A result of Hermite He outlined a proof of the fact that there are only finitely many algebraic number fields with a given discriminant.
I wish to know whether this fact is somehow related to the assertion that, for any C>0, the number of complex quadratic fields with class number less than C is finite...
Thanks in advance for your answers.
 A: Hermite's theorem (which, incidentally, only shows that there are only finitely many number fields of bounded discriminant and given degree), has very little to do with the "class number one problem". The class number one problem does not assert that there aren't too many fields with small discriminant over $\mathbf{Q}$, but rather that there are quite a lot of relative extensions with small discriminant, i.e. there are many pairs $L \supsetneq K \supsetneq \mathbf{Q}$ with $K$ imaginary quadratic and the discriminant of $L$ over $K$ equal to 1. So in a sense they are even "in opposite directions"
As I commented above, from the standpoint of Hermite's theorem it's not obvious why imaginary quadratic fields should be all that different from real quadratic fields, and it's conjectured that there are infinitely many positive $d$ such that $\mathbf{Q}(\sqrt{d})$ has class number 1.
So while Hermite's theorem is extremely important in all sorts of parts of algebraic number theory, it is not terribly relevant to the solution of the class number one problem by Heegner-Stark-Baker (which is a much more recent and far harder result)
