Linked Questions

3
votes
2answers
351 views

When is a Unique Factorization Domain a Principal Ideal domain [duplicate]

"Let $R$ be a Unique Factorization Domain and $(a,b)=(c)$ for $a,b,c \in R$. Show that $R$ is a Principal Ideal domain." To be honest I found this very hard, here is my naive try: Lets assume $R$ is ...
1
vote
1answer
61 views

Why isn't $\{0\}$ being prime ideal is not maximal in $\mathbb{Z}$? [duplicate]

I am confused with the following arguments :- $\mathbb{Z}$ is a Euclidean Domain with the evaluation map $\phi(r)=|r|$ and so it is a PID. The ideal $\{0\}$ is a prime ideal in $\mathbb{Z}$ since $...
0
votes
1answer
93 views

Let $D$ be a UFD such that Bezout's Identity holds. Then Every ideal is finitely generated implies that $D$ is PID [duplicate]

My proof $I=<a_1,\dots,a_n>=Da_1+Da_2+\cdots +Da_n$ let $g=gcd(a_1,\dots,a_n)$ Then since Bezout's identity holds and the binary operator gcd is associative, $g\in I$. Also any element of $I$ is ...
0
votes
1answer
31 views

Quadratic Rings which are UFDs but not PIDs [duplicate]

D=-1,-2,-3,-7 and-11 are the only five negative square free integers for which the corresponding quadratic Rings are Euclidean domains. Also D=-19,-41,-43 and-167 are the only four square free ...
0
votes
0answers
35 views

the condition when a UFD is a PID [duplicate]

Prove that a $UFD$ is a $PID$ if and only if every prime ideal is maximal. I know a proof but used Zorn's lemma.Is there a proof that doesn't require Zorn's lemma?
10
votes
3answers
2k views

Sufficient conditions for being a PID

Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, ...
14
votes
3answers
289 views

Given $d \equiv 5 \pmod {10}$, prove $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ never has unique factorization

With the exception of $d = 5$, which gives $\mathbb{Z}[\phi]$, of course (as was explained to me in another question). I'm not concerned about $d$ negative here, though that might provide a clue I ...
-2
votes
4answers
2k views

Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$. [closed]

Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$. How to prove? I really have no idea... Thank you a lot.
8
votes
2answers
499 views

What information do we gain from PIDs

I am self-learning some algebraic number theory and my question is regarding the advantages to studying PIDs. I have seen that Euclidean Domains $\subseteq$ Principal Idea Domains $\subseteq$ Unique ...
2
votes
3answers
550 views

$K[x_1,x_2, …, x_n]$ is not a PID if $\,n> 1$

Every ideal in $K[x]$ is of the form $(p(x))$ for some $p(x) \in K[x]$. But the result is not valid for the ring $K[x_1,...x_n]$. Comments: I was able to solve the first case using the division ...
4
votes
1answer
438 views

Is there an example of a non-noetherian one-dimensional UFD?

Or in the contrapositive form Is every one-dimensional UFD noetherian? I know how to construct a non-noetherian UFD (polynomials in infinite number of variables over a field) and I know that it is ...
0
votes
2answers
213 views

$R/Rg$ is a field iff $g\in R$ is irreducible.

Let $R$ be a PID and $g\in R$. I want to show: $R/Rg$ is a field iff $g\in R$ is irreducible. I.e. I want to show that all $a\notin Rg$ are invertible modulo $g$ iff $g$ is irreducible. So if I ...
4
votes
2answers
110 views

Constructing nonprincipal ideals in a non-UFD

It's well-known that all PIDs are UFDs, i.e. all non-UFDs are not PIDs. Now, it seems to me that there are two ways that a ring $R$ could fail to be a UFD: Some element $x$ has no factorisation into ...
1
vote
1answer
246 views

Can we always write $gcd(x,y)$ as $ax+by$ in UFD?

Let $R$ be a commutative ring with unity. Now assume that $R$ is Unique Factorization Domain, but not necessarily Principal Ideal Domain. Question: Let $x,y\in R$ be such that their GCD exists in $R$...
1
vote
1answer
53 views

$f(x),g(x)$ co-prime in $R[X]$ with $R$ a UFD implies $f(x), g(x) $ co-prime in $\operatorname{Frac}(R)[X]$

Let $R$ be a UFD and suppose $f(x),g(x)$ are co-prime in $R[X]$. I want to show that $f(x), g(x)$ are then also co-prime in $\operatorname{Frac}(R)[X]$, where $\operatorname{Frac}(R)$ is the field of ...

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