Finite dimensional fibres I can't understand why the following two statements are equivalent:


*

*the fibre of $\mbox{Spec}(B) \to \mbox{Spec}(A)$ of any field valued point $\mbox{Spec}(K) \to \mbox{Spec}(A)$ is $0$-dimensional or empty

*every fibre $\mbox{Spec}(B) \to \mbox{Spec}(A)$ is a finite set


Looks like it is a corollaries from finitness of base change (if $A \to B$ is finite then for any map $A \to A'$ base change $A' \to B \otimes_A A'$ is finite too) but I can't see how implicitly we can obtain such two properties from that fact.

It is clear that the first statement is stronger than the second one. So we have to show that the second one implies the first.
Let $K$ be a field with a map $A \to K$. The kernel $\mathfrak p$ of this map is prime, because the image is an integral domain. The map factors as
$$A \to A/\mathfrak p \to \operatorname{Frac}(A/\mathfrak p) \to K,$$
hence the base change $K \to B \otimes_A K$ can be obtained from $A \to B$ via the intermediate step $\operatorname{Frac}(A/\mathfrak p) \to B \otimes_A \operatorname{Frac}(A/\mathfrak p)$
The second statement says that $\operatorname{Frac}(A/\mathfrak p) \to B \otimes_A \operatorname{Frac}(A/\mathfrak p)$ is finite, hence $K \to B \otimes_A K$ is also finite as a base change.
