I would like to write down local equation(s) for a surface $S\subset \mathbb C^4$ whose singular locus is itself a singular curve $C$, say $C=\{z=w=x^2-y^2=0\}$, where $(x,y,z,w)$ are the coordinates on $\mathbb C^4$ and which, around smooth points of $C$ has transversal singularities, say ordinary double points. By a transversal singularity I mean that if $p\in C$ is a smooth point of $C$, then there exists a neighborhood $U=U(p)$ of $p$ in $S$ such that $S\cap U \cong (C\cap U) \times\{(z,w)\in \mathbb C^2\mid z^2-w^2=0\}$.
My first approach was to consider the variety $V=\{x^2-y^2=z^2-w^2=0\}\subset \mathbb C^4$. But if I compute the singular locus of $V$, I obtain only the point $(x,y,z,w)=(0,0,0,0)$ as singular locus. I then tried to look at $V=\{(x^2-y^2)(z^2-w^2)=0\}$ which seems to be closer to what I want but still the singular locus is wrong; for example $\{x=z=w=0\}$ is part of the singular locus.
How do I find local equations? Is there a general procedure for producing a transversal singularity along a (singular) curve?
Thanks in advance for any kind of help!
Complete intersections are not difficult. We will do this in $\mathbb{C}^3$ with $w=0$. Your curve $C$ is then given by $z=0=x^2-y^2$ and then the hypersurface $X$ given by $F: z^2-(x^2-y^2)^3=0$ will do the trick. You can easily check that the singular locus is precisely $C$. The non-singular points of $C\subset X$ are covered by the two open sets $x\neq 0, y\neq 0$.
We look at the open set $x\neq 0$, the other being similar. The ring is given by $k[x,y,z,x^{-1}]/F =k[x,y,z,x^{-1}]/ z^2-x^6(1-(yx^{-1})^2)^3=k[x,y,z,x^{-1}]/ ((zx^{-3})^2-(1-(yx^{-1})^2)^3$. Changing variables, let $z'=zx^{-3}, y'=yx^{-1}$, we see that the ring is $k[x,y',z',x^{-1}]/(z'^2-(1-y'^2)^3=(k[y'.z']/z'^2-(1-y'^2)^3)[x,x^{-1}]$ which is what you want.