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0 votes
0 answers
38 views

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 ...
Adam Rubinson's user avatar
2 votes
0 answers
106 views

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 ...
Eric Wheeler's user avatar
3 votes
2 answers
677 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 ...
abeliangrape's user avatar
0 votes
1 answer
80 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?
zjy's user avatar
  • 55
0 votes
1 answer
645 views

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 ...
Kishalay Sarkar's user avatar
1 vote
2 answers
364 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 ...
jskattt797's user avatar
  • 1,751
3 votes
3 answers
759 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.
Promit Mukherjee's user avatar
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 ...
Mark's user avatar
  • 9
2 votes
1 answer
881 views

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 ...
GreginGre's user avatar
  • 15.3k
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 ...
Alex C's user avatar
  • 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 ...
carciofo21's user avatar
1 vote
1 answer
91 views

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}=...
AfterMath's user avatar
  • 852
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 ...
Snate's user avatar
  • 192
3 votes
1 answer
479 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 ...
user313212's user avatar
  • 2,236
3 votes
1 answer
385 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 ...
Erdbeer0815's user avatar

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