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 .