About the localization of a UFD - Mathematics Stack Exchange
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Understanding the definition of Unique Factorization Domains
2015年10月16日 · Actually, you should think in this way. UFD means the factorization is unique, that is, there is only a unique way to factor it. For example, in $\mathbb{Z}[\sqrt5]$ we have $4 =2\times 2 = (\sqrt5 -1)(\sqrt5 +1)$. Here the factorization is not unique.
Is every PID a UFD? - Mathematics Stack Exchange
2020年1月31日 · As you noted, being a UFD requires the uniqueness of such a factorization and this leads to another definition: an AP-domain is a domain where every atom is prime. So we have the following characterization of UFDs: $$\text{UFD}\iff \text{Atomic domain}+\text{AP-domain}.$$
abstract algebra - Gauss' Lemma prove $\mathbb {Z} [x]$ UFD ...
2018年8月16日 · I would direct you to my Monthly paper with David McKinnon for a general discussion on Gauss's Lemma and unique factorization.
UFDs are integrally closed; so too are GCD & Dedekind domains.
The well-known simple proof of RRT = Rational Root Test remains valid in a UFD or GCD domain. The sought result is simply the special case when $\:\!\rm f(x)\:\!$ is monic (lead coef $=1$ ). Generalization: $ $ we prove RRT in a form that works for both $\rm\color{#0a0}{gcds}$ and cancellable $\rm\color{#c00}{ideals}$ by using only universal ...
abstract algebra - Does $A$ a UFD imply that $A[T]$ is also a UFD ...
Etc. The same holds for polynomials over a field, since a field is trivially a UFD (or because you know ...
number theory - When are quadratic rings of integers unique ...
2018年11月28日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Is a polynomial ring over a UFD in countably many variables a UFD?
Cohn proved that every gcd domain can be inertly embedded in a Bezout domain, and every UFD can be inertly embedded in a PID. There are a few variations on the notion of inertness that prove useful when studying the relationship between factorizations in base and extension rings, e.g. a weaker form where $\, d = ab\,\Rightarrow\, au, b/u\in D ...
abstract algebra - Prove that a UFD is a PID if and only if every ...
Let R be a UFD in which every nonzero prime ideal is maximal. Then R is a PID. Lemma 1. If p is a nonzero irreducible element of R, then (p) is a prime ideal of R. Proof. Let a and b be nonzero elements of R such that ab ∈ (p), i.e. ab = pf for some nonzero element f ∈ R.
Is $\mathbb Z[ \sqrt{-3}]$ UFD? - Mathematics Stack Exchange
FYI, $\mathbb{Z}[\sqrt{-3}]$ is not only not a UFD, but it's the unique imaginary order of a quadratic ring of algebraic integers that has the half-factorial property (Theorem 2.3)--ie any two factorizations of a nonzero nonunit have the same number of irreducibles.