localization in the D + M construction
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.