the reason for my question is the following:
the two-dimensional canonical singularities are the ADE-singularities, which all are quotients of either affine space or another ADE-singularity by finite abelian Groups and even more, quotients of affine 2-space by finite solvable Groups but one - namely $E\_8$, which is factorial and a quotient of $A\_1$ by the perfect Group $A\_5$.


Now a generalization of the ADE-singularities in Dimension 3 are the compound du Val singularities, which are canonical and are formally equivalent to $f(x,y,z)+tg(x,y,z,t)$, where f is the polynomial of an ADE-singularity. 


Now some of these are again quotients of affine 3-space, while a lot more than the one before ($E\_8$) are factorial. For example the one in the heading. Which means in particular, that they cannot be a Quotient of affine space by a solvable Group. 


As you may see now, it would be very nice, if these singularities were also some quotients by any Groups. Any idea? I am quite sure that it can't be a finite Group and affine 3-space. 


Thank you and looking forward to your suggestions,
Lukas



Is the factorial cDV-singularity $T_1^2 + T_2^3 + T_3^4T_4$ any quotient of any affine space by any group?
Let $B=k[x,y,z,u]$ be a polynomial ring in four variables over a field $k$ of characteristic $0$, and let $X$ be 4-dimensional affine space over $k$. A locally nilpotent derivation $D$ of $B$ induces an algebraic action of the additive group $G=(k,+)$ on $X$ via the exponential mapping ${\rm exp} (tD)$, $t \in k$. The ring of invariants of the group action is equal to the kernel of the derivation, which is a UFD of transcendence degree $3$ over $k$ (It is not known whether the kernel must be affine).


Consider the triangular derivation $D: u \to z \to y \to x^n$ and $x \to 0$, where $n$ is a positive integer. The kernel of $D$ is of the form $k[x,f,g,h]$, where $x^{2n}u+f^3+g^2=0$. The corresponding algebraic variety $X/G$ is therefore of the type you are looking for.


