Kummer surfaces are smooth Let $X$ be the Kummer surface associated to an abelian surface $A$. I will denote by $\epsilon : \tilde{A} \rightarrow A$ the blow-up of $A$ at the 16 fixed points of the involution $i : A \rightarrow A$ sending $a \mapsto -a$, by $\sigma : \tilde A \rightarrow \tilde A$ the extension of $i$ to $\tilde A$, and by $\pi : \tilde A \rightarrow \tilde A/\sigma = X$ the projection map. I am attempting to understand the proof in Beauville's book that $X$ is a smooth. The problem is that I have only recently completed introductory courses in algebraic geometry, and haven't done many examples on this sort of thing.
So, Beauville first remarks that it is only necessary to prove smoothness at points $\pi(q)$, where $q \in E_i$ - one of the 16 exceptional divisors on $\tilde A$. Then he says that writing $A$ as $V/\Gamma$, where $V = \mathbb{C}^2$, one gets local coordinates $(x,y)$ on $A$, in a neighbourhood of $p_i \in A$ (where $p_i$ is the fixed point of $i$ corresponding to the exceptional divisor $E_i$), such that $i^*(x)=-x$ and $i^*(y)=-y$. Now, I don't understand this last statement. What is $i^*$ in this context? And where does this $i^*(x)=-x$ and $i^*(y)=-y$ come from? Then he proceeds to set $x' := \epsilon^*x$ and $y' := \epsilon^*y$, and says that we can assume that $x'$ and $t := \frac{y'}{x'}$ are local coordinates on $\tilde A$ near $q$. How do we know that they generate the maximal ideal of the local ring $\mathcal{O}_{\tilde A, q}$? He concludes by saying that $\sigma^*(x') = -x'$ and $\sigma^*t = t$, so that $t$ and $u := x'^2$ are local coordinates near $\pi(q)$ on $X$, but again, I don't see why they generate the maximal ideal of $\mathcal{O}_{X, \pi(q)}$. 
Thanks for your help.
 A: You have your abelian surface $A$ and the involution $i$ has $16$ fixed points. When you blow-up $A$ you get $\tilde A$, and of course you can extend the involution $i$ on $A$ to an involution on $\tilde A$, let us denote it by $\tau$. Then $\tau$ acts just as $i$ on points outside the exceptional divisors and point-wise fixes these divisors. 
You want to check the quotient of $\tilde A$ by $\tau$ is smooth, so we may work locally. Around points which are not fixed by $\tau$ this is obvious (the projection restricted to a small disk would be an isomorphism). So we take $p\in E\subset\tilde A$ a point lying on $E$, anyone of the 16 exceptional divisors, and work locally around $p$. 
Let us think analytically: we may choose coordinates $(x,y)$ such that $p$ is the origin of $\Bbb{C}^2$, $E$ is the $y$-axis $\{x=0\}$, and the action of $\tau$ is the reflection wrt $E$, that is $(x,y)\to(-x,y)$. In other words, we have locally linearized and diagonalized the action of $\tau$ as
$$\tau_\mathrm{lin}=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$
Now, in the quotient $\Bbb{C}^2/\langle\tau_\mathrm{lin}\rangle$ the points $(x,y)$ and $(-x,y)$ are identified, thus $(x^2,y)$ is a honest pair of local charts: the quotient is smooth. 
Algebrically, I think one would say that the invariant ring $\Bbb{C}[x,y]^{\tau_\mathrm{lin}}$ has two generators $x^2$ and $y$ and $\Bbb{C}[x^2]$ is a polynomial ring; hence $\Bbb{C}[x^2,y]\simeq \Bbb{C}[x^2][y]$ is a polynomial ring, i.e. its Spec is smooth.
