All Questions
19
questions
3
votes
2
answers
682
views
Definition of UFD and the fact that UFDs are integrally closed
I am trying to understand the proof of the fact that UFDs are integrally closed. In addition to the lecture notes I have, there are at least two solutions on MSE:
One is here: How to prove that UFD ...
- 603
0
votes
1
answer
82
views
An example showing $\mathbb{Z}[\sqrt[3]{7}]$ is not a UFD [closed]
It cannot be a UFD because it's the ring of integers of $\mathbb{Q}(\sqrt[3]{7})$ and has class number 3. How can we give an example showing this?
- 55
1
vote
2
answers
365
views
If $R$ is a UFD, then $R[x]$ is Noetherian?
If $R$ is Noetherian, then $R[x]$ is Noetherian. However, if $R$ is a UFD, then $R$ might not be Noetherian. $\DeclareMathOperator{\Frac}{Frac}$
I want to show that if $R$ is a UFD, then $R[x]$ is a ...
- 1,751
3
votes
3
answers
764
views
Is $\mathbb{Z}[√13] $ a Unique factorization domain?
I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.
- 241
0
votes
1
answer
94
views
Disproving unique factorization of $\mathbb{Q}(\sqrt{-6})$
For homework, we were given a problem that asked to explain why $(\sqrt{-6})(-\sqrt{-6})$ implies $\mathbb{Q}(\sqrt{-6})$ does not have a unique factorization.
I understand the unique part in this ...
- 9
1
vote
0
answers
117
views
Factorization in a principal ideal ring/rng
It is known that every PID is a UFD.
Is it true that every element of a commutative principal ideal ring (PIR) or rng that is not zero and not a unit is a product of a finite number of irreducible ...
- 1,120
0
votes
2
answers
138
views
Number of irreducible divisors in a UFD
The following fact is totally obvious, but I cannot find a way to prove it.
Let $R$ be a UFD and $a \in R$ be non zero and non invertible. Factor it as a product of irreducible elements: $a=p_1 \cdot ...
- 579
1
vote
1
answer
177
views
Ring with infinitely reducible elements
Can you give or construct an elementary example of a factorial ring with elements which are product of infinitely many irreducible elements? i.e. there are reducible elements that can't be written as ...
- 192
3
votes
1
answer
480
views
Why a certain integral domain is not a UFD.
Let
$$\mathbb{Z}[q]^{\mathbb{N}} = \varprojlim_j \mathbb{Z}[q]/((1-q)\cdots (1- q^j))$$
Why isn't $\mathbb{Z}[q]^{\mathbb{N}}$ a unique factorization domain?
The author proposes a proof whose ...
- 2,236
3
votes
1
answer
386
views
Number of prime elements in a ring
Is there any way to count the prime elements in a ring?
More precisely a way to count prime elements in a UFD? Which would be the same as counting irreducibles. Are there even UFD with finitely many ...
- 547
17
votes
1
answer
963
views
What is the correct notion of unique factorization in a ring?
I was recently writing some notes on basic commutative ring theory, and was trying to convince myself why it was a good idea to study integral domains when it comes to unique factorization.
If $R$ is ...
- 997
1
vote
2
answers
136
views
Justifying the representation of a monic polynomial over a UFD
Let $D$ be a UFD and $f(x) \in D[x]$ be monic. The book I'm reading from claims that $$f(x) = p_1(x)^{e_1} \cdots p_n(x)^{e_n}$$
where $p_i(x)$ are distinct, irreducible, and monic, and $e_i >0$.
...
- 4,039
2
votes
1
answer
306
views
A simple algebraic ring extension of a UFD having no prime elements
Let $D$ be a UFD over a field $k$ of characteristic zero.
Assume that $w$ is algebraic over $D$.
Denote $R=D[w]$.
Observe that $R$ is not necessarily a UFD.
Can one find an example in which $R$ ...
- 6,705
3
votes
1
answer
84
views
A non-UFD $B$ such that $A \subset B \subset C$, where $A \cong C$ are UFD's
Let $A \subset C$ be two isomorphic unique factorization domains (UFD's).
Is it possible to find an integral domain $B$, $A \subset B \subset C$,
such that $B$ is not a UFD?
I have tried to ...
- 6,705
2
votes
1
answer
154
views
Proof help involving prime factorization, exponents
Let $a,b\in\mathbb{N}$. Moreover, let $p_1,p_2,...,p_k$ be the collection of all primes which divide $a$ or $b$ or both. We'll write $a=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ and $b=p_1^{\...
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