Integral domain satisfying a.c.c. on radical ideals and with algebraically closed fraction field 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 . 
 A: As requested, I am making my comment an answer.  For every integer $n\geq 1$, denote by $R_n$ the power series ring, $$R_n := \mathbb{C}[[z_n]],$$ where $z_n$ is a variable.  For every pair of integers $m,n\geq 1$ with $m$ dividing $n$, denote by $f_{m,n}$ the local $\mathbb{C}$-algebra homomorphism, continuous for the $\mathfrak{m}$-adic topologies, $$f_{m,n}: R_m \to R_n, \ \ f_{m,n}(z_m) = z_n^{n/m}.$$ This is a filtering directed system of integral domains.  Denote the colimit by $R$, $$(f_m:R_m\to R)_{m\geq 1}.$$  As a filtering colimit of integral domains, also $R$ is an integral domain, i.e., the zero ideal is a prime ideal.  Since every homomorphism $f_{m,n}$ is finite, every homomorphism $f_m$ is integral.  Thus, the induced map of fraction fields is an algebraic field extension, $$\text{Frac}(f_m):\text{Frac}(R_m) \to \text{Frac}(R).$$  The fundamental theorem of Puiseux series is that  $\text{Frac}(R)$ is an algebraically closed field.
For every integer $n\geq 1$, there are precisely two prime ideals of $R_n$: the zero ideal and the maximal ideal $\mathfrak{m}_n$ that is the kernel of the local $\mathbb{C}$-algebra homomorphism, $$g_n:R_n \to \mathbb{C}, \ \ g_n(z_n) = 0.$$  Note that $g_n\circ f_{m,n}$ equals $g_m$ for every pair of integers $m,n\geq 1$ with $m$ dividing $n$.  Thus, there exists a unique $\mathbb{C}$-algebra homomorphism, $$g:R\to \mathbb{C},$$ such that $g\circ f_m$ equals $g_m$ for every integer $m\geq 1$.  In particular, since $g_m$ is surjective, also $g$ is surjective.  Thus the kernel $\mathfrak{m}$ of $g$ is a maximal ideal of $R$.
Finally, for every prime ideal $\mathfrak{p}$ of $R$, if $\mathfrak{p}$ is nonzero, then there exists an integer $n\geq 1$ such that $f_n^{-1}(\mathfrak{p})$ is nonzero.  As a nonzero prime ideal in $R_n$, this prime ideal equals $\mathfrak{m}_n$.  Thus, $f_{1}^{-1}(\mathfrak{p})$ equals $f_{1,n}^{-1}(f_n^{-1}(\mathfrak{p}))$, and this equals $f_{1,n}^{-1}(\mathfrak{m}_n)$.  So $f_1^{-1}(\mathfrak{p})$ equals $\mathfrak{m}_1$.  Thus, for every integer $n\geq 1$, the prime ideal $f_n^{-1}(\mathfrak{p})$ pulls back under $f_{1,n}$ to $\mathfrak{m}_1$, hence the prime ideal $f_n^{-1}(\mathfrak{p})$ equals $\mathfrak{m}_n$.  Thus $\mathfrak{p}$ equals the colimit of the direct system of prime ideals $(\mathfrak{m}_n)$, i.e., $\mathfrak{p}$ equals $\mathfrak{m}$.  Therefore the only prime ideals in $R$ are the zero ideal and $\mathfrak{m}$.
