Every $mathbb{Z}/6mathbb{Z}$-module is projective












5












$begingroup$



I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.




I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).

My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.






abstract-algebra modules projective-module







share|cite|improve this question











$endgroup$

















    5












    $begingroup$



    I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.




    I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).

    My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.






    abstract-algebra modules projective-module







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      2



      $begingroup$



      I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.




      I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).

      My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.






      abstract-algebra modules projective-module







      share|cite|improve this question











      $endgroup$





      I have to prove that every $mathbb{Z}/6mathbb{Z}$-module is projective.




      I've found already this question Prove that every $mathbb{Z}/6mathbb{Z}$-module is projective and injective. Find a $mathbb{Z}/4mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).

      My idea is to use the fact that $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} bigoplus mathbb{Z}/3mathbb{Z}$, these two submodules are projective and they are fields, hence all $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}/3mathbb{Z}$-modules are free. But I don't know if taken any $mathbb{Z}/6mathbb{Z}$-module I can "decompose" it in a sum of a $mathbb{Z}/2mathbb{Z}$-module and a $mathbb{Z}/3mathbb{Z}$-module.






      abstract-algebra modules projective-module




      abstract-algebra modules projective-module






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 22:46









      user26857

      39.5k124283




      39.5k124283










      asked Jan 1 at 21:10









      user289143user289143

      985313




      985313






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.



          If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
          so $2Mcap 3M=0$.



          On the other hand, $m=3m-2m$, so $2M+3M=M$.



          Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.



          Thus the claim you wanted to prove is true.



          Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @user289143 Yes of course, thanks for catching that :)
            $endgroup$
            – jgon
            Jan 1 at 21:45



















          0












          $begingroup$

          Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            “the proof should be based just on the equivalent definitions of projective modules”
            $endgroup$
            – egreg
            Jan 1 at 23:26










          • $begingroup$
            @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
            $endgroup$
            – Pedro Tamaroff♦
            Jan 1 at 23:39










          • $begingroup$
            Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
            $endgroup$
            – jgon
            Jan 2 at 3:07












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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.



          If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
          so $2Mcap 3M=0$.



          On the other hand, $m=3m-2m$, so $2M+3M=M$.



          Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.



          Thus the claim you wanted to prove is true.



          Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @user289143 Yes of course, thanks for catching that :)
            $endgroup$
            – jgon
            Jan 1 at 21:45
















          3












          $begingroup$

          Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.



          If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
          so $2Mcap 3M=0$.



          On the other hand, $m=3m-2m$, so $2M+3M=M$.



          Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.



          Thus the claim you wanted to prove is true.



          Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @user289143 Yes of course, thanks for catching that :)
            $endgroup$
            – jgon
            Jan 1 at 21:45














          3












          3








          3





          $begingroup$

          Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.



          If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
          so $2Mcap 3M=0$.



          On the other hand, $m=3m-2m$, so $2M+3M=M$.



          Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.



          Thus the claim you wanted to prove is true.



          Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.






          share|cite|improve this answer











          $endgroup$



          Let $M$ be a $Bbb{Z}/6Bbb{Z}$ module. We'll show that $M=2Moplus 3M$.



          If $min 2Mcap 3M$, then $m=2m'=3m''$ for some $m',m''in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$
          so $2Mcap 3M=0$.



          On the other hand, $m=3m-2m$, so $2M+3M=M$.



          Thus $M=2Moplus 3M$. Note that $2M$ is a $newcommandZZ{Bbb{Z}}ZZ/3ZZ$ module and $3M$ is a $ZZ/2ZZ$ module.



          Thus the claim you wanted to prove is true.



          Then $2M$ is a direct sum of copies of $ZZ/3ZZ$ and $3M$ is a direct sum of copies of $ZZ/2ZZ$, and as you've noted $ZZ/3ZZ$ and $ZZ/2ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 21:45

























          answered Jan 1 at 21:40









          jgonjgon

          16.1k32143




          16.1k32143












          • $begingroup$
            @user289143 Yes of course, thanks for catching that :)
            $endgroup$
            – jgon
            Jan 1 at 21:45


















          • $begingroup$
            @user289143 Yes of course, thanks for catching that :)
            $endgroup$
            – jgon
            Jan 1 at 21:45
















          $begingroup$
          @user289143 Yes of course, thanks for catching that :)
          $endgroup$
          – jgon
          Jan 1 at 21:45




          $begingroup$
          @user289143 Yes of course, thanks for catching that :)
          $endgroup$
          – jgon
          Jan 1 at 21:45











          0












          $begingroup$

          Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            “the proof should be based just on the equivalent definitions of projective modules”
            $endgroup$
            – egreg
            Jan 1 at 23:26










          • $begingroup$
            @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
            $endgroup$
            – Pedro Tamaroff♦
            Jan 1 at 23:39










          • $begingroup$
            Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
            $endgroup$
            – jgon
            Jan 2 at 3:07
















          0












          $begingroup$

          Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            “the proof should be based just on the equivalent definitions of projective modules”
            $endgroup$
            – egreg
            Jan 1 at 23:26










          • $begingroup$
            @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
            $endgroup$
            – Pedro Tamaroff♦
            Jan 1 at 23:39










          • $begingroup$
            Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
            $endgroup$
            – jgon
            Jan 2 at 3:07














          0












          0








          0





          $begingroup$

          Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.






          share|cite|improve this answer









          $endgroup$



          Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 1 at 23:19









          BenBen

          4,313617




          4,313617








          • 1




            $begingroup$
            “the proof should be based just on the equivalent definitions of projective modules”
            $endgroup$
            – egreg
            Jan 1 at 23:26










          • $begingroup$
            @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
            $endgroup$
            – Pedro Tamaroff♦
            Jan 1 at 23:39










          • $begingroup$
            Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
            $endgroup$
            – jgon
            Jan 2 at 3:07














          • 1




            $begingroup$
            “the proof should be based just on the equivalent definitions of projective modules”
            $endgroup$
            – egreg
            Jan 1 at 23:26










          • $begingroup$
            @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
            $endgroup$
            – Pedro Tamaroff♦
            Jan 1 at 23:39










          • $begingroup$
            Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
            $endgroup$
            – jgon
            Jan 2 at 3:07








          1




          1




          $begingroup$
          “the proof should be based just on the equivalent definitions of projective modules”
          $endgroup$
          – egreg
          Jan 1 at 23:26




          $begingroup$
          “the proof should be based just on the equivalent definitions of projective modules”
          $endgroup$
          – egreg
          Jan 1 at 23:26












          $begingroup$
          @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
          $endgroup$
          – Pedro Tamaroff♦
          Jan 1 at 23:39




          $begingroup$
          @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair?
          $endgroup$
          – Pedro Tamaroff♦
          Jan 1 at 23:39












          $begingroup$
          Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
          $endgroup$
          – jgon
          Jan 2 at 3:07




          $begingroup$
          Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question.
          $endgroup$
          – jgon
          Jan 2 at 3:07


















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