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 Post subject: Exact functor
PostPosted: Fri May 27, 2016 3:50 pm 
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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

\(\displaystyle{S^{-1}\,M' \xrightarrow{S^{-1}\,f} S^{-1}\,M \xrightarrow {S^{-1}\,g} S^{-1}\,M''}\)


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