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Let $R$ be a UFD ring, $S$ a nonempty subset of $R$, $S$ closed under $R$ multiplication. Considering fractional ring $S^{-1}R$, why is $S^{-1}R$ a UFD ring?

Ottavio
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Mathematics Qi
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    The intuition is this: if you invert $a$ then you are inverting every irreducible factor of $a$, e.g., in $\mathbf Z[1/6]$, you are creating inverse of 2 and 3 since $1/2 = 3/6$ and $1/3 = 2/6$. So basically all that is happening is you are making the unit group larger and killing the irreducible factors of the numbers in $S$. Thus what's left is a UFD whose irreducible factors are the irreducibles not dividing a number in $S$. – KCd Dec 15 '20 at 01:30
  • It’s not a UFD if $S$ contains $0$. Otherwise, yes. – Arturo Magidin Dec 15 '20 at 01:38

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