
> 
> Let $f(z\_1,z\_2,\dots ,z\_n)$ be an analytic function in $\mathbb{C}[[z\_1,z\_2,\dots ,z\_n]]$ whose leading term defines an isolated singularity at the origin. 
>  If we have the following types of singularities, then it is called a simple(ADE) singularity.
> 
> 
> $A\_n:z\_1^{n+1}+\sum\_2^{n} z\_i^2=0$ $(n\ge 1)$
> 
> 
> $D\_n:z\_1^{n-1}+z\_1z\_2^2+\sum\_3^{n} z\_i^2=0$ $(n\ge 4)$
> 
> 
> $E\_6:z\_1^4+z\_2^3+\sum\_3^{n} z\_i^2=0$
> 
> 
> $E\_7:z\_1^3z\_2+z\_2^3+\sum\_3^{n} z\_i^2=0$
> 
> 
> $E\_8:z\_1^5+z\_2^3+\sum\_3^{n} z\_i^2=0$.
> 
> 
> 


I have many (homogeneous) equations which define hypersurfaces with a singularity(in some pojective space). I would like to know if these singularities are simple. 


For example, a equation is (locally) given by:


$(f(y,z)-y)x^2+2g(y,z)x+(y^2+h(y,z))=0$, where $f,g\in \mathbb{C}[[y,z]]\_{\ge 2}$ and $h\in \mathbb{C}[[y,z]]\_{\ge 3}$


1. Since the term of degree $2$ is $y^2$, it's not of type $A\_n$. right?
2. Why is the singularity of type $D\_n$ only if the coefficient of $z^2$ in $g$ does not vanish?
What happen if the coefficient of $z^2$ in $g$ vanish? Can't it be of type $E\_n$?
3. Is there an effective way to determine whether this singularity is simple or not?



How can one determine if a singularity is simple?
The easiest way is to use the Newton polyhedron (after a linear change of variables): once you get nondegenerate principal part, just compare to the list. See Arnold, Gussein-Zade, Varchenko (volume 1) for details.


