Definition of degree of finite morphism plus context Let $f: X \rightarrow Y$ be a finite morphism of schemes, defined here,
http://en.wikipedia.org/wiki/Finite_morphism
I always assumed that the degree of $f$ was the degree of the induced field extension
$$
[K(Y):K(X)],
$$
of course only defined if $f$ is dominant (hence surjective).
However, is this standard?
Or is the degree of $f$ defined as: Let $f$ locally be given by maps of rings $B_i \rightarrow A_i$. The degree of $f$ is the minimal number of elements of $A_i$ that generate it as a $B_i$ module?
I'm almost sure this does not work.
Q1: Why not? (excuse my lack of commutative algebra knowledge) (i am more than happy with just a sketchy answer to this!)
Q2: Does it work if we assume $X$ and $Y$ to be varieties over $\mathbb{C}$? If so/not, why?
Q3: What is indeed the commonly agreed definition of the degree of a finite map?
That's quite a lot, thanks for your answer!

I just want to note that you don't need any finiteness condition on $f$ to define the degree.  Any dominant morphism $f : X \to Y$ of integral schemes $X$ and $Y$ maps the generic point $x$ of $X$ to the generic point $y$ of $Y$, hence there is a canonical homomorphism $\mathscr{O}_{Y,y} \to \mathscr{O}_{X,x}$, which gives a field extension $R(Y) \hookrightarrow R(X)$.  (Thanks for Georges Elencwajg for simplifying my original argument.)
So we don't need finiteness conditions on $f$, or on $X$ and $Y$, to talk about the field extension $R(Y) \hookrightarrow R(X)$ or its degree $\deg(f) =  [R(X) : R(Y)]$.  However if one wants to ensure this degree is finite, it suffices to assume $f$ is locally of finite type and
(i) $f$ is closed, the dimensions of $X$ and $Y$ are equal (see (Stacks, 02JX) and (Stacks, 02NX)),
or
(ii) $f$ is separated or quasi-compact (e.g. proper or even just universally closed), and there exists an affine open $V \subset Y$ such that the restriction $f^{-1}(V) \to V$ is finite (Stacks, 02NX).
