Given a finitely generated $\def\CC{\mathbb C}\CC$-algebra $R$ and a $\CC$-point (maximal ideal) $p\in Spec(R)$, I define the singularity type of $p\in Spec(R)$ to be the isomorphism class of the completed local ring $\hat R_p$, as a $\CC$-algebra.

Do there exist non-algebraic singularity types? That is, does there exist a complete local ring with residue field $\CC$ which is formally finitely generated (i.e. has a surjection from some $\CC[[x_1,\dots, x_n]]$), but is not the complete local ring of a finitely generated $\CC$-algebra at a maximal ideal?

Googling for "non-algebraic singularity" suggests that the answer is yes, but I can't find a specific example. I would expect that it should be possible to write down a power series in two variables $f(x,y)$ so that $\CC[[x,y]]/f(x,y)$ is non-algebraic.

What is a specific formally finitely generated non-algebraic singularity?

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    $\begingroup$ One idea I had is to take a non-algebraic variety $X$, and consider the completed local ring at the cone point of a cone on $X$. The problem is that making a cone on $X$ involves choosing a very ample line bundle on $X$. But as soon as $X$ is projective, it is algebraic by GAGA. $\endgroup$ Aug 29, 2011 at 6:46
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    $\begingroup$ Regarding your expectation, any analytic plane curve singularity is in fact algebraic. See, for example, Corollary 7.7.3 of the book by Casas-Alvero "Singularities of plane curves", London Mathematical Society Lecture Note Series, 276. I'm not sure though whether a similar result holds for formal planar singularities but it seems possible to me. $\endgroup$
    – naf
    Aug 29, 2011 at 12:41
  • $\begingroup$ Not an answer to your question, but maybe you find something relevant in Angelos answer here: mathoverflow.net/questions/51530/… $\endgroup$ Aug 29, 2011 at 14:48
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    $\begingroup$ @S. Carnahan: that information is automatically included, since the topology is induced by the maximal ideal. $\endgroup$ Aug 29, 2011 at 15:58
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    $\begingroup$ If the singularity is isolated (that is if $\mathbb{ℂ}[[x,y]]/f(x,y)$ minus its closed point is formally smooth over $\mathbb{C}$), then it is algebraic. This is Theorem 3.8 in Artin's paper on "Algebraic approximation of structures over Henselian local rings", at IHES, hence easily available on numdam.org. It uses Artin's approximation theorem plus works of Hironaka $\endgroup$
    – Joël
    Sep 23, 2011 at 23:32

3 Answers 3


I got this example from Frank Loray. I'll explain the analytic version, but the formal variant works just as well.

Let $U\subset \mathbb{C}$ be open. Choose two holomorphic functions $f$, $g$ which are algebraically independent over $\mathbb{C}$ (e.g. $f(z)=z$, $g(z)=e^z$). For simplicity, assume that $f$, $g$, $0$, $1$ never coincide (pairwise) on $U$. Now define $X\subset U\times\mathbb{C}^2$ (with coordinates $z$, $x$, $y$) as the union of $x=0$, $y=0$, $x=y$, $y=f(z)\,x$, and $y=g(z)\,x$. Thus, if we freeze $z$, we get five lines in the plane, with slopes $\infty$, $0$, $1$, $f(z)$ and $g(z)$. Globally, $X$ is the union of five copies of $U\times\mathbb{C}$ meeting along $Z:=U\times\mathbb{0}$.

Now, the point is that the cross-ratio of four (ordered) lines through the origin in the plane is an intrinsically defined invariant. In particular, independently of the coordinates, we can recover $f$ and $g$ as holomorphic functions on the singular locus $Z$. If $X$ were isomorphic to a complex open subset of an algebraic variety, $f$ and $g$ would have to be algebraically dependent because $\dim Z=1$: contradiction.

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    $\begingroup$ Beautiful! ${}$ $\endgroup$ Sep 26, 2011 at 18:44
  • $\begingroup$ Ok, I buy it. I was confused for a bit because I thought you needed to include the information of how to fix $z$ in order to recover $f$ and $g$. But of course the whole point of the cross ratio is that it's the same no matter how you slice it. Thanks for the example! $\endgroup$ Sep 27, 2011 at 23:52
  • $\begingroup$ one question about this example: the trick is that since the cross ratio stays invariant under "scanning by lines" and thus it's invariant under trafos of coordinates, we can fully reconstruct $f$ and $g$ from their values on $U \times 0$ independently of the choice of local coordinates,this is the message, right? the only point that I still not understand is why $\dim Z=1$ implies that $f$ and $g$ are already algebraically dependent, if we assume that $X$ is an open subset of an algebraic variety? $\endgroup$
    – user267839
    Nov 20, 2019 at 4:04
  • $\begingroup$ @TimGrosskreutz: If $X$ is algebraic, then $f$ and $g$ must be algebraic over $\mathbb{C}(z)$. $\endgroup$ Nov 20, 2019 at 13:00
  • $\begingroup$ ...you mean it in the sense, that if $X$ is algebraic, then it's closed subset $Z$ is also algebraic, and $f$ and $g$ are algebraic over $K(Z)=\mathbb{C}(z)$, right? $\endgroup$
    – user267839
    Nov 20, 2019 at 14:39

The main question of the PI has been beautifully answered by Moret-Bailly, but not the secondary question arisen from his expectation: "I would expect that it should be possible to write down a power series in two variables f(x,y) so that ℂ[[x,y]]/f(x,y) is non-algebraic." though we got quite close in comments.

So for the record: this is not possible. Indeed, such a singularity would be analytic by a result of Michael Artin ( "On the solutions of analytic equations", Invent. Math. 5 1968, 277–291, cf. Angelo's answer to Analytic vs. formal vs. étale singularities) and then algebraic by Ulrich's comment (that is by Corollary 7.7.3 of the book by Casas-Alvero "Singularities of plane curves", London Mathematical Society Lecture Note Series, 278).

This is of course consistent with the fact that the example quoted by Moret-Bailly is in three variables.


This is not answer, but a complement to the comments above about isolated singularities of hypersurfaces.

Every isolated hypersurface singularity, $R=\mathbb C[[x_1, \ldots, x_n]]/f(x_1,\ldots,x_n)$, is not just algebraic but also $k$-determined, for some $k \in \mathbb N$.

If $\mathfrak m \subset \mathbb C[[x_1,\ldots, x_n]]$ then we say that $f$ is $k$-determined if for every $g\in \mathbb C[[x_1, \ldots, x_n]]$ satisfying $f-g \in \mathfrak m^k$ there exists an automorphism $\varphi :\mathbb C[[x_1,\ldots, x_n]] \to \mathbb C[[x_1, \ldots, x_n]]$ such that $\varphi(g)= f$.

Moreover, always assuming we have isolated singularities, the natural number $k$ can be easily determined from $f$. For instance, in this book you will find the following result:

If $\mathfrak m^{k+1} \subset \mathfrak m^2 J(f)$, where $J(f)$ is the Jacobian ideal of $f$, then $f$ is $k$-determined.


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