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

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4
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1answer
80 views

Is $\mathbb{Z}_p[\mathbb{Z}_p]$ a PID?

Is $\mathbb{Z}_{p}[G]$ a PID, where $G=(\mathbb{Z}_{p},+)$ is the additive group of the $p$-adics $\mathbb{Z}_{p}$? I am studying a paper where the authors implicitly use that claim, but it is ...
1
vote
1answer
44 views

Is $\mathbb{Z}[\mathbb{Z}/(p)]$ a PID?

As the title suggests, I'm interested whether $\mathbb{Z}[\mathbb{Z}/(p)]$ a PID or not. Assume $p$ is prime. My feeling is that it is a PID, since $\mathbb{Z}/(p)$ is cyclic an morally if an ideal ...
1
vote
1answer
29 views

Decomposition in a PID

I am working towards understanding the statement and proof of the following theorem: Let $R$ be a PID and $a_1,...,a_r \in R \setminus \{0_R\}$. Then, there are $q_1,...,q_s \in R\setminus \{0_R\}$...
5
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0answers
142 views

Applications of the Dedekind-Hasse criterion

It is a fact that an integral domain $R$ is a principal ideal domain if and only if there is a Dedekind-Hasse function $|R|\setminus\{0\}\xrightarrow{\ \ \delta\ \ }\mathbb{N}$ on $R$, i.e. a function ...
1
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1answer
26 views

$M$ is the (possibly infinite) direct sum of its $p$-primary components as $p$ runs over all primes of $R$.

Let $R$ be PID, let $M$ be a torsion $R$-module and $p$ be prime in $R$. Prove that $M$ is (possibly infinite) direct sum of its $p$-primary components as $p$ runs over all primes of $R$. Take $m \in ...
5
votes
2answers
65 views

Is $\mathbf{Z}[X]/(2,X^2+1)$ a field/PID?

I've been asked to determined whether the following are fields, PIDs, UFDs, integral domains: $$\mathbf{Z}[X],\quad \mathbf{Z}[X]/(X^2+1),\quad \mathbf{Z}[X]/(2,X^2+1)\quad \mathbf{Z}[X]/(2,X^2+X+1)$$ ...
0
votes
2answers
61 views

Is it possible to find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i)=(\alpha)$? [closed]

Is it possible to find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i)=(\alpha)$? Is anyone could give me a full explication in ''Answer the question''?
3
votes
1answer
45 views

Prime ideal in R[x] is either principal or $\mathfrak p = (q,f)$

$R$ is a PID and $\mathfrak p$ a prime ideal of $R[x]$. Show that $\mathfrak p$ is principal or $\mathfrak p = (q,f)$ for some $q\in R$ prime and $f \in R[x]$ monic. I can't figure out this problem. ...
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2answers
52 views

Show that I is a principal ideal

Let $\mathbb{Z}[\sqrt{-13}]$ be the smallest subring of $\mathbb{C}$ containing $\mathbb{Z}$ and $\sqrt{-13}$ and let the ideal $I = \left<2,\sqrt{-13}\right>$. Show that $I$ is a principal ...
2
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2answers
59 views

Find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i) = (\alpha)$ [closed]

Let $\mathbb{Z}[i] = \{a+bi : a,b \in \mathbb{Z}\}$ and $\mathbb{Q}(i) = \{a+bi : a,b \in \mathbb{Q}\}$. Find $\alpha \in \mathbb{Z}[i]$ such that $(3+5i,1+3i) = (\alpha)$ Definition : If $A$ ...
3
votes
3answers
46 views

$R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ , $R/I$ is finite ; then is $R$ a PID or at least Noetherian?

Let $R$ be an infinite commutative ring with unity such that for every non-zero ideal $I$ of $R$ , $R/I$ is finite; then is $R$ a PID or at least Noetherian ? I can only prove that $R$ must be an ...
1
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1answer
34 views

Torsion elements with relatively prime orders

Let $x$ and $y$ be torsion elements in an R-module $M$ having orders $a$ and $b$. With $a,b \in R$, $a$ and $b$ are relatively prime, and $R$ is a PID. I want to show that $x + y$ has order $ab$. So ...
5
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0answers
83 views

Elementary divisors for chains of submodules

Given free modules $N \le M$ of finite rank over a PID $R$, it's well-known that there is a basis $\{x_1,\ldots,x_n\}$ of $M$ and there are $e_1,\ldots,e_n \in R$ such that $\{e_ix_i\mid e_i \neq 0\}$...
3
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1answer
57 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?
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votes
1answer
42 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?
8
votes
1answer
143 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
20 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
39 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
38 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 \...
2
votes
4answers
183 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
108 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 ...
2
votes
2answers
63 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, r_2/...
0
votes
2answers
24 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 ...
2
votes
1answer
38 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
67 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 $$R^*=\{cX^i\text{mod}(XY-1):c\in\mathbb{R}...
2
votes
1answer
59 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
62 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
144 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 Z}^{n}(M,...
2
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0answers
39 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
44 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
38 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
43 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
66 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
61 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 $\mathbb{Z}...
2
votes
1answer
80 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 ...
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vote
1answer
14 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 see ...
8
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2answers
430 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 ...
0
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1answer
20 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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2answers
93 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. My ...
4
votes
2answers
48 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 $n$
2
votes
1answer
54 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
114 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 \mathbb{Z}...
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0answers
41 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 ...
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vote
2answers
157 views

Is a localization of a UFD at a prime ideal a PID? [duplicate]

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 R\...
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votes
2answers
65 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
29 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 ...
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0answers
39 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
votes
0answers
62 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 $f=...
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0answers
98 views

Dummit & Foote <Abstract Algebra> Exercise 8.2.4 [duplicate]

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 $...
2
votes
0answers
78 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 ...