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Sep 12, 2016 at 15:10 history edited Jeremy Rickard CC BY-SA 3.0
minor typo
Sep 12, 2016 at 11:44 vote accept Xiaosong Peng
Sep 12, 2016 at 9:59 comment added Xiaosong Peng Get it! Thank you very much for your help.
Sep 12, 2016 at 9:54 comment added Jeremy Rickard @Penson $\gamma(A)$ is the submodule of $G^{(I)}$ generated by $\gamma(1)$, and $\gamma(1)\in G^{(J)}$ for some finite $J$. (The important point is that $A$ is finitely generated as an $A$-module: if a finitely generated module is contained in a direct sum then it is contained in a finite subsum.)
Sep 12, 2016 at 9:49 comment added Xiaosong Peng @ Jeremy Rickard Since A can be seen as a projective A-module, so we can take a splitting map $\gamma : A \rightarrow G^{(I)}$, but if the index set I is infinite, why the image of $\gamma$ is contained in $G^{(J)}$ for some finite J?
Sep 12, 2016 at 9:23 history answered Jeremy Rickard CC BY-SA 3.0