Let $T$ be an integral domain, and let $K$ be a subfield of $T$ and $M$ a maximal ideal of $T$ such that $T = K + M$. Let $D \subseteq K$ be an integral domain, and set $R = D + M$. My question is: if $m \in M$ such that $m \neq 0$, is $R_m = T_m$?
In this context, $R_m$ is the localization of $R$ at the multiplicative set $S = \{m^k \mid k \in \mathbb{N} \cup \{0\}\}$.
I was reading the paper "Overrings and dimensions of general D + M constructions" by Costa, Mott, and Zafrullah (sorry I don't have a link), and in the second page they use this fact, but I do not see why this is true. Am I missing something obvious?
The answer is yes. Since $R\subseteq T$, also $R_m\subseteq T_m$; for the other direction, it suffices to show that $T\subseteq R_m$, so fix any $a\in T$. Then $am\in M$, since $M$ is an ideal of $T$ and $m\in M$. Hence, since $M\subseteq R$, we have $am\in R$. Thus $a=am/m\in R_m$, as desired.