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!

If you want to talk of function fields, it is prudent to consider only finite dominant morphisms $f\colon X\to Y$ between (locally noetherian) integral schemes.
The degree of $f$ is then defined as the degree of the corresponding extension of function fields $n=[k(X):k(Y)]$.
Qing Liu in his wonderful book asks you to prove (Chapter5, Exercise 1.25) that if $f$ is flat, then all fibers $f^{-1}(y)$ of $f$ have $n$ points, if you count them suitably: $\dim_{k(y)}\mathcal O(f^{-1}(y))=n$ .

Warning: The result is always false if $f$ is not flat!
This is part of the exercise mentioned above.
For example if you normalize the node $N=V(y^2-x^2-x^3)\subset \mathbb A^2_k$, you obtain the finite morphism $$f\colon \mathbb A^1_k\to N\colon t\mapsto (t^2-1,t^3-t)$$ of degree $$[k(\mathbb A^1_k):k(N)]=[k(t):k(t^2-1,t^3-t)]=[k(t):k(t)]=1$$
However for the fiber over $P=(0,0)\in N$ you have $\dim_{k(P)}\mathcal O(f^{-1}(P))=2$ because $f^{-1}(P)$ consists of the two simple points $t=-1, t=+1$.

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).