Does there exist a Noetherian domain which is not a field whose field of fractions is (isomorphic with) $\mathbb C$ ?
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Perhaps related to this unanswered question, where the same is asked for $\mathbb{R}$ instead of $\mathbb{C}$. – Pierre-Guy Plamondon Aug 29 '16 at 12:13
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If there is such a Noetherian domain $R$ of positive dimension, we may assume it to be one-dimensional by passing to a localization at a height one prime. Passing further to the integral closure of $R$ within ${\mathbb C}$ (which is again Noetherian by the Krull-Akizuki Theorem), we may even assume $R$ to be normal, hence a discrete valuation ring corresponding to a discrete valuation $\eta: {\mathbb C}^{\times}\to {\mathbb Z}$.
However, there is no discrete valuation on ${\mathbb C}$, because as explained in every field of characteristic 0 has a discrete valuation ring? the value group of each valuation of ${\mathbb C}^{\times}$ needs to be divisible due to the presence of arbitrarily high roots in ${\mathbb C}$.
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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 aa 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 12 '17 at 14:39
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could you please check if my above argument works ? I think the only problem it might face is that some square root of square root of ... $a$ might turn out to be a unit ... can we somehow make the argument work ? – Apr 12 '17 at 14:41