Show That Function on Ideal is Just Multiplication Map

Question

If $R$ is a Noetherian domain, $_RM$ a torsion-free $R$-module and $I \leq R$ a left ideal. If we consider $f: I \rightarrow M$, we can write $f(x) = x m(x)$ for some $m(x) \in S^{-1}M$ where $S = R \setminus \{0\}$. How do I now see that $m(x) = m(y)$ for all $x,y \in I$?

Answer 1

Let $R$ be a left noetherian domain and $S=R\setminus\{0\}$, then $R$ is a left Ore domain, and $Q(_RR)=S^{-1}R$ the maximal left quotient ring of $R$. There are two important properties of $Q(_RR)$: (1) $Q(_RR)$ is module isomorphic to the injective hull of $_RR$, and (2) every left $Q(_RR)$--module is an injective left $R$--module. In addition, $S^{-1}M=Q(M)=Q(_RR)\otimes_RM$ is an injective left $R$--module.

In this context, if we consider the map $h:I\stackrel{f}{\longrightarrow}{M}\hookrightarrow{Q(M)}$, ($M$ is torsionfree), there exists a map $g:R\longrightarrow{Q(M)}$ such that $h=g_{\mid{I}}$.

Finally, for every $x\in{I}$ we have $f(x)=h(x)=g(x)=xg(1)$, being $g(1)\in{Q(M)}$.

Answer 2

Let $x,y\in{I\setminus\{0\}}$, if you start with $f(xy)$, you obtain $x^{-1}f(x)=y^{-1}f(y)$, which is the answer to your question.