Let $A$ be an integral domain, is it true that $$A=\cap_{ht(p)=1}A_p=\cap_{p \text{ maximal}}A_p=\cap_{p \text{ prime }}A_p$$
My idea:
Claim1: $$A=\cap_{p \text{ prime}}A_p$$ '$\subset$': Since $A$ is an integral domain $A\hookrightarrow S^{-1}A$ for any localization, in particular $A\hookrightarrow A_p$, $\forall p$.
'$\supset$': Let $\frac{a}{s}\in K:=Frac(A)$ such that $\frac{a}{s}\in A_p$, $\forall p$ prime. Consider the ideal $$D(\frac{a}{s})=\{x\in A:x\cdot\frac{a}{s}\in A\}$$ Then one can see $D(\frac{a}{s})=A\Leftrightarrow \frac{a}{s}\in A$ and $$\frac{a}{s}\in A_p\Leftrightarrow \exists t\in A\setminus p, b\in A \text{ such that }at=sb\Leftrightarrow D(\frac{a}{s})\not\subset p$$ $\Rightarrow$ of the very last equivalence is because $t\not\in p$ and $t\in D(\frac{a}{s})$ because $t\cdot \frac{a}{s}=\frac{sb}{s}=a\in A$.
$\Leftarrow$ of the very last equivalence is because if $\exists x\in A\setminus p$ such that $x\cdot \frac{a}{s}\in A$ one can put $t:=x$ and $b:=x\cdot \frac{a}{s}\in A$ then $\frac{a}{s}= \frac{x\cdot\frac{a}{s}}{x}=\frac{b}{t}$ hence $at=sb$.
Then $\frac{a}{s}\not\in A\Leftrightarrow D(\frac{a}{s})\subsetneq A\Rightarrow \exists p \text{ prime such that $D(\frac{a}{s})\subset p$}\Leftrightarrow \exists \text{ prime $p$ such that }\frac{a}{s}\not\in A_p\Rightarrow \text{ contradicts assumption that }\frac{a}{s}\in A_p, \forall p$.
Similarly one can also prove $$A=\cap_{p \text{ maximal}}A_p$$
Claim2:$$A=\cap_{ht(p)=1}A_p$$ '$\subset$' is proven in claim 1.
'$\supset$': As $A$ is integral domain, we know $p\subset q \Rightarrow A_p\supset A_q\Rightarrow \cap_{ht(p)= 1}A_p=\cap_{ht(p)\geq 1}A_p=\cap_{p \text{ prime}} A_p$
But in claim 1 we have already proven that $A=\cap_{p \text{ prime}} A_p$
