Exact functor
Posted: Fri May 27, 2016 3:50 pm
Let \(\displaystyle{S}\) be a multiplicative subset of the ring \(\displaystyle{A}\) and let \(\displaystyle{M}\)
be an \(\displaystyle{A}\) - module.
The functor \(\displaystyle{M\rightsquigarrow S^{-1}\,M}\) is exact. In other words, if the sequence of \(\displaystyle{A}\) - modules
\(\displaystyle{M' \xrightarrow{f} M \xrightarrow {g} M''}\) is exact, then so also is the sequence of
\(\displaystyle{S^{-1}\,A}\) - modules
\(\displaystyle{S^{-1}\,M' \xrightarrow{S^{-1}\,f} S^{-1}\,M \xrightarrow {S^{-1}\,g} S^{-1}\,M''}\)
be an \(\displaystyle{A}\) - module.
The functor \(\displaystyle{M\rightsquigarrow S^{-1}\,M}\) is exact. In other words, if the sequence of \(\displaystyle{A}\) - modules
\(\displaystyle{M' \xrightarrow{f} M \xrightarrow {g} M''}\) is exact, then so also is the sequence of
\(\displaystyle{S^{-1}\,A}\) - modules
\(\displaystyle{S^{-1}\,M' \xrightarrow{S^{-1}\,f} S^{-1}\,M \xrightarrow {S^{-1}\,g} S^{-1}\,M''}\)