Q: 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.

A: 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.
