Let $X$ an affine normal scheme of finite type over a field $k$ of characteristic zero.
Let $G$ a finite group acting on $X$ and $Y=X/G=Spec(K[X]^{G})$.


We assume that $Y=\mathbb{A}^{n}=k[f\_{1},\dots,f\_{n}]$ with $f\_{i}\in K[X]^{G}$.


Then we have a finite flat surjective morphism $\pi:X\rightarrow Y$ generically étale of group $G$.


Do we have that $\pi$ is a complete intersection?



quotient by finite group actions that are smooth
No, there is no reason that $X$ should be a complete intersection. For instance, begin with $Z=\mathbb{A}^2\_k$ with coordinates $(z\_0,z\_1)$. Consider the action on $Z$ of the group of $n^{\text{th}}$ roots of unity, $\mu\_n$, by $\zeta\cdot(z\_0,z\_1) = (\zeta z\_0,\zeta z\_1)$. The ring of invariants, $k[z\_0,z\_1]^{\mu\_n}$, is generated by the monomials $y\_{a,b} = z\_0^az\_1^b$ for all pairs $(a,b)$ of nonnegative integers with $a+b=n$. So the finitely generated $k$-algebra $k[z\_0,z\_1]^{\mu\_n}$ has Krull dimension $2$ and has $n+1$ elements $y\_{a,b}$ in a minimal set of generators. Yet, already the vector space of quadratic relations among these generators has dimension roughly $n^2/2$. Indeed, it is generated by the binomials,
$$ Q\_{(a\_0,b\_0),(a\_1,b\_1),(a\_2,b\_2),(a\_3,b\_3)} = y\_{a\_0,b\_0}y\_{a\_1,b\_1}-y\_{a\_2,b\_2}y\_{a\_3,b\_3},$$
where $a\_0+a\_1=a\_2+a\_3$ (which then forces the same identity for the $b$ indices). Of course these generators are not linearly independent. However, using a monomial ordering, there are roughly $n^2/2$ leading monomials that occur among these binomials. The dimension of the vector space is at least as large as the number of leading monomials, thus roughly $n^2/2$. When $n$ is large, $k[z\_0,z\_1]^{\mu\_n}$ cannot possibly be a complete intersection of dimension $2$.


Let $Y$ be the quotient of $Z$ by this action of $\mu\_n$. The point is that there is a larger group, $\mu\_n\times\mu\_n$, that acts on $Z$, e.g., $(\zeta\_0,\zeta\_1)\cdot(z\_0,z\_1) = (\zeta\_0 z\_0,\zeta\_1 z\_1).$ Of course the original action comes from the diagonal subgroup $\Delta(\mu\_n) \leq \mu\_n \times \mu\_n$. Since this group is Abelian, there is an induced action of the quotient group $G=(\mu\_n\times \mu\_n)/\Delta(\mu\_n)$ on the quotient variety $Y = Z/\Delta(\mu\_n)$. The quotient of $Y$ by this induced action is the same as the quotient of $Z$ by the full action of $\mu\_n\times \mu\_n$. But, of course, $k[z\_0,z\_1]^{\mu\_n\times \mu\_n}$ equals $k[z\_0^n,z\_1^n]$, which is a polynomial ring.


**Edit.** Of course you can compute the precise dimension of the vector space of quadratic relations is $n(n-1)/2$. Thus, already for $n=3$, $Y$ is not a complete intersection.


