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### Non-normal locus

Let us start with a plane -singularity given by in , which can easily be done by hand.

We get , as test ideal and as a non-zerodivisor of . Now

hence , as expected. If we do the same with the -surface singularity , we get , ,

and, finally, which is true, since any isolated hypersurface singularity of dimension is normal.

Let us compute the nonnormal locus of two transversal cusps in the plane, using the procedure nnlocus from normal.lib in SINGULAR.


ring S = 0,(x,y),dp;
ideal I = (x2-y3)*(x3-y2);
ideal NN = nnlocus(I);

The radical of the singular locus is computed as

==> J[1]=xy2-y  J[2]=x2y-x  J[3]=y4-x  J[4]=x4-y

is chosen as a non-zerodivisor in and is

==> _[1]=xy2-y  _[2]=y4-x2y  _[3]=x3y-y3  _[4]=x4-y

Typing NN; we get as result the following ideal defining the non-normal locus (equal to , but with different generators):

==> NN[1]=y3-x2  NN[2]=xy2-y  NN[3]=x2y-x  NN[4]=x3-y2


Christoph Lossen
2001-03-21