ux$? Working with ...
0
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0answers
38 views

Proving the Gaussian Integers are a Principal Ideal Domain

Is there a good way to show that the Gaussian integers are a Principal Ideal Domain without using the fact that they are a Euclidean Domain? It seems like a lot of extra structure to need to prove ...
3
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0answers
59 views

$\mathbb{Z}[i]$ is principal. And what are the units

I have elements of the form $a+bi$. I have attempted to consider arbitrary ideals in $\mathbb{Z}[i]$. If $N$ is ideal and $N=\{0\}$ then it is generated by $0$. If $N$ is not trivial, then exists ...
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0answers
89 views

Dummit & Foote <Abstract Algebra> Exercise 8.2.4

I'm trying to solve Exercise 8.2.4 in Dummit & Foote Let $R$ be an integral domain. Prove that if the following conditions hold then R is a Principal domain (1) any two nonzero elements ...
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0answers
48 views

Rank of a matrix over a principal ideal domain

I apologize if my question is stupid but I'm not very familiar with matrices over a principal ideal domain $R$ (For example, $R=\mathbb{Z}$ or $R=\mathbb{R}[X]$). I was wondering how to define the ...
0
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1answer
38 views

What is the simplest non-principal ideal?

Let's restrict ourselves to commutative rings (not necessarily with unity). Is there a simpler example of a non-principal ideal than $\langle a,x\rangle$ in $R[x]$, where $a\in R$ is not a unit (and ...
2
votes
1answer
67 views

Proof that $f\in R[X]$ with $f(u)=u^{-1}$ exists for commutative ring $R$ [closed]

Let $R$ be a commutative ring and $U\subset R^*$ a finite subset. How do I prove that there exists an $f\in R[X]$ with $f(u)=u^{-1}$ for all $u\in U$?
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1answer
70 views

Prime ideal being maximal ideal and PID

Q1. Does there exist an ID R in which every non zero prime ideal of type pR is maximal ideal but R is not PID? Q2. Does there exist an ID R in which every non zero prime ideal is maximal ideal but R ...
4
votes
1answer
87 views

Structure theorem (PIDs) from Smith Normal Form

How exactly does the structure theorem follow from Smith Normal Form? (Wikipedia statement) It is said that a presentation (map from relations to generators) is put into Smith Normal form. Now, I see ...
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1answer
34 views

Problem of Module on a PID

Suppose that $R$ is a $\mathrm{PID}$. Suppose that $a$ and $b$ are nonzero elements of $R$ which are relatively prime. Let $M$ be an $R$ module so that $abM=\{0\}.$ Show that $aM=M_b$ and $bM=M_a$ ...
0
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0answers
73 views

Finitely generated modules over principal ideal domain

Let $A$ be principal ideal domain with field of fractions $K$. $L$ is finite separable extension of $K$ and $B$ is the integral closure of $A$ in $L$. It is obvious that there exists a constant $d$ in ...
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0answers
49 views

Is any Direct Summand of a Free Module over a PID also Free?

Let $R$ be a PID and $F$ be a free module over $R$. Suppose we have $F=A\oplus B$ for some $R$-modules $A$ and $B$. Then are $A$ and $B$ necessarily free? If $F$ is finitely generated, then I ...
0
votes
1answer
30 views

If $A \in M_n(R)$, with $R$ a P.I.D., can $A$ be put in Jordan form iff all the roots of the characteristic polynomial are in $R$?

If $A \in M_n(R)$, with $R$ a P.I.D., can $A$ be put in Jordan form iff all the roots of the characteristic polynomial are in $R$? If this is false in general, is it possibly true for nilpotent ...
7
votes
1answer
79 views

Is this ring a PID? [closed]

Let $R$ be the $k$-subalgebra of $k(t)$ generated by the set $k[t]$, of all polynomials, and a pair of rational functions: ${1\over{t-1}}$ and ${1\over{t-2}}$. Is the ring $R$ a PID?
4
votes
0answers
37 views

Are the ring of integers of the constructible numbers a Euclidean domain?

I suspect that since Euclid uses the Euclidean Algorithm to perform division on constructible numbers in Elements, the ring of integers of the constructible numbers are a Euclidean Domain, but I have ...
0
votes
1answer
41 views

A direct sum of cyclic modules over a PID

Let $R$ be a PID, let $p$ be a prime of $R$, and let $M$ be the $R$-module $R/Rp^{e_1}\oplus \cdots \oplus R/Rp^{e_n}$ where the $e_i$ and $n$ are positive integers. Define $M(p)=\{m: pm=0\}$ and ...
1
vote
1answer
27 views

Prove that $I^2$ is principal.

Consider the ideal $I=(2,\sqrt{-10})$ of $\mathbb{Z}[\sqrt{-10}]$. Prove that $I^2$ is principal. My Try: $I^2=(4,-10,2\sqrt{-10})$. I tried to prove that $I^2=(\sqrt{-10})$. But failed. Is my ...