Let (R,m) be a commutative local ring. Recall that this means that R is a commutative ring with unique maximal ideal m. Prove the following: a) If M is a finitely generated R-module and mM = M, then M = 0. b) If M is a finitely generated R-module, N containted in M is a submodule, and N + mM = M, then N=M. c) If M is a finitely generated R-module and {m1,....,mn} in M are elements of M whose images {n1,....nn} form a generating set for the R-module M/mM, then {m1,....,mn} form a generating set for M.
Let (R,m) be a commutative local ring. Recall that this means that R is a commutative ring with unique maximal ideal m. Prove the following: a) If M is a finitely generated R-module and mM = M, then M = 0. b) If M is a finitely generated R-module, N containted in M is a submodule, and N + mM = M, then N=M. c) If M is a finitely generated R-module and {m1,....,mn} in M are elements of M whose images {n1,....nn} form a generating set for the R-module M/mM, then {m1,....,mn} form a generating set for M.
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How to prove that: If M is a finitely generated R-module and {m1,....,mn} in M are elements of M whose images {n1,....nn} form a generating set for the R-module M/mM, then {m1,....,mn} form a generating set for M.
by Bartleby ExpertHow to show that: If M is a finitely generated R-module and mM = M, then M = 0.
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