Integral closure

Remarks

The set $\overline{A}$ is a subring of $B$ that contains $A$, and it is integrally closed in $B$ by transitivity of integrality.

Example: the integral closure of $\Z$ in $\Q(i)$ is $\Z[i]$. If $a + bi$ with $a, b \in \Q$ is integral, then so is its conjugate, so $2a$ and $a^2 + b^2$ are rational and integral over $\Z$, hence integers, and this forces $a, b \in \Z$.

Non-example: $\Z[\sqrt{5}]$ is not integrally closed in $\Q(\sqrt{5})$. The element $(1 + \sqrt{5})/2$ is a root of $x^2 - x - 1$, so it is integral over $\Z$, but it is not in $\Z[\sqrt{5}]$. The integral closure of $\Z$ in $\Q(\sqrt{5})$ is $\Z[(1 + \sqrt{5})/2]$.