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I am trying to prove the following result:

Let $R$ be a UFD and let $S \subset R$ be a multiplicative set. Then if $q \in R$ is irreducible, either $q/1$ is a unit or is irreducible in $S^{-1}R$.

I thought I had figured out a proof, but then I realized that I was implicitly assuming that all fractions $a/b \in S^{-1}R$ could be taken in ``reduced form'', i.e., $\gcd(a, b) = 1$. Although this is certainly true for, say, the field of fractions over $R$, I do not see why it would generally true for an arbitrary localization. It requires the property that if $a \in R$ and $b \in S$, then $b / \gcd(a, b) \in S$, and I am not convinced that this needs to be true.

My question: Does anybody know a proof of this result that does not assume reduced forms exist in $S^{-1}R$ (this would be the most desirable answer). Alternatively, is there some proof that reduced forms always exist (I find this doubtful)? Or, another possible alternative, does the result fail to hold for some $R$ and $S$ such that reduced forms do not exist?

user26857
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Doug
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  • See this question for a counterexample. – rogerl Nov 06 '14 at 00:58
  • If you are talking about the example with $\mathbb{Z}[\sqrt{-5}]$, this ring is not a UFD. I am considering localizations of UFDs. It is, apparently, a standard fact that localizations of UFDs are UFDs, and, apparently, a standard claim in proving this fact that an irreducible in $R$ becomes either an irreducible or a unit when passed to the localization. – Doug Nov 06 '14 at 01:16
  • Sorry, I missed the fact that you were considering UFDs. – rogerl Nov 06 '14 at 01:26
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    Isn't an irredicible a prime element in UFDs? When you localize, prime ideals either become the unit ideal or stay as a prime ideal in the localization. – Youngsu Nov 06 '14 at 09:06
  • @Youngsu: Your statement is equivalent to $$q \in R \text{ irreducible } \Longrightarrow q \text{ unit or } q \text{ prime in } S^{-1}R,$$ which is just a slightly weaker version of my question, and an equivalent version given the fact that any localization of a UFD is a UFD (which I am not assuming). – Doug Nov 09 '14 at 00:18

2 Answers2

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Suppose $x$ is irreducible in $R$. Then if $x$ divides some element of $S$, say $xy = s$ for $s\in S$, then $x$ clearly becomes a unit in $S^{-1}R$ since $$\frac{x}{1}\cdot\frac{y}{s} = \frac{xy}{s} = 1.$$ in $S^{-1}R$. Alternatively, if $x$ does not divide an element of $S$, then suppose that $x$ factors in $S^{-1}R$, say $x = \frac{y}{s'}\cdot\frac{z}{s''}$. Then $xs's'' = yz$. By unique factorization, $x$ must divide either $y$ or $z$. It cannot divide both since there is only one factor of $x$ on the LHS. Suppose it divides $y$. Then all irreducibles in the factorization of $z$ are divisors of $s'$ or $s''$ and thus divide some element of $S$. So $\frac{z}{s''}$ must be a unit (by the first case) and thus $x$ is irreducible.

rogerl
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A multiplicative subset $S$ of $R$ is called saturated if whenever $st\in S$ then $s,t\in S$. If $S$ is any multiplicative subset of $R$ then the set $\bar{S}$ of all those elements dividing some element of $S$ is saturated. It is easy to see that $S^{-1}R=\bar{S}^{-1}R$. Now it is clear that all fractions in $\bar{S}^{-1}R$ can be taken in reduced form.

Diego
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  • Certainly, if $S$ is properly contained in $\bar{S}$, then $S^{-1}R \neq \bar{S}^{-1}R$. You are saying that these are isomorphic as rings? – Doug Nov 09 '14 at 00:47
  • If you see every thing as subrings of the quotient ring of $R$ then $S^{-1}R=\bar{S}^{-1}R$. In fact, the image of $\bar{S}$ in $S^{-1}R$ is contained in the group of units and therefore the universal property provides an inclusion $\bar{S}^{-1}R\rightarrow S^{-1}R$ – Diego Nov 18 '14 at 21:30