Dec 17, 2014 · I was wondering, is the localization of a UFD also a UFD? How would one go about proving this? It seems like it would be kind of messy to prove if it is true. If it is not true, …

Oct 16, 2015 · The point of a UFD is that any element can be rewritten as a product of irreducible factors, where any other product of irreducible factors is just a rearrangement of the exact …

Jan 31, 2020 · So we have the following characterization of UFDs: $$\text {UFD}\iff \text {Atomic domain}+\text {AP-domain}.$$ Therefore you can prove that a PID is a UFD showing that is …

The fact that $A$ is a UFD implies that $A [X]$ is a UFD is very standard and can be found in any textbook on Algebra (for example, it is Proposition 2.9.5 in these notes by Robert Ash).

Nov 28, 2018 · For example, the Gaussian integers $R_ {-1}$ are a $UFD$ whereas the ring $R_ {-5}$ is not. There are several ways to show this, including computing the class number of …

Mar 14, 2019 · In general, a subring of a UFD need not be a UFD. As mentioned in an answer to this question, the easiest counterexamples are constructed by choosing an integral domain …

Prove that if R is a UFD then R [x] is also a UFD. [duplicate] Ask Question Asked 8 years, 10 months ago Modified 8 years, 10 months ago

Aug 16, 2018 · To show that $\mathbb {Z} [x]$ is a UFD I must show that every non-zero elements has a unique factorisation into a product of (not necessarily distinct) primes and a …

This is norm-Euclidean and in particular a UFD. Note that all UFDs are integrally closed, so this is another way to see that $\mathbf {Z} [\sqrt {-3}]$ does not have unique factorization.

Dec 28, 2021 · However, in a UFD it is common to prove the Law and Lemma using prime factorizations - which yields less generality and, further, is less constructive - since in many …