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I am working through a proof that if we have an integral domain $D$ and a multiplicative subset $S$ where $0 \notin S$ and $D$ is a UFD then $S^{-1}D$ is a UFD.

I am looking at Arturo's proof here:

About the localization of a UFD

I understand the first claim he makes and I am stuck on something with the second claim he makes. Arturo defines the following sets:

"Let T be the set of all irreducibles that divide an element of S, and let M be the set of all irreducibles not in S."

This is the claim he makes: "If $p\in M$, then the image of $p$ in $S^{−1}D$ is irreducible."

The part I am stuck on is in order for the image of $p$ to be irreducible in $S^{-1}D$, $ps/s$ cannot be a unit, which is what I am trying to show now.

I am trying to find a contradiction with the hypothesis being $p \in M$, that is $p \notin S$ but $ps/s$ is also a unit.

Essentially why does $M \cap T =\varnothing$?

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  • There's probably a typo in his proof. Essentially $M$ is supposed to be complement of $T$, i.e. the irreducibles of $D$ not in $T$. I commented on his question. I have done this proof before, and "my $M$" was the complement of $T$. Follow the discussion there. – Patrick Da Silva Feb 10 '14 at 05:36
  • I have added a counter example to explain that his choice of $M$ was wrong. – Patrick Da Silva Feb 10 '14 at 05:44
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    I edited Arturo's answer myself since Arturo seems to be gone forever (see his profile). Try to follow the argument with the modification I made (i.e. $M$ is the set of irreducibles not in $T$). – Patrick Da Silva Feb 11 '14 at 02:24
  • I've shown so far that every element can be written as a product of irreducible elements and now I am trying to show that the factorization is unique up to associates. –  Feb 11 '14 at 04:40
  • Feel free to post your argument (written by yourself) if you want some comments! – Patrick Da Silva Feb 11 '14 at 08:24

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This community wiki solution is intended to clear the question from the unanswered queue.

It appears that there was a typo in the post you referred to: $M$ was supposed to be defined as the complement of $T$ (relative to the set of all irreducibles), which immediately gives you that $$T \cap M = \emptyset,$$ which was what you were trying to show.

Robert Cardona
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