Let $K$ be an algebraically closed field and let $A=K[x_1,\dots,x_n]/I$ be a $K$-algebra of finite type which has only an isolated singularity at the origin. Let $\mathfrak{m}=(x_1,\dots,x_n)$ and consider the localization $A_{\mathfrak{m}}$ and its $\mathfrak{m}$-adic completion $B$.

Is $B$ also an isolated singularity? The localization at $\mathfrak{m}$ should not be a problem, but completion might. In general, the completion of an isolated singularity need not to be isolated. However, I was wondering whether in this situation the answer might be positive. I am willing to assume some stronger hypothesis. Can anybody point out any results in this direction?

Thank you in advance.


1 Answer 1


I believe that in your situation, $B$ indeed has an isolated singularity at the maximal ideal $\mathfrak{n} \subseteq B$. Let me first give two possible definitions for “isolated singularity”; please let me know if there are standard definitions for these notions, and I will edit this answer accordingly!

Definition. Let $X$ be a locally noetherian scheme.

  1. We say that $X$ has an isolated singularity at a closed point $x \in X$ if there is an open neighborhood $U \ni x$ such that the scheme $U \smallsetminus \{x\}$ is regular.

  2. Suppose that $X$ is a scheme over a field $k$. We say that $X$ has a geometric isolated singularity at a closed point $x \in X$ if there is an open neighborhood $U \ni x$ such that the scheme $U \smallsetminus \{x\}$ is geometrically regular over $k$.

Adopting either definition, the ring $B$ in your situation has a isolated singularity (resp. geometric isolated singularity) at the maximal ideal $\mathfrak{n} \subseteq B$. We can in fact prove the following more general result:

Proposition. Let $X$ be a locally noetherian scheme (resp. locally noetherian scheme over a field $k$) with an isolated singularity (resp. geometric isolated singularity) at a closed point $x \in X$, such that $\mathcal{O}_{X,x}$ is a $G$-ring. Then, $\operatorname{Spec}\widehat{\mathcal{O}}_{X,x}$ has an isolated singularity (resp. geometric isolated singularity) at the unique closed point $\widehat{x}$.

Note that this proposition applies to your situation, since $A$ is of fintie type over a field, hence a $G$-ring in the sense of [Mat89, p. 256] by [Mat89, Cor. to Thm. 32.6] ($A$ is also excellent, but we only need the $G$-ring part of the definition of excellence).

Proof. Let $U \subseteq \operatorname{Spec} \mathcal{O}_{X,x}$ be the intersection of $\operatorname{Spec} \mathcal{O}_{X,x}$ and a neighborhood of $x$ in $X$ satisfying the condition in the definition above. Let $f\colon \operatorname{Spec} \widehat{\mathcal{O}}_{X,x} \to \operatorname{Spec} \mathcal{O}_{X,x}$ be the morphism induced by the completion homomorphism. The morphism $$f^{-1}(U) \smallsetminus \{\widehat{x}\} \longrightarrow U \smallsetminus \{x\}\tag{1}\label{eq:completion}$$ is regular in the sense of [Mat89, p. 255] since the morphism $f$ is regular by the $G$-ring condition, and regular morphisms are preserved under base change [EGAIV$_2$, Prop. 6.8.3(iii)].

For isolated singularities, we apply [Mat89, Thm. 23.7] to the (localizations of the) morphism \eqref{eq:completion} to conclude that $f^{-1}(U) \smallsetminus \{\widehat{x}\}$ is regular.

For geometric isolated singularities, the composition $$f^{-1}(U) \smallsetminus \{\widehat{x}\} \longrightarrow U \smallsetminus \{x\} \longrightarrow \operatorname{Spec} k$$ is regular by [Mat89, Thm. 32.1], i.e., $f^{-1}(U) \smallsetminus \{\widehat{x}\}$ is geometrically regular over $k$. $\blacksquare$


[EGAIV$_2$] Alexander Grothendieck and Jean Dieudonné. “Éléments de géométrie algébrique. IV. Étude locale des schémas et des morphismes de schémas. II.” Inst. Hautes Études Sci. Publ. Math. (1965), no. 24, 1–231. DOI: 10.1007/BF02684322. MR: 199181.

[Mat89] Hideyuki Matsumura. Commutative ring theory. Second ed. Translated from the Japanese by M. Reid. Cambridge Stud. Adv. Math. 8. Cambridge Univ. Press, Cambridge, 1989. DOI: 10.1017/CBO9781139171762. MR: 1011461.

  • $\begingroup$ Dear Takumi, thank you for your nice and precise answer. It looks correct to me. The key point here is really the fact that finitely generated algebras over a field are $G$-rings. I have not thought about it. Concerning your question about the "standard" definition of isolated singularity. Personally, when I say isolated singularity I mean the first one. Or more algebraically, I say that a local ring $(R,\mathfrak{m})$ has an isolated singularity if $R_{\mathfrak{p}}$ is regular for any prime ideal $\mathfrak{p}\neq\mathfrak{m}$. I am not sure whether this is "standard" or not. $\endgroup$
    – Alessio
    May 27, 2019 at 16:29
  • $\begingroup$ Dear @Alessio, thank you for the kind words! You are correct that the key point is the $G$-ring condition; I have tried to make this clearer by explicitly stating a result only requiring this condition. Thank you for clarifying the definition of "isolated singularity" as well; my confusion is always that over imperfect fields, I don't know whether "singular" should mean "not regular" or "not smooth". $\endgroup$ May 28, 2019 at 13:47
  • $\begingroup$ You're welcome. I think it really depends whom do you ask to. Being more a commutative algebraist guy I'd like to think "singular" = "not regular", but I guess other people would prefer "not smooth". $\endgroup$
    – Alessio
    May 30, 2019 at 15:40

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.