Finitely generated free module is projective.












1














Call a $R$-module projective if every short exact sequence $0 to Astackrel{f} to Bstackrel{g} to C to 0$ of $R$-modules splits.



Call a short exact sequence as above split, if it admits a section. i.e. an $R$-linear map $h:Crightarrow B$ such that $gcirc h= text{id}_C$.



I wish to show that a finitely generated free module is projective. So I need to produce a section for the above short exact sequence where $C$ is finitely generated and free.



My thoughts:




  1. I can write down an $R$-linear map $h:B/Arightarrow B$. But no information is given about the map $g$ (except I know that it is surjective). How could I verify that the required composition is the identity on $C$?


  2. Perhaps, to show that the sequence splits, I can show that $Bcong A oplus C$. I'm not sure how I would proceed to do this though. I know that $C$ has a finite basis. Maybe this helps?



This is a homework question. Please don't provide complete solutions. Hints are appreciated. Thanks!






modules homological-algebra exact-sequence projective-module free-modules







share|cite|improve this question





























    1














    Call a $R$-module projective if every short exact sequence $0 to Astackrel{f} to Bstackrel{g} to C to 0$ of $R$-modules splits.



    Call a short exact sequence as above split, if it admits a section. i.e. an $R$-linear map $h:Crightarrow B$ such that $gcirc h= text{id}_C$.



    I wish to show that a finitely generated free module is projective. So I need to produce a section for the above short exact sequence where $C$ is finitely generated and free.



    My thoughts:




    1. I can write down an $R$-linear map $h:B/Arightarrow B$. But no information is given about the map $g$ (except I know that it is surjective). How could I verify that the required composition is the identity on $C$?


    2. Perhaps, to show that the sequence splits, I can show that $Bcong A oplus C$. I'm not sure how I would proceed to do this though. I know that $C$ has a finite basis. Maybe this helps?



    This is a homework question. Please don't provide complete solutions. Hints are appreciated. Thanks!






    modules homological-algebra exact-sequence projective-module free-modules







    share|cite|improve this question



























      1












      1








      1







      Call a $R$-module projective if every short exact sequence $0 to Astackrel{f} to Bstackrel{g} to C to 0$ of $R$-modules splits.



      Call a short exact sequence as above split, if it admits a section. i.e. an $R$-linear map $h:Crightarrow B$ such that $gcirc h= text{id}_C$.



      I wish to show that a finitely generated free module is projective. So I need to produce a section for the above short exact sequence where $C$ is finitely generated and free.



      My thoughts:




      1. I can write down an $R$-linear map $h:B/Arightarrow B$. But no information is given about the map $g$ (except I know that it is surjective). How could I verify that the required composition is the identity on $C$?


      2. Perhaps, to show that the sequence splits, I can show that $Bcong A oplus C$. I'm not sure how I would proceed to do this though. I know that $C$ has a finite basis. Maybe this helps?



      This is a homework question. Please don't provide complete solutions. Hints are appreciated. Thanks!






      modules homological-algebra exact-sequence projective-module free-modules







      share|cite|improve this question















      Call a $R$-module projective if every short exact sequence $0 to Astackrel{f} to Bstackrel{g} to C to 0$ of $R$-modules splits.



      Call a short exact sequence as above split, if it admits a section. i.e. an $R$-linear map $h:Crightarrow B$ such that $gcirc h= text{id}_C$.



      I wish to show that a finitely generated free module is projective. So I need to produce a section for the above short exact sequence where $C$ is finitely generated and free.



      My thoughts:




      1. I can write down an $R$-linear map $h:B/Arightarrow B$. But no information is given about the map $g$ (except I know that it is surjective). How could I verify that the required composition is the identity on $C$?


      2. Perhaps, to show that the sequence splits, I can show that $Bcong A oplus C$. I'm not sure how I would proceed to do this though. I know that $C$ has a finite basis. Maybe this helps?



      This is a homework question. Please don't provide complete solutions. Hints are appreciated. Thanks!






      modules homological-algebra exact-sequence projective-module free-modules




      modules homological-algebra exact-sequence projective-module free-modules






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      share|cite|improve this question













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      edited Nov 25 at 6:13









      Lord Shark the Unknown

      100k958131




      100k958131










      asked Nov 25 at 5:10









      tangentbundle

      408211




      408211






















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          You want a homomorphism $h:Cto B$ with certain properties. As $C$ is finitely
          generated projective it has a basis $c_1,ldots,c_n$. Given $b_1,ldots,b_nin B$,
          there is a unique homomorphism $h:Cto B$ with $h(c_i)=b_i$ for all $i$.



          If you can choose the $b_i$ so that $g(b_i)=c_i$, then $(hcirc g)(b_i)=c_i$.
          That would imply $hcirc g=text{id}_C$.






          share|cite|improve this answer





















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            You want a homomorphism $h:Cto B$ with certain properties. As $C$ is finitely
            generated projective it has a basis $c_1,ldots,c_n$. Given $b_1,ldots,b_nin B$,
            there is a unique homomorphism $h:Cto B$ with $h(c_i)=b_i$ for all $i$.



            If you can choose the $b_i$ so that $g(b_i)=c_i$, then $(hcirc g)(b_i)=c_i$.
            That would imply $hcirc g=text{id}_C$.






            share|cite|improve this answer


























              2














              You want a homomorphism $h:Cto B$ with certain properties. As $C$ is finitely
              generated projective it has a basis $c_1,ldots,c_n$. Given $b_1,ldots,b_nin B$,
              there is a unique homomorphism $h:Cto B$ with $h(c_i)=b_i$ for all $i$.



              If you can choose the $b_i$ so that $g(b_i)=c_i$, then $(hcirc g)(b_i)=c_i$.
              That would imply $hcirc g=text{id}_C$.






              share|cite|improve this answer
























                2












                2








                2






                You want a homomorphism $h:Cto B$ with certain properties. As $C$ is finitely
                generated projective it has a basis $c_1,ldots,c_n$. Given $b_1,ldots,b_nin B$,
                there is a unique homomorphism $h:Cto B$ with $h(c_i)=b_i$ for all $i$.



                If you can choose the $b_i$ so that $g(b_i)=c_i$, then $(hcirc g)(b_i)=c_i$.
                That would imply $hcirc g=text{id}_C$.






                share|cite|improve this answer












                You want a homomorphism $h:Cto B$ with certain properties. As $C$ is finitely
                generated projective it has a basis $c_1,ldots,c_n$. Given $b_1,ldots,b_nin B$,
                there is a unique homomorphism $h:Cto B$ with $h(c_i)=b_i$ for all $i$.



                If you can choose the $b_i$ so that $g(b_i)=c_i$, then $(hcirc g)(b_i)=c_i$.
                That would imply $hcirc g=text{id}_C$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 6:20









                Lord Shark the Unknown

                100k958131




                100k958131






























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