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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.

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  • $\begingroup$ What is the "link" of a local ring? $\endgroup$ – Armando j18eos Dec 9 '18 at 12:49
  • $\begingroup$ @Armandoj18eos : thanks for your comment, I meant the link of $X$ at $x$, I edited. $\endgroup$ – student Dec 10 '18 at 12:23
  • $\begingroup$ Why not make clear what you mean for $n=2$ $\endgroup$ – reuns Dec 10 '18 at 12:46
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    $\begingroup$ @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. $\endgroup$ – student Dec 11 '18 at 14:35
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    $\begingroup$ 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. $\endgroup$ – Dori Bejleri Dec 16 '18 at 3:19

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