Finite implies Quasi-finite Please help if you know a proof or a good reference for the following fact (exercise 3.5, Hartshorne's text).
Fact. A finite morphism of schemes $f: X \rightarrow Y$ is quasi-finite.
Here, the definition of quasi-finite is taken as $f^{-1}(q)$ is finite for all $q \in Y$.
 A: Hartshorne indicates at some point that the fibre corresponds to $X \times_Y \operatorname{Spec} k(q)$. The morphism from this to $\operatorname{Spec} k(q)$ is still finite, so you just need to know that if $k$ is a field and $A$ a finite-dimensional $k$-algebra then $\operatorname{Spec} A$ is finite.
You could quote some theorem about Artinian rings, but the geometric language helps us give a proof. Some steps: (i) the points of $\operatorname{Spec} A$ are closed (ii) the irreducible components are just points (iii) $A$ is Noetherian.
A: Let me show in an elementary way, as a complement to TTS's fine answer, that a finite dimensional algebra $A$ over a field $k$ has finite spectrum $\operatorname {Spec} A$, i.e. that $A$ has only finitely many prime ideals.  
a) Every prime ideal $\mathfrak p\subset A$ is maximal
Indeed $A/\mathfrak p$ is an integral finite dimensional algebra over $k$, and is thus a field (here is a proof), so that $\mathfrak p$ is maximal. 
b) There are only finitely many maximal ideals in $A$
If $\mathfrak m_1, \cdots,  \mathfrak m_r\subset A $ are distinct maximal ideals, the Chinese remainder theorem (Atiyah-Macdonald, Proposition 1.10) implies that $A\to A/\mathfrak m_1\times \cdots \times A/\mathfrak m_r \;$ is surjective, so that $\operatorname { dim }\: A\geq \sum \operatorname {dim}(A/\mathfrak m_i)\geq r$.   
Hence we deduce that $A$ has at most $\operatorname { dim } A\: $  maximal ideals and, taking b) into account,  at most $\operatorname { dim }\: A $   prime ideals.
