Wanted: example of a non-algebraic singularity

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.