I have a ring $R=k[S]$ which defines an affine toric variety $X_\sigma$, where $S=M\cap \sigma^\vee$ is the semigroup from a rational polyhedral cone $\sigma$. Let $I$ be the ideal for the toric boundary divisor of $X_\sigma$, in other words, $I$ is obtained from the toric ideal of $S$ that is $>0$ on the closure of $\sigma$. I want to know if the $I$-adic completion $\hat{R}$ of $R$ is still an integral domain?

A simple example is $R=k[x,y]$, and $I$ is the principal ideal $(xy)$. Is the completion an integral domain?

  • $\begingroup$ For any excellent noetherian ring $A$ (which includes any ring finitely generated over a field), by EGA IV$_2$ 7.8.3(v) it follows that for any ideal $I$ of $A$ and the $I$-adic completion $\widehat{A}$ of $A$, the flat map $X = {\rm{Spec}}(\widehat{A}) \rightarrow {\rm{Spec}}(A) = Y$ is regular (i.e., all fibers $X_y$ are regular and remain so after finite extension on $k(y)$). In particular, $\widehat{A}$ is normal if $A$ is normal, so $\widehat{A}$ is a domain in such cases precisely when its spectrum is connected, which is to say ${\rm{Spec}}(A/I)$ is connected. $\endgroup$ – user76758 Feb 22 '14 at 17:01
  • $\begingroup$ Thanks for the answer. This is exactly the thing I am looking for. Now I understand $\hat{A}$ is normal, and as a result it is a product of integrally closed domains. Then could you please explain why $\hat{A}$ is a domain provided that the closed subscheme Spec$(A/I)$ is connected? How is Spec$(A/I)$ related to the set of closed points in Spec$(\hat{A})$? $\endgroup$ – user38276 Feb 23 '14 at 20:49

The completion of $R=k[x,y]$ w.r.t. to $(xy)$ (or any principal ideal) is indeed a domain. It is isomorphic to $R[[t]]/(t-xy)$, so we must prove that the ideal $(t-xy)$ is prime. To do that we can localize along $\mathfrak{m}:=(x,y)$; then the local ring $R_{\mathfrak{m}}[[t]]$ is regular, hence factorial, so we must show that $t-xy$ is irreducible in $R_{\mathfrak{m}}[[t]]$; this is clear since it is irreducible in $k[[x,y,t]]$.

  • $\begingroup$ Thanks for the explanation. I proved that in this case, $\hat{R}_I$ is an integral domain by showing that it is subring of the integral domain $k[y][[x]]$. But yours is a much simpler proof. $\endgroup$ – user38276 Feb 23 '14 at 20:12

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