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Second isomorphism theorem for modules, doubts with proof


Further explanation on proof that associated primes are precisely those belonging to primary modules in reduced decomposition of $0$.Error in proof that submodules of f.g. modules are f.g.Module isomorphism, simple modules, and quotientsRank-nullity theorem for modulesIsomorphism between modules and submodulesProblem with first isomorphism theorem for modulesPrimary decomposition of modules - uniqueness proofIsomorphism of modules and projectionProof of the Second Isomorphism Theorem for modulesTo prove the Second Isomorphism Theorem for modules from the opposite direction:













0












$begingroup$


Theorem:
Let $S$ and $T$ be submodules of $M$, and let $S+T=s+t,sin S,t in T$. Then $S+T$ and $S∩T$ are submodules of M, and:



$S/(S∩T)≃(S+T)/T$.



I saw the proof but don't we have to check also $S cap T$ is submodule of $S$ and $T$ is submodule of $S+T$? As I remember these terms have to be satisfied due to definition. Or maybe it is obvious for everyone?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    Theorem:
    Let $S$ and $T$ be submodules of $M$, and let $S+T=s+t,sin S,t in T$. Then $S+T$ and $S∩T$ are submodules of M, and:



    $S/(S∩T)≃(S+T)/T$.



    I saw the proof but don't we have to check also $S cap T$ is submodule of $S$ and $T$ is submodule of $S+T$? As I remember these terms have to be satisfied due to definition. Or maybe it is obvious for everyone?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      Theorem:
      Let $S$ and $T$ be submodules of $M$, and let $S+T=s+t,sin S,t in T$. Then $S+T$ and $S∩T$ are submodules of M, and:



      $S/(S∩T)≃(S+T)/T$.



      I saw the proof but don't we have to check also $S cap T$ is submodule of $S$ and $T$ is submodule of $S+T$? As I remember these terms have to be satisfied due to definition. Or maybe it is obvious for everyone?










      share|cite|improve this question









      $endgroup$




      Theorem:
      Let $S$ and $T$ be submodules of $M$, and let $S+T=s+t,sin S,t in T$. Then $S+T$ and $S∩T$ are submodules of M, and:



      $S/(S∩T)≃(S+T)/T$.



      I saw the proof but don't we have to check also $S cap T$ is submodule of $S$ and $T$ is submodule of $S+T$? As I remember these terms have to be satisfied due to definition. Or maybe it is obvious for everyone?







      abstract-algebra modules






      share|cite|improve this question













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      asked Mar 17 at 13:35









      ToidiToidi

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          $begingroup$

          It does need to be checked. It's rather easy though: They are subsets and they are modules, therefore they are submodules.






          share|cite|improve this answer









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            $begingroup$

            It does need to be checked. It's rather easy though: They are subsets and they are modules, therefore they are submodules.






            share|cite|improve this answer









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              $begingroup$

              It does need to be checked. It's rather easy though: They are subsets and they are modules, therefore they are submodules.






              share|cite|improve this answer









              $endgroup$















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                3








                3





                $begingroup$

                It does need to be checked. It's rather easy though: They are subsets and they are modules, therefore they are submodules.






                share|cite|improve this answer









                $endgroup$



                It does need to be checked. It's rather easy though: They are subsets and they are modules, therefore they are submodules.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 17 at 13:45









                ArthurArthur

                120k7120203




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