Let $A$ be a commutative, integrally closed, noetherian ring and as $\mathfrak p$ ranges over height one prime ideals, we have: $$A = \bigcap_\mathfrak p A_{\mathfrak p}.$$
The proof I have seen proceeds so:
It is clear that $A \subset \bigcap_\mathfrak p A_{\mathfrak p}$. For the other way, assume $\phi \in \bigcap_\mathfrak p A_{\mathfrak p}$ and define the ideal: $$I_\phi = \{x \in A: x\phi \in A\}.$$ Since $\phi \in A_\mathfrak p$, there exists $f,g \in A$ and $g\not\in \mathfrak p$ such that $\phi = f/g$. This implies that $g \in I_\phi$ and therefore $I_\phi$ is not contained in any height one prime ideal which implies that $I_\phi = A$(See Edit).
Therefore, $1 \in I_\phi \iff \phi = \phi.1 \in A$ and this finishes the proof.
My question is: Where were the assumptions that $A$ is noetherian and integrally closed used?
EDIT: I realize now that one cannot conclude that $I_\phi = A$. I would like to know if this proof can be modified to finish it off. However maybe this approach is failed and that would be an acceptable answer too.