Given $f:X\to Y$ a finite morphism of schemes, with $Y$ locally noetherian, let's take a point $q\in Y$, and an affine noetherian open set $$q\in U=Spec(B)\subseteq Y$$ Then $$f^{-1}(U)=V=Spec(A)\subseteq X$$ is affine since a finite morphism is affine, and $A$ is a finitely generetade $A$-module. If $$K(q)= B_q/qB_q$$ we know that $$f^{-1}(q)\cong Spec(A\otimes_BK(q))$$ and this ring (that we'll call $A_q$) is a finitely generated $K(q)$ vector space.
Is it true that the cardinality of the fiber $f^{-1}(p)$ is equal to the dimension as a vector space of $A_q$?
We know that $A_q$ is an artinian ring, but this doesn't seem to be enough, since $\mathbb C$ has only one prime ideal, but it's a 2 dimension vector space on $\mathbb R$..
This isn't true. Consider a finite Galois extension $K$ of $\mathbf{Q}$ in which the prime $p$ is inert. Look at the morphism $\mathrm{Spec}(\mathscr{O}_K)\to\mathrm{Spec}(\mathbf{Z})$, which is finite. The fiber over $(p)$ is the spectrum of $\mathscr{O}_K\otimes_\mathbf{Z}\mathbf{F}_p=\mathscr{O}_K/p\mathscr{O}_K$. The dimension of $\mathscr{O}_K/p\mathscr{O}_K$ over $\mathbf{F}_p$ is $[K:\mathbf{Q}_p]$, but the cardinality of its spectrum is just $1$, since it's a field.