if the fibers of all points are finite then will the map be finite? If $f:X\to Y$ is a regular map between affine varieties then we say $f$ is finite if $k[X]$ is integral over $k[Y]$. If $f$ is finite then fibers of all points are finite. I think the converse of this statement is false but I can't give a counter example. Can you give a counterexample?
 A: Exercise II.3.5 in Hartshorne tells you that you are correct:
"Show by example that a surjective, finite-type, quasi-finite morphism need not be finite,"
where $f:X \to Y$ is quasi-finite if $f^{-1}(y)$ is a finite set for every $y\in Y$. 
For a specific example, the distinction you want to keep in mind is finitely generated as an algebra versus finitely generated as a module. The solution I've seen before (not mine) is the map
$$
\text{Spec}\left( k[t,t^{-1}]\oplus k[t,(t-1)^{-1}]\right) \to \text{Spec}\left(k[t]\right)
$$
for your choice of field $k$. 
A: An open immersion has finite fibres but is generally not finite. In fact I think an open immersion is finite if and only if it is also a closed immersion, hence an isomorphism onto a connected component.
A: It is not true that a map $f \colon X \to Y$ which is quasi-finite (i.e. has finite fibres) is finite - you need to also assume that $f$ is universally closed (see p.100 of Hartshorne). 
To complement Derek's answer, here is another example where this fails: consider the map on spectra induced by the ring map $k[x] \hookrightarrow \frac{k[x,y]}{(xy-1)}$, for a field $k = \overline{k}$. Geometrically, we have the hyperbola $xy=1$ and we are projecting this onto the $x$-axis. The fibres either consist of one point or are empty (the fibre above $x=0$).
However, this map is not finite: indeed, a finite morphism is closed, but the image of this map is the complement of the point $x=0$ in $\mathbb{A}^1_k$, which is open. 
