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Non-normal locus

Let us start with a plane $ A_k$-singularity given by $ I = \langle x^2\!\!\!\;+\!\!\:y^{k+1}\rangle$ in $ \mathbf{Q}[x,y]$, which can easily be done by hand.

We get $ I_{Sing} = \langle x,y^k\rangle$, as test ideal $ J = \sqrt{I_{Sing}} = \langle x,y\rangle $ and $ u = x$ as a non-zerodivisor of $ S/I = \mathbf{Q}[x,y]/\langle
x^2\!\!\!\;+\!\!\:y^{k+1}\rangle$. Now

$\displaystyle (uJ+I):J \,=\, \langle x^2,xy,y^{k+1}\rangle : \langle x,y\rangle
\,=\, \langle x,y^k\rangle\,, $

hence $ I_{\text{\it NN}} = \langle
:\langle x,y^k\rangle = \langle x,y\rangle $, as expected. If we do the same with the $ A_k$-surface singularity $ I =
\langle x^2\!\!\!\;+\!\!\:z^2\!\!\!\;+\!\!\:y^{k+1}
\rangle\subset\mathbf{Q}[x,y,z]$, we get $ J = \langle
x,y,z\rangle$, $ u = z$,

$\displaystyle (uJ+I):J = \langle z^2,yz,xz,y^{k+1}\!\!\!\;+\!\!\:x^2\rangle
:\langle x,y,z\rangle = \langle z,y^{k+1}\!\!\!\;+\!\!\:x^2\rangle $

and, finally, $ I_{\text{\it NN}} = \langle u,I\rangle:((uJ+I):J)
= \langle 1\rangle $ which is true, since any isolated hypersurface singularity of dimension $ \geq 2$ 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 $ J$ of the singular locus is computed as
    ==> J[1]=xy2-y  J[2]=x2y-x  J[3]=y4-x  J[4]=x4-y
$ u = xy^2\!\!\!\;-\!\!\:y$ is chosen as a non-zerodivisor in $ J$ and $ (uJ+I):J$ 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 $ J$, but with different generators):
    ==> NN[1]=y3-x2  NN[2]=xy2-y  NN[3]=x2y-x  NN[4]=x3-y2

Christoph Lossen