Localization of an integral extension at a prime need not be integral

An integral extension $A\subset B$ need not give an integral extension $A_{P\cap A}\subset B_P$. For example, take $\mathbb{Z}\subset \mathbb{Z}[i]$ and $P=(2+i)$, whose contraction is $(5)$. The element $1/(2-i)$ belongs to $\mathbb{Z}[i{]}_P$ but is not integral over ${\mathbb{Z}}_{(5)}$: multiplying a hypothetical monic equation by a power of $2-i$ and reducing modulo the other Gaussian prime ideal $(2-i)$ gives $1=0$ in ${\mathbb{F}}_5$. Localizing both rings by the same multiplicative subset does preserve integrality.