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 .

  • 2
    $\begingroup$ Let $R$ be the ring of Puiseux series, i.e., the integral closure of $\mathbb{C}[[x]]$ in the algebraic closure of $\mathbb{C}((x))$. There are only two prime ideals in this ring: the zero ideal and the maximal ideal. Thus, there are also only two radical ideals -- these same two prime ideals. $\endgroup$ – Jason Starr Mar 10 '18 at 15:51
  • $\begingroup$ @Jason Starr : Could you please elaborate your comment as an answer ? $\endgroup$ – user111492 Mar 12 '18 at 14:30

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 $\mathfrak{m}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy