Tagged Questions

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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 ...
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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? ...
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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 ...
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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 ...
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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 ...
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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 ...
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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 $B$.
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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 ...
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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?
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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).
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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 $\forall a \in R$, ...
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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 ...
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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 ...
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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 ...
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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.
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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 ...
347 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 ...
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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 ...
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? ...
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 $B$ ...