1
vote
1answer
45 views

Difference between PID and principal ideal ring

All rings are commutative, associative and with 1. Wikipedia states that the difference between PID and Principal Ideal Ring is that the former has to be integral domain while the latter does not. ...
1
vote
2answers
147 views

Is $\mathbb{C}[x,y] / (y^2-x^3)$ a PID?

First, I'd like to show $\mathbb{C}[x,y] / (y^2-x^3)$ is an integral domain. Then I need to find out whether or not it is a PID. For the first part, I want to show $y^2-x^3 \: | \: fg \implies ...
18
votes
2answers
353 views

Quotient of polynomials, PID but not Euclidean domain?

While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$ is ...
2
votes
1answer
214 views

Is any UFD also a PID?

Is there any counterexample that will disprove that every unique factorization domain (UFD) is also a principal ideal domain (PID)? I mean, any PID is a UFD, does the converse hold? Thanks in ...
0
votes
2answers
58 views

Nonprincipal prime ideals contain two relatively prime elements

Let $R$ be a principal ideal domain and let $P$ be a nonprincipal prime ideal of $R[x]$. I'm having trouble seeing why $P$ must contain two elements with no common divisor. Can anyone help me? ...
4
votes
1answer
149 views

A question on valuation overrings of a PID

Let $A$ be a PID and let $K$ be its quotient field. Let $V$ be a valuation ring of $K$ containing $A$ and assume $V\neq K$. Show that $V$ is a local ring $A_{(p)}$ for some prime element $p$. I ...
3
votes
2answers
54 views

If $N\cap rM=rN$ for all $r\in R$, then is $M=N\oplus K$ for some $K$?

Suppose $M$ is a finitely generated free module over a principal ideal domain $R$, and $N$ a submodule. Why does the condition $N\cap rM=rN$ for all $r\in R$ implies that $M=N\oplus K$ for some ...
1
vote
1answer
109 views

Infinite direct product of rings free.

Let $A$ be a commutative ring (viewed as an $A$-module over itself) that is not a field. Are there some conditions that guarantee that $\prod_{k=0}^\infty A$ is free? What if $A=\mathbf{Z}$ or more ...
2
votes
1answer
76 views

Intersection of a PID and a field

There is an exercise in Bourbaki about the intersection $A = k (x,y) [z] \cap k (z, x + yz)$. These are two subrings of $k (x,y,z)$. The first is PID, the second is a field. Bourbaki requests to prove ...
2
votes
1answer
58 views

Finite intersection of DVRs

Let $K$ be a field and $R_1,\dots,R_n$ DVRs of $K$ with $m_i$ the maximal ideal of $R_i$ and $R_i \not\subseteq R_j$ for $j\neq i$ . Define $A=\bigcap_{i=1}^n R_i$. Then $A$ is semilocal with maximal ...
0
votes
0answers
24 views

Example of principal v-ideal

I am confuse in constructing counter example that If $A⊆B$ is an extension of domains such that $J$ is a principal $v$-ideal in $A$, then $JB$ is also principal in $B$.
3
votes
1answer
102 views

PID modulo a non-zero ideal is a semilocal ring

Let $R$ be a commutative ring, $\mathfrak{m}\subset R$ a maximal ideal and $f$ a monic polynomial in $R[x]$. I want to show that $A:=\frac{R[x]}{\mathfrak{m}[x]+(f)}$ is a semilocal ring, where ...
2
votes
2answers
289 views

radical of sum of two ideals

$I$ and $J$ are ideals in $k[x_1,\cdots,x_n]$. Show that $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$. I have no idea how to prove it. Can someone help?
5
votes
3answers
820 views

Ring of trigonometric functions with real coefficients

Let $R$ be the ring of functions that are polynomials in $\cos t$ and $\sin t$ with real coefficients. Prove that $R$ is isomorphic to $\mathbb R[x,y]/(x^2+y^2-1)$. Prove that $R$ is not a unique ...
8
votes
4answers
1k views

An integral domain whose every prime ideal is principal is a PID

Does anyone has a simple proof of the following fact: An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
3
votes
2answers
262 views

A question about Euclidean Domain

This is a problem from Aluffi's book, chapter V 2.17. "Let $R$ be a Euclidean Domain that is not a field. Prove that there exists a nonzero, nonunit element $c$ in $R$ such that $\forall a \in R$, ...
0
votes
3answers
1k views

Show that the ring of all rational numbers, which when written in simplest form has an odd denominator, is a principal ideal domain.

Show that the ring of all rational numbers $m/n$ with $n$ an odd integer is a principal ideal domain. We haven't really discussed principal ideal domains. I've heard that this is easy, but I just ...
14
votes
2answers
1k views

A subring of the field of fractions of a PID is a PID as well.

Let $A$ be a PID and $R$ a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the field of fractions of $A$. How to show $R$ is also a PID? Any ...
2
votes
1answer
248 views

Rings such that $A[x]$ is a principal ideal domain

Let $A$ be a commutative ring. Then the following assertions are equivalent. $A$ is a field; $A[x]$ is a Euclidean domain; $A[x]$ is a principal ideal domain; $A[x]$ is a unique factorization ...
5
votes
2answers
211 views

Characterization of primary ideals in a principal ideal domain

On the commutative algebra wiki, a table of properties lists that "for a PID, the primary ideals coincide with the powers of prime ideals." I played around with it, couldn't produce a proof, ...
0
votes
4answers
3k views

Show that every ideal of the ring $\mathbb Z$ is principal

Let $\mathbb Z$ be the ring of integers. The question asks to show that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.
1
vote
1answer
139 views

Queries on proof that every PID is a factorisation domain

I'm reading a proof from C. Musili's Rings and Modules that every PID is a factorisation domain. The author defines a factorisation domain as a commutative integral domain $R$ with a unit such that ...
10
votes
1answer
378 views

Are all subrings of the rationals Euclidean domains?

This is a purely recreational question -- I came up with it when setting an undergraduate example sheet. Let's go with Wikipedia's definition of a Euclidean domain. So an ID $R$ is a Euclidean domain ...
8
votes
1answer
671 views

Ring of Polynomials is a Principal Ideal Ring implies Coefficient Ring is a Field?

I read this proof that if $D$ is an integral domain and $D[X]$ is a principal ideal domain, then $D$ is a field. My question is if the requirements can be relaxed a bit, namely: Is it true that ...
3
votes
1answer
139 views

Do a matrix and its transpose have the same invariant factors over a PID?

I suspect this is true since it holds in the case over a field. But suppose $A\in M_{m\times n}(R)$ where $R$ is a PID. Does it still hold that $A$ and $A^{T}$ have the same invariant factors? ...
10
votes
1answer
348 views

Finitely generated modules over PID

Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus $ $B$ $\cong$ $C\oplus $ $D$ and $A\oplus $ $D$ $\cong$ $C\oplus $ $B$ . Prove that $A$ $\cong$ $C$ and $B$ ...