Every R-module $M$ has a maximal linearly independent subset $A$. If submodule $M_A$ is generated by $A$,...











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I am stuck at proving




Every R-module $M$ has a maximal linearly independent subset $A$. If a submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.




I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?










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    up vote
    1
    down vote

    favorite












    I am stuck at proving




    Every R-module $M$ has a maximal linearly independent subset $A$. If a submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.




    I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am stuck at proving




      Every R-module $M$ has a maximal linearly independent subset $A$. If a submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.




      I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?










      share|cite|improve this question













      I am stuck at proving




      Every R-module $M$ has a maximal linearly independent subset $A$. If a submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.




      I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?







      abstract-algebra modules free-modules






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 16 at 5:46









      Eric Curtis

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