# Linked Questions

16 questions linked to/from One-dimensional [Noetherian] UFD is a PID
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### 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 ...
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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 $... 1answer 93 views ### Let$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$is ... 1answer 31 views ### Quadratic Rings which are UFDs but not PIDs [duplicate] D=-1,-2,-3,-7 and-11 are the only five negative square free integers for which the corresponding quadratic Rings are Euclidean domains. Also D=-19,-41,-43 and-167 are the only four square free ... 0answers 35 views ### the condition when a UFD is a PID [duplicate] Prove that a$UFD$is a$PID$if and only if every prime ideal is maximal. I know a proof but used Zorn's lemma.Is there a proof that doesn't require Zorn's lemma? 3answers 2k views ### Sufficient conditions for being a PID Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, ... 3answers 289 views ### Given$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$negative here, though that might provide a clue I ... 4answers 2k views ### Let$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]$. How to prove? I really have no idea... Thank you a lot. 2answers 499 views ### What information do we gain from PIDs I am self-learning some algebraic number theory and my question is regarding the advantages to studying PIDs. I have seen that Euclidean Domains$\subseteq$Principal Idea Domains$\subseteq$Unique ... 3answers 550 views ###$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]$. Comments: I was able to solve the first case using the division ... 1answer 438 views ### Is there an example of a non-noetherian one-dimensional UFD? Or in the contrapositive form Is every one-dimensional UFD noetherian? I know how to construct a non-noetherian UFD (polynomials in infinite number of variables over a field) and I know that it is ... 2answers 213 views ###$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$is irreducible. So if I ... 2answers 110 views ### Constructing nonprincipal ideals in a non-UFD It's well-known that all PIDs are UFDs, i.e. all non-UFDs are not PIDs. Now, it seems to me that there are two ways that a ring$R$could fail to be a UFD: Some element$x$has no factorisation into ... 1answer 246 views ### Can we always write$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$... 1answer 53 views ###$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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