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If $R$ is an integral domain satisfying acc on radical ideals (i.e. Noetherian spectrum) and if the fraction field of $R$ is algebraically closed, then is $R$ a field ?

If $R$ is normal (integrally closed in its fraction field) and a factorization domain and the fraction field of $R$ is algebraically closed, then I can show that every element of $R$ is a perfect square, hence it has no irreducibles, so $R$ is a field. Using this and Kull-Akizuki theorem, I can show that a Noetherian domain with algebraically closed fraction field, is itself a field. But I don't know what happens if I weaken the condition and assume only $Spec (R)$ is Noetherian .

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  • $\begingroup$ Let $A$ be an integral domain with fraction field $K$, let $\overline{K}$ be an algebraic closure of $K$, and let $\overline{A}$ be the integral closure of $A$ in $\overline{K}$. Then the localizations of $\overline{A}$ at maximal ideals are local domains with fraction field $\overline{K}$. $\endgroup$ – Minseon Shin Sep 10 '18 at 0:41
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This is not true. A counterexample is given by the ring of integral Puiseux series over $\mathbb{C}$, that is, $R$ is the ring of expressions $\sum a_n x^{q_n}$, with $a_n\in\mathbb{C}$ and $q_n\in \mathbb{Q}_{\geq 0}$ with bounded denominators such that $q_n\to\infty$. The fraction field of $R$ is an algebraic closure of the field of Laurent series over $\mathbb{C}$. The spectrum of $R$ consists of just two points: the $0$ ideal, and the unique maximal ideal consisting of series with $a_n\in\mathbb{Q}_{>0}$.

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