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Let $R$ be a ring, and let $D$ be a multiplicatively closed subset of $R$. Is it the case that if $D^{-1}R$ is a field, then $R$ must be an integral domain?

user26857
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Nishant
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This is not true, one reason being that being an integral domain is not a local property. For a concrete example take $\mathbb{Z}/6\mathbb{Z}$, this is not an integral domain, but if we localise at the set $\{ 1, 3, 5\} = \mathbb{Z}/6\mathbb{Z} \setminus (2)$ we obtain a finite integral domain and hence a field. You can find some details about determining this localisation here.

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Alex J Best
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    Wait, all those elements are already units, so won't that just give back the original ring? – Nishant Jul 01 '14 at 13:17
  • 3 isn't a unit! – Alex J Best Jul 01 '14 at 13:17
  • Oh, oops. My bad. – Nishant Jul 01 '14 at 13:19
  • So this means that the contraction of a maximal ideal in a localized ring is not necessarily a prime ideal in the original ring? – Nishant Jul 01 '14 at 13:20
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    +1. Generalizing Alex's example: for any von Neumann regular ring, localizations at primes are all fields, but the ring itself is a domain iff it is a field (and there are many examples that aren't fields.) – rschwieb Jul 01 '14 at 13:21
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    @Nishant Well a maximal ideal must be prime, and the contraction of a prime ideal is always prime again. In this case our local ring is $\mathbb{Z}/2\mathbb{Z}$ with maximal ideal $(0)$ which then contracts to $2\mathbb{Z}/6\mathbb{Z}$ which is prime. – Alex J Best Jul 01 '14 at 13:27
  • Wait, so if $D^{-1}Q$ is a maximal ideal of $D^{-1}B$, then $D^{-1}B/D^{-1}Q$ is a field, but this is equal to $D^{-1}(B/Q)$, so if $Q$ is in fact prime, doesn't $B/Q$ have to be an integral domain? – Nishant Jul 01 '14 at 13:36
  • @Nishant, I'm not sure I quite understand, if $Q$ is a prime of $B$ then $B/Q$ is always an integral domain. – Alex J Best Jul 01 '14 at 13:45
  • Right, but then the localization of $B/Q$ is a field, so I was trying to conclude from this that $B/Q$ must be an integral domain, but I guess I could have done it directly? – Nishant Jul 01 '14 at 13:54
  • @Nishant, ah I see, yes if you knew (or could work out) that $Q$ was a prime of $B$ then it follows straight away that $B/Q$ is an integral domain, without having to try and use the statement in the question. – Alex J Best Jul 01 '14 at 14:02
  • Yeah, I was trying to follow the answer here: https://math.stackexchange.com/questions/10701/if-every-prime-ideal-is-a-contracted-ideal-does-it-imply-that-the-induced-map-b?rq=1

    In particular, I was trying to find the "standard argument" that $Q$ is prime.

     – Nishant Jul 01 '14 at 14:08