All Questions
Tagged with prime-factorization unique-factorization-domains
35
questions
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Do all polynomials $p(x)$ with integer coefficients output infinitely many composite numbers for integer $x$? [duplicate]
Inspired by this question, I was wondering if there was a simple proof that if $p(x)$ is a polynomial of degree $\geq 1$ and with integer coefficients, then $p(x)$ is a composite number for infinitely ...
2
votes
0
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106
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What are the integers that lack unique factorization in $\mathbb Z[\zeta_n],$ $n= 23$? [closed]
In the cyclotomic integers $\mathbb{Z}[\zeta_n]$, $n=23$ has class number $3$ and so unique factorization fails (https://en.wikipedia.org/wiki/Cyclotomic_field ). Can anyone give me examples of such ...
3
votes
2
answers
677
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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 ...
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1
answer
80
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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?
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645
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Prime elements of the ring $\mathbb Z[i]$.
I want to find the prime elements of the ring $\mathbb Z[i]$ of Gaussian integers.Define $d(a+ib)=a^2+b^2$ for all $a+ib\in \mathbb Z[i]-\{0\}$.I think if $d(a+ib)$ is prime in $\mathbb Z$ then and ...
1
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2
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364
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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 ...
3
votes
3
answers
759
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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.
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1
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94
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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 ...
2
votes
1
answer
881
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Does a prime ideal contains an irreducible element?
The context. Let $R$ be an integral domain. It is known that a domain $R$ is a UFD if and only if any nonzero prime ideal contains a prime element.
It is also known that $R$ is a UFD if and only if ...
1
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0
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117
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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 ...
0
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2
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138
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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 ...
1
vote
1
answer
91
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Showing irreducibility is invariant under conjugation in quadratic integers
I found the following statement in a paper I'm reading on (the integers in) a quadratic field:
If $\alpha =a+b\sqrt{D}$ is an irreducible element of $\mathbb{Z}[\sqrt{D}]$, then so is $\bar{\alpha}=...
1
vote
1
answer
177
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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 ...
3
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1
answer
479
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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 ...
3
votes
1
answer
385
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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 ...