meaning of normalization I have seen the following construction and I would be very happy if someone could explain its meaning to me. 
We start from a smooth projective algebraic variety $Z$ over the complex numbers and a reduced effective divisors with simple normal crossings $D$. Let $V=Z-D$. Let $U \to V$ be an étale cover.  
What's the meaning of taking $\pi: Y \to X$ the normalization of $Z$ in the function field $\mathbb{C}(U)$? 
Does it mean that $U \hookrightarrow Y$ and that the complement is a divisor with normal crossings laying above $D$? 
Thanks for your help
 A: Your $X$ should be $Z$. Suppose for simplicity that $U$ is integral. Then the normalization $\pi: Y\to Z$ of $Z$ in $\mathbb C(U)$ is a finite morphism such $Y$ is normal and birational to $U$. 
More precisely, for any non-empty affine open subset $W$ of $Z$, $\pi^{-1}(W)$ is an affine open subset of $Y$ and $O_Y(\pi^{-1}(W))$ is the integral closure of $O_Z(W)\subset \mathbb C(W)=\mathbb C(V)$ in the finite extension $\mathbb C(U)/\mathbb C(V)$. This construction makes sense for any morphism of integral schemes $U\to V$ (but then $\pi$ will not be necessarily a finite morphism). 
In your situation, the complement of $U$ in $Y$ is not necessarily a divisor with normal crossings ($Y$ needs not be smooth). For example, let $Z$ be the affine plane with coordinates $x,y$, let $D=\{ xy=0\}$ and consider the étale cover $t^2=xy$ over $U=Z\setminus D$. Then the normalization of $Z$ in $\mathbb C(U)$ is just the affine surface defined by $t^2-xy=0$ in $\mathbb C^3$. This surface is normal but singular at $x=y=t=0$. The pre-image of $D$ is $t=0$.
However, the morphism $Y\to Z$ is a so called tamely ramified cover and its local structure is well understood (see SGA 1, Exposé XIII, § 5). 
