Localization of integral ring extension is not integral, in general

Question

Consider $R\subseteq S$ an integral ring extension and $1\in U\subseteq S$ a multiplicatively closed subset of $S$ containing $1$. Set $T=U\cap R$. I have already proved that $T$ is a multiplicatively closed subset of $R$ containing $1$, but I would like to find a counter-example in order to prove that, in general, the ring extension $T^{-1}R\subseteq U^{-1}S$ is NOT integral (where $T^{-1}R$ denotes the localisation of the ring $R$ at the multiplicatively closed subset $T$). Examples with $R=\mathbb{Z}$ and (for example) $S=\mathbb{Z}[\sqrt{2}]$ do not seem to work. Do you have any simple counter-example that would work here?

Answer 1

$T^{−1}R⊆U^{−1}S$ is not a ring extension since the canonical morphism is not necessarily injective. I suggest you to add the condition integral domains.

For a counterexample localize $\mathbb Z[i]$ at $(2+i)$ and $\mathbb Z$ at $(5)$. The element $\frac{1}{2−i}\in\mathbb Z[i]_{(2+i)}$ is not integral over $\mathbb Z_{(5)}$.

A similar counterexample can be found if one localizes $\mathbb Z[\sqrt 2]$ at $(3+\sqrt2)$. Then $\frac{1}{3-\sqrt2}$ is not integral over $\mathbb Z_{(7)}$.