This is actually not a problem, but it's a statement which is taken for granted and I don't know how to prove it. Hope some one can help me. I really appreciate:
Suppose $A$ is subring of ring $B$, and $A$, $B$ are integral domains. If $B$ is integral over $A$, prove that field of fraction of $B$ is algebraic over field of fraction of $A$. Does the converse remain true?
On
[I'm using $K(\phantom{x})$ to denote taking the fraction field.]
The elements of $B$ are certainly algebraic over $K(A)$, and for a general algebraic extension $L/K$ the elements of $L$ algebraic over $K$ form a field. If this fact is unknown to you then it would be a good exercise.
If I'm understanding you correctly, the last part has no chance of being true. A silly example: $\mathbb{Z} \subset \mathbb{Z}[\tfrac12]$. The fraction fields are the same!
It is a well-known fact from field theory that if $a$ is algebraic, then $a^{-1}$ is also algebraic. And for field extensions algebraic = integral. This shows that every integral extension of integral domains induces an algebraic extension of their fraction fields. Of course the converse is not true (consider $\mathbb{Z} \subseteq \mathbb{Q}$ for example).