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 . 



Integral domain satisfying a.c.c. on radical ideals and with algebraically closed fraction field
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 $\mathfrak{m}$.


