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

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

hence $I_{\text{\it NN}} = \langle x,x^2\!\!\!\;+\!\!\:y^{k+1}\rangle :\langle x,y^k\rangle = \langle x,y\rangle$, 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 $J = \langle x,y,z\rangle$, $u = z$,

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

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