Number of points in a general fiber for a dominant rational map In Joe Harris's book "Algebraic Geometry: A First Course", we find the following proposition: 
Let $f:X\longrightarrow Y$ be a dominant rational map, for $X$ and $Y$ be two varieties. 
Proposition 7.16. The general fiber of the map $f$ is finite if and only if the inclusion $f^*$ expresses the field $K(X)$ as a finite extension of the field $K(Y)$. In this case, if the characteristic of $K$ is $0$, the number of points in a general fiber of $f$ is equal to the degree of the extension. 
My question is very objective: what does "general fiber" mean? In other words, what characterizes $f^{-1}(y)$, for $y \in Y$, as a general fiber? 
Thanks.
 A: You cannot characterize the subset $Z \subseteq Y$ of points over which the fibre is not finite by looking at the field extension, since by the function fields $K(Y)$ and $K(Y-Z)$ are the same. In fact, the function field is a birational invariant, not a biregular one.
If you want to figure out what "general" means in this context, you can argue as follows. 
Let us assume that $X$ and $Y$ are irreducible. If we assume that $K(X)$ is a finite extension of $K(Y)$, then necessarily $\dim X = \dim Y$. Supposing that you have resolved the singularities of the map so that $f \colon X \to Y$ is defined everywhere, you can look at the differential map $$df_x \colon T_xX \longrightarrow T_{f(x)}Y.$$ Then the subset $Z \subseteq Y$ of points where $f$ is finite is contained in the subset $Z'$ of points where $f$ is  unramified, namely the complement the image of the set of critical points of $df$ (a point $x \in X$ is critical when $df_x$ is not of maximal rank). 
So being unramified (in characteristic $0$) is a general property, since the locus of critical values has zero measure in $Y$ (for instance, by Sard lemma).
