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.



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> What is a specific formally finitely generated non-algebraic singularity?
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Wanted: example of a non-algebraic singularity
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.


