quotient by finite group actions that are smooth 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?
 A: 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.
