When does the link of an algebraic singularity determines it algebraic type?

Let $$X \subset \Bbb C^n$$ be an algebraic hypersurface with an isolated singularity $$x$$ which is locally irreducible, i.e the local ring $$\mathcal O_{X,x}$$ is an integral domain (this is a necessary hypothesis, see Dori Bejleri's comment).

I know that when $$X$$ is a curve, $$\mathcal O_{X,x}$$ is determined by the link of $$X$$ at $$x$$ (considered as a topological space embedded in $$S^3$$).

Who proved this for the first time ?

I am also interested to know if it holds for a surface as well.

• What is the "link" of a local ring? – Armando j18eos Dec 9 '18 at 12:49
• @Armandoj18eos : thanks for your comment, I meant the link of $X$ at $x$, I edited. – student Dec 10 '18 at 12:23
• Why not make clear what you mean for $n=2$ – reuns Dec 10 '18 at 12:46
• @Armandoj18eos : So if $X$ is a topological space embedded in some real affine space, its link at $x$ is the intersection of a small sphere around $x$ with $X$. If e.g $X$ is an algebraic set this is well defined. For a smooth complex curve $X$ it always give a circle embedded trivially in $S^3$ but you get a knot when $x$ is singular, e.g you get the trefoil for a cusp. If $X$ is not locally irreducible you get a link (union of circles) hence the name. This is explained in Milnor's book "Singular points ...". However I don't think there is the statement I want inside that book. – student Dec 11 '18 at 14:35
• This is false. For example if the singularity is 4 smooth branches through the origin, the isomorphism type of $\mathcal{O}_{X,x}$ depends on the cross ratio of the tangent directions but the links are all the same. – Dori Bejleri Dec 16 '18 at 3:19