A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are

Assume $A$ is integrally closed, and let $\mathfrak{p}$ be a prime ideal. If $x\in \mathrm{Frac}(A)$ is integral over $A_{\mathfrak{p}}$, then [L2] gives some $s\mathrm{\notin }\mathfrak{p}$ with $sx$ integral over $A$. By [L1], $sx\in A$, so $x=(sx)\mathrm{/}s\in A_{\mathfrak{p}}$. Hence every prime localisation is integrally closed.

Assume every maximal localisation $A_{\mathfrak{m}}$ is integrally closed, and let $x\in \mathrm{Frac}(A)$ be integral over $A$. Because $A$ is a domain, it is nonzero, so [L3] applies and makes the $A$-algebra $M:=A[x]$ a finite $A$-module. For each maximal ideal $\mathfrak{m}$, the same element $x$ is integral over $A_{\mathfrak{m}}$ by [L2], so the hypothesis gives $x\in A_{\mathfrak{m}}$ and therefore $M_{\mathfrak{m}}=A_{\mathfrak{m}}$. Thus the localisation of the inclusion $i:A\hookrightarrow M$ at every maximal ideal is surjective.

Every maximal ideal is prime, so step 1.1 implies that if all prime localisations are integrally closed, then all maximal localisations are integrally closed.

By [L4], the map $i:A\hookrightarrow M$ is surjective. Hence $M=A$, so $x\in A$. By [L1], the domain $A$ is integrally closed.

Step 1.1 proves $(1)\Rightarrow (2)$, step 2.1 gives $(2)\Rightarrow (3)$, and step 2.2 proves $(3)\Rightarrow (1)$. Therefore the three conditions are equivalent.