Finite type + integral = finite Let $A \subseteq B$ be rings (comm. with unity). 
I am struggling to see why the following equivalence holds for $B$ interpreted as a $A$-Algebra: 
$A \rightarrow B$ is of finite type and $A\subseteq B$ is integral $\Leftrightarrow$ $A \rightarrow B$ is finite. 
It should be possible to conclude this by using $b \in B$ integral over $A \Leftrightarrow$ $A[b]$ finitely generated as an $A$-module and one or two other elementary statements. 
 A: To complete the hints in the comments, note that 
$B = A[b_1, ..., b_n]$ for certain $b_i \in B$ since $A\rightarrow B$ is of finite type. 
Now the statement follows via induction on $n$.
A: Let $A$, $B$ be rings, let $B$ be an $A$-algebra with a ring homomorphism $f:A\to B$.
"finite type + integral = finite" (in terms of $f$)
is equivalent to
"$B$ is finitely generated as an $A$-algebra + $B$ is integral over $f(A)$ = $B$ is finitely generated as an $A$-module" (in terms of $B$)
Let $B$ be a finitely generated $A$-module. Then $B$ is a finitely generated $A$-algebra, by the definition.
For any $x\in B$,  $f(A)[x]$ is contained in $B$ implies that $x$ is integral over $f(A)$, (by Prop 5.1 in Atiyah's book)
so the claim follows.
Conversely, let $B$ be a finitely generated $A$-algebra + integral over $f(A)$.
Then, by the def, there exists a surjective $A$-algebra homomorphism $g:A[t_1,\cdots,t_n]\to B$.
Let $x_1,\cdots,x_n\in B$ be the images of indeterminants $t_1,\cdots, t_n$ under $g$.
Since the $x_i$ are integral over $f(A)$,
$f(A)[x_1,\cdots,x_n]$ is finitely generated as a $f(A)$-module, by Cor 5.2., so as an $A$-module.
Since $g$ is surjective, $B=f(A)[x_1,\cdots,x_n].$
