Finite Morphisms between Varieties have Finite Fibers

affine-varietiescommutative-algebra

My question refers to following previous thread:
Geometric interpretation of the Noether normalization lemma

Obviously the core problem is: If $X= Spec(A), Y= Spec(R)$ are affine varieties and $\phi: X \to Y$ is a finite surjective morphism (so equivalently $A$ is a finite $R$-module)

why $\phi$ has finite fibers, so if $P \in Y$ why is $\phi^{-1}(P)$ finite?

Using again the language of commutative algebra the problem boils down to following:

Let $R$ wlog local with maximal ideal $p$ (in general case: localize $R$ by at $p$). We then get a morphism $R/p \to A/pA$.

I want to know why is $A/pA$ is finite dimensional $R/p$-vector space.

Following the thread The fibers of a finite morphism of affine varieties are all finite

it suffice to prove that. But I don't see why it holds, so why is $A/pA$ is finite dimensional $R/p$-vector space?

Best Answer

Since $A$ is a finite $R$-module, it is generated as an $R$-module by some elements $a_1, \dots, a_n$. Then $A/pA$ is spanned as an $R/p$-vector space by the images of the $a_i$.

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