*Question seems simple, but I just can't find the solution.
Let A/B be an integral ring extension and let P be a prime ideal of B. By going-up theorem, there is Q, a prime ideal of A, lying over P. Then the ring of fractions of A localised on Q is still integral on that of B localised on P?*

Thank Karl and Matt, the two exemples from you are brilliants. But in the special case for the field of fractions, I think this proposition is true. I don't know if my proof is exact or not.

Let $A/B$ be an integral ring extension, $K=FracA$ the field of fractions for $A$, $k=FracB$ the field of fractions for $B$. Then $K$ is algebraic on $k$.

I have a proof: Let $S=A- \lbrace 0\rbrace, s=B- \lbrace 0\rbrace$ then $S^{-1}A=K, s^{-1}B=k,\mbox{then } k \subseteq{s^{-1}A} \subseteq K.$ $\forall \mbox{ } a/S_0\in K, a\in A, \mbox{ }S_0\in S,$ there is $S_0/s_0X-a/s_0 \in k(S_0/s_0, a/s_0)[X]$ where $s_0 \in s \subseteq B$ so that $a/S_0$ is its root. So $a/S_0$ is algebraic on the field $k(S_0/s_0,a/s_0)$ (In fact, it just lies in this field). Besides, $a/s_0$ and $S_0/s_0$ are algebraic on $k$, since $s^{-1}A$ is integral on $k$. So $k(S_0/s_0,a/s_0)/k$ is finite, then $a/S_0$ is algebraic on $k$.

domainslocalized at the origin, but that's just because only one prime of $A$ lies above $\langle 0_B \rangle$. $\endgroup$ – Karl Schwede Dec 20 '13 at 2:36