induced sequence exact

Question

If $D$ is a multiplicatively closed subset of $R$. I'm trying to come up with an example where

$$0\to L \to M \to N \to 0$$

is not exact, but the induced sequence

$$0 \to D^{-1}L \to D^{-1}M \to D^{-1}N \to 0$$

is exact.

Answer 1

Let $r\in R$ be nilpotent and $D=\{r^n:n\geq 0\}$. Then $D$ is a multiplicative system and $D^{-1}R$ is the zero ring. Since every module over the zero ring is isomorphic to the zero module, applying $D^{-1}(-)$ to any sequence $$ 0\to L\to M\to N\to 0 $$ in $_{R}\mathsf{Mod}$ gives the sequence $$ 0\to0\to0\to0\to0 $$ in $_{D^{-1}R}\mathsf{Mod}$.

More generally, if $D$ is any multiplicative system in $R$ with $0\in D$ then the above still holds.

Answer 2

Consider the special case $N=0$. Then

$$0 \to L \to M \to 0 \to 0$$

is exact precisely when $L \to M$ is an isomorphism. So, is it possible that $D^{-1}(L) \to D^{-1}(M)$ is an isomorphism, yet $L \to M$ is not?

Certainly. Let $R = \mathbb{Z}$. Let $L \to M$ be $\mathbb{Z} \xrightarrow{2} \mathbb{Z}$, and let $D$ be generated by $2.$