Tagged Questions
4
votes
3answers
81 views
Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring.
Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring. I know the definition of principal ideal ring is that every ideal is ...
2
votes
1answer
74 views
which of the following statements are true and why?
which of the following statements are true and why?
Any two irreducibles in any UFD are associates.
If $D$ is a PID, then $D[x]$ is a PID.
In any UFD, if $p|a$ for an irreducible $p$, then $p$ ...
0
votes
2answers
152 views
Show that $\mathbb{Z}[\theta]$ (where $\theta = (1 + \sqrt{19}i)/2$) is a principal ideal domain.
I'm having difficulties with a homework problem from Algebra by Hungerford.
Let $R$ be the following subring of the complex numbers: $R = \{a + b(1 + \sqrt{19}i)/2 \mid a, b \in \mathbb{Z}\}$. ...
3
votes
1answer
115 views
Units in $\mathbb{Z}[\sqrt[3]{2}]$ : $\pm(1+\sqrt[3]{2}+(\sqrt[3]{2})^2)^n$?
The subring $\mathbb{Z}[\sqrt[3]{2}]\subset\mathbb{C}$ is a PID. I remember reading somewhere that the units in $\mathbb{Z}[\sqrt[3]{2}]$ are precisely the elements ...
4
votes
1answer
86 views
Is $(x^3-x^2+2x-1)$ prime in $\mathbb{Z}/(3)[x]$?
This is somewhat of a follow up on this question: Why is $(3,x^3-x^2+2x-1)$ not principal in $\mathbb{Z}[x]$?
I'm curious, is $\mathbb{Z}[x]/I$ a domain, with $I=(3,x^3-x^2+2x-1)$? I know $I$ is not ...
2
votes
1answer
129 views
Ring of analytic functions on the circle
Let $A = C^\omega(S^1)$ (resp. $C^\omega_{\mathbb C}(S^1)$) the ring of real-analytic real-valued (resp. complex valued) functions on the circle.
These rings have maximal ideals $\mathfrak m_p = ...
2
votes
4answers
134 views
If two elements in a ED have the same Euclidean norm, they are associates?
Is it very obvious that on a Euclidean Domain, two elements $x$ and $y$ have the same Euclidean norm $\nu(x) = \nu(y)$ then they are associates?
Can someone give me a proof of this?
1
vote
1answer
181 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 ...
2
votes
1answer
117 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 ...
8
votes
1answer
265 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 ...
5
votes
3answers
479 views
Dedekind domain with a finite number of prime ideals is principal
I am reading a proof of this result that uses the Chinese Remainder Theorem on (the finite number of) prime ideals $P_i$. In order to apply CRT we should assume that the prime ideals are coprime, i.e. ...
2
votes
1answer
145 views
Number of ideals of a PID modulo an ideal
Let $R$ be a Principal Ideal Domain and $(a)\neq(0)$ an ideal of $R$. Prove $R/(a)$ has a finite number of ideals.
0
votes
1answer
173 views
$R$ is PID, so $R/I$ is PID, and application on $\mathbb{Z}$ and $\mathbb{N}$
I'm supposed to show in a part of an exercise that if we have a ring $R$ that is a principal ideal domain, then for any ideal $I$ in $R$, $R/I$ will also be a PID.
So $I=(i)$ for some $i \in R$, and ...
5
votes
2answers
379 views
How to show that $R/I$ is Artinian when R is PID
I'm working through some of Hungerfords "Algebra", and having trouble with Excercise VIII 1.2.:
Show that if $I$ is a non-zero ideal in a principal ideal domain (PID) $R$, then the ring $R/I$ is ...
2
votes
1answer
353 views
Norm-Euclidean rings?
For which integer $d$ is the ring $\mathbb{Z}[\sqrt{d}]$ norm-Euclidean?
Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a,b \in \mathbb{Z}\}$, not the ring of integers of ...
1
vote
1answer
160 views
Can we consider set of all composite integers as an ideal? And if yes, why then Z is a PID?
In this wikipedia article it is said that set Z is a principal ideal domain, i.e. each one of its ideals can be generated by a single element.
But if we consider set C of all composite integer numbers ...
