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Does there exist a Noetherian domain (but not a field) whose field of fractions is the field of real numbers $\mathbb{R}$ ?

Any help will be appreciated. Thanks

user26857
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    Since you ask for a noetherian example, you might want to include a non-noetherian example, you are aware of. – MooS Jul 27 '16 at 14:06
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    If yes then $\mathbb R$ is the field of fractions of a noetherian local domain, hence of a DVR. This shows that the multiplicative group $\mathbb R^\times$ has a quotient isomorphic to $\mathbb Z$. Is this possible? – user26857 Jul 27 '16 at 19:51
  • why the domain will be local ? –  Jul 27 '16 at 23:51
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    http://math.stackexchange.com/questions/497256/when-is-a-field-a-nontrivial-field-of-fractions?rq=1 A non-noetherian example is given here. – Alex Wertheim Jul 27 '16 at 23:57
  • oh ..i understand, localizing at any non-zero prime ideal will give that Noetherian local domain whose field of fraction will be the same as that of the first domain. –  Jul 28 '16 at 09:36
  • @user26857 why should the local Noetherian domain be a DVR? Or How do you obtain a DVR from a general Noetherian local domain? – Pavel Čoupek Nov 09 '16 at 17:03
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    @PavelČoupek See here. One can also use the Mori-Nagata Theorem: the integral closure of a noetherian domain is a Krull domain, and then localize at a height one prime. – user26857 Nov 09 '16 at 18:18
  • @user26857 Thanks. – Pavel Čoupek Nov 09 '16 at 18:40
  • @Panja : if such a Noetherian domain , which is not a field , exists then it must have positive dimension , then passing to a height one prime ideal we have a Noetherian domain of dimension 1 and then taking its integral closure , since the field of fractions remains the same , and the integral closure is still Noetherian (by say Krull-Akizuki theorem) we may assume our Noetherian domain is normal also . Now let $a$ be a positive real number which irreducible in our normal Noetherian domain , then $(a)\subset (\sqrt a) \subset (\sqrt {\sqrt a} ) \subset ...$ gives a non-ending ascending chain –  Apr 11 '17 at 17:40

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