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

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43 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 $\mathbb{Q}$.
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3answers
30 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 ...
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1answer
29 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
63 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 ...
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2answers
58 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 ...
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1answer
78 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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1answer
12 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 ...
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2answers
427 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
44 views

prove that in PID every ideal has a unique factorization into prime ideals [on hold]

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 ...
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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 ...
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2answers
89 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
46 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 $n$
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1answer
49 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
105 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
37 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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2answers
125 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 ...
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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 $\mathbb{Z}[X]/(5X,X^2)$. ...
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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
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 ...
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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 ...
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0answers
97 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 ...
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0answers
60 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 ...
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1answer
45 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 ...
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1answer
69 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
78 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
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1answer
113 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
35 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 $bM=M_a$ ...
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74 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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58 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 ...
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1answer
33 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 ...
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1answer
85 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?
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0answers
40 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
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1answer
47 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 ...
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1answer
28 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 ...
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2answers
71 views

$R$ a local ring, also a PID. $I,J$ ideals from $R$. Show that $I \subseteq J$ or $J \subseteq I$

$R$ a local ring, also a PID. $I,J$ ideals from $R$. Show that $I \subseteq J$ or $J \subseteq I$ My brief attempt to try use Bezout theorem at a PID. but unsuccess.. Thanks any help.
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1answer
127 views

Proof: $\mathbb{Z}[\zeta_6]$ is a PID.

I am reading through A First Course in Modular Forms. In Proposition 2.2.3 they claim that $\mathbb{Z}[\zeta_6]$ is known to be a principal ideal domain. Does anyone have a reference for the proof of ...
2
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1answer
33 views

Decomposition of Torsion Module

Let $k$ be a field, $k[X]$ the polynomial ring in one variable and $M$ a torsion $k[X]$-module (not necessarily finitely generated). Consider the submodules \begin{equation*} M_1 = \{a \in M \mid ...
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2answers
169 views

Does there exist such an invertible matrix?

Let $n \geq 1$ and $A = \mathbb{k}[x]$, where $\mathbb{k}$ is a field. Let $a_1, \dots, a_n \in A$ be such that $$Aa_1 + \dots + Aa_n = A.$$ Does there exist an invertible matrix $\|r_{ij}\| \in ...
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1answer
52 views

Finite (cardinality) modules over a PID

Let $R$ be a principal ideal domain with $|R|=\infty$. Suppose $M$ is an $R$-module such that $2 \le |M| < \infty$. What properties does this imply about $R$? Background: I was hoping that $R\cong ...
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1answer
191 views

Principal ideal domains that are not integral domains

In the usual definition, a principal ideal domain $R$ is also assumed to be an integral domain? However, the property that every ideal is generated by a single element does not seem to immediately ...
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2answers
169 views

Prime ideals in $R[x]$, $R$ a PID

Let $R$ be a PID. Show that if $\mathfrak p \subset R[x]$ is a prime ideal, $(r) = \left\{h(0) \colon h(x) \in \mathfrak p \right\}$, and $$\mathfrak p = (r, f(x), g(x)),$$ where $f(x), g(x) \in ...
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1answer
55 views

Why is $(2,x)$ non-principal in $\Bbb Z[x]$? [duplicate]

Why is $(2,x)$ non-principal in $\Bbb Z[x]$? Apparently this is the case, I just read it on wiki, as a counter example to $\Bbb Z[x]$ being a PID. What is $x$ here? I mean $2$ can surely generate ...
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2answers
118 views

Proof about finitely generated torsion-free R-module M is free, where R is a PID

Proof about finitely generated torsion-free $R$-module $M$ is free, where $R$ is a PID. Can I prove by the following way? Proof by induction on $n$, where $M=\langle v_1,...,v_n\rangle$. If ...
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1answer
63 views

Principal ideal domain, $\forall x=(x_1,x_2)^t \in R^2~~\exists G \in SL_2(R) : Gx=(\gcd(x_1,x_2), 0)^t$

Let $R$ be a principal ideal domain. Prove that for every $x=(x_1,x_2)^t \in R^2$ exists a matrix $G \in SL_2(R)$ for which $Gx=(\gcd(x_1,x_2), 0)^t$. I think it's easy, but do not know how to ...
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2answers
40 views

Distributive law of ideals in $\mathbf Z$ relating $\cap$ & $+$

Let $\mathfrak a, \mathfrak b$ and $\mathfrak c$ be ideals in $\mathbf Z$. Then show that $$ \mathfrak a \cap (\mathfrak b + \mathfrak c) = \mathfrak a \cap \mathfrak b + \mathfrak a \cap \mathfrak c ...
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1answer
48 views

Proving that for an integral domain $R$, $y\in (x)\iff (y)\subseteq (x)$.

I am trying to prove the following statement. Let $R$ be a integral domain. Then for all $x,y\in R$ we have $$x\mid y\iff y\in(x)\iff (y)\subseteq (x).$$ Note that $(x)$ denotes the principal ...
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2answers
57 views

Prove that $(5x^3+9x^2-27x+3)$ is a maximal ideal in $\mathbb{Q}[x]$

We have shown that $Q[x]$ is a Euclidean Domain, and thus is a Principal ideal domain. A principal ideal $(f)$ is maximal $\iff$ $f$ is irreducible in $Q[x]$. But how do I show that \begin{equation*} ...
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1answer
40 views

all abelian groups with 625 elements with 24 elements of order 5

Let $R$ be a principal ideal domain, $p \in R$ a prime element and $M$ a finitely generated $p$-torsion module of the form $M = R/(p^{e_1}) \oplus \cdots \oplus R/(p^{e_t})$. Let $_pM = \{m \in M: p ...
0
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1answer
37 views

vector space constructed through a torsion module

Let $R$ be a principal ideal domain, $p \in R$ a prime element and $M$ a finitely generated $p$-torsion module of the form: $$ M = R/(p^{e_1}) \oplus \dots \oplus R/(p^{e_t}). $$ Let now be $_pM = ...
0
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0answers
41 views

properties of finitely generated torsion-modules and their submodules over a PID

Let $R$ be a principal ideal domain, $M$ a finitely generated $R$-torsion module, and $N \subseteq M$ a submodule. I want to show that there exist free R-modules $F, F', F''$ and module homomorphisms ...