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[SOLVED] Algebra - modules
I'm reading this proof from D&F and there's something I don't get. It is theorem 1 of chapter 12 on modules over principal ideal domains. The theorem is the following
"Let R be a ring and let M be a left R-module. Then,
(Ever nonempty set of submodules of M contains a maximal element under inclusion) ==> (Every submodule of M is finitely generated)"
The authors prove this assertion by letting N be a submodule of M and denoting S the collection of all finitely generated submodules of N. The hypothesis guarantees the existence of a maximal element N' of S, and we'd be happy if N'=N. So suppose N' is different from N and consider an element x in N\N'. Then the submodule of N generated by {x}uN' is finitely generated, thus violating the maximality of N'.
How is this so? Notice that R is not assumed to have a 1. So there is nothing I can see that guarantees that N' is contained, let alone properly contained, in <{x}uN'>.
Homework Statement
I'm reading this proof from D&F and there's something I don't get. It is theorem 1 of chapter 12 on modules over principal ideal domains. The theorem is the following
"Let R be a ring and let M be a left R-module. Then,
(Ever nonempty set of submodules of M contains a maximal element under inclusion) ==> (Every submodule of M is finitely generated)"
The authors prove this assertion by letting N be a submodule of M and denoting S the collection of all finitely generated submodules of N. The hypothesis guarantees the existence of a maximal element N' of S, and we'd be happy if N'=N. So suppose N' is different from N and consider an element x in N\N'. Then the submodule of N generated by {x}uN' is finitely generated, thus violating the maximality of N'.
The Attempt at a Solution
How is this so? Notice that R is not assumed to have a 1. So there is nothing I can see that guarantees that N' is contained, let alone properly contained, in <{x}uN'>.