Proving $Ind^G_HM:=Mbigotimes_RHRG$, where $M$ is projective, is projective.

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Let $Hleq G$ be a subgroup, $R$ a ring and $M$ be a projective $H$-module. Prove that the induced module $Ind^G_HM:=Mbigotimes_RHRG $ is projective.




homological-algebra




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    Let $Hleq G$ be a subgroup, $R$ a ring and $M$ be a projective $H$-module. Prove that the induced module $Ind^G_HM:=Mbigotimes_RHRG $ is projective.




    homological-algebra




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      favorite









      up vote
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      down vote

      favorite











      Let $Hleq G$ be a subgroup, $R$ a ring and $M$ be a projective $H$-module. Prove that the induced module $Ind^G_HM:=Mbigotimes_RHRG $ is projective.




      homological-algebra




      share|cite|improve this question












      Let $Hleq G$ be a subgroup, $R$ a ring and $M$ be a projective $H$-module. Prove that the induced module $Ind^G_HM:=Mbigotimes_RHRG $ is projective.





      homological-algebra





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      asked Aug 27 at 15:47









      Rhoswyn

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          1 Answer
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          This is a special case of the general result: let $P$ be a projective
          right $A$-module for some ring $A$, and let $B$ be a ring with a compatible
          left $A$-module structure, that is $a(bb')=(ab)b'$ for $ain A$, $b$, $b'in B$.
          Then $Potimes_A B$ is a projective right $B$-module.



          To prove this, note there's a projective $A$-module $Q$ with $Poplus Q=F$
          free. Then $(Potimes B)oplus(Qotimes B)cong Fotimes B$ is free over $B$
          and so $Potimes B$ is projective over $B$
          (as a summand of a free module).






          share|cite|improve this answer




















          • So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
            – Rhoswyn
            Aug 27 at 16:14










          • $A=RH$, $B=RG$.
            – Lord Shark the Unknown
            Aug 27 at 16:14










          • Ahh yes, of course! Thank you.
            – Rhoswyn
            Aug 27 at 16:16










          Your Answer




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          1 Answer
          1






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          oldest

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          active

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          up vote
          1
          down vote













          This is a special case of the general result: let $P$ be a projective
          right $A$-module for some ring $A$, and let $B$ be a ring with a compatible
          left $A$-module structure, that is $a(bb')=(ab)b'$ for $ain A$, $b$, $b'in B$.
          Then $Potimes_A B$ is a projective right $B$-module.



          To prove this, note there's a projective $A$-module $Q$ with $Poplus Q=F$
          free. Then $(Potimes B)oplus(Qotimes B)cong Fotimes B$ is free over $B$
          and so $Potimes B$ is projective over $B$
          (as a summand of a free module).






          share|cite|improve this answer




















          • So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
            – Rhoswyn
            Aug 27 at 16:14










          • $A=RH$, $B=RG$.
            – Lord Shark the Unknown
            Aug 27 at 16:14










          • Ahh yes, of course! Thank you.
            – Rhoswyn
            Aug 27 at 16:16














          up vote
          1
          down vote













          This is a special case of the general result: let $P$ be a projective
          right $A$-module for some ring $A$, and let $B$ be a ring with a compatible
          left $A$-module structure, that is $a(bb')=(ab)b'$ for $ain A$, $b$, $b'in B$.
          Then $Potimes_A B$ is a projective right $B$-module.



          To prove this, note there's a projective $A$-module $Q$ with $Poplus Q=F$
          free. Then $(Potimes B)oplus(Qotimes B)cong Fotimes B$ is free over $B$
          and so $Potimes B$ is projective over $B$
          (as a summand of a free module).






          share|cite|improve this answer




















          • So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
            – Rhoswyn
            Aug 27 at 16:14










          • $A=RH$, $B=RG$.
            – Lord Shark the Unknown
            Aug 27 at 16:14










          • Ahh yes, of course! Thank you.
            – Rhoswyn
            Aug 27 at 16:16












          up vote
          1
          down vote










          up vote
          1
          down vote









          This is a special case of the general result: let $P$ be a projective
          right $A$-module for some ring $A$, and let $B$ be a ring with a compatible
          left $A$-module structure, that is $a(bb')=(ab)b'$ for $ain A$, $b$, $b'in B$.
          Then $Potimes_A B$ is a projective right $B$-module.



          To prove this, note there's a projective $A$-module $Q$ with $Poplus Q=F$
          free. Then $(Potimes B)oplus(Qotimes B)cong Fotimes B$ is free over $B$
          and so $Potimes B$ is projective over $B$
          (as a summand of a free module).






          share|cite|improve this answer












          This is a special case of the general result: let $P$ be a projective
          right $A$-module for some ring $A$, and let $B$ be a ring with a compatible
          left $A$-module structure, that is $a(bb')=(ab)b'$ for $ain A$, $b$, $b'in B$.
          Then $Potimes_A B$ is a projective right $B$-module.



          To prove this, note there's a projective $A$-module $Q$ with $Poplus Q=F$
          free. Then $(Potimes B)oplus(Qotimes B)cong Fotimes B$ is free over $B$
          and so $Potimes B$ is projective over $B$
          (as a summand of a free module).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 27 at 15:59









          Lord Shark the Unknown

          88.4k955115




          88.4k955115











          • So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
            – Rhoswyn
            Aug 27 at 16:14










          • $A=RH$, $B=RG$.
            – Lord Shark the Unknown
            Aug 27 at 16:14










          • Ahh yes, of course! Thank you.
            – Rhoswyn
            Aug 27 at 16:16
















          • So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
            – Rhoswyn
            Aug 27 at 16:14










          • $A=RH$, $B=RG$.
            – Lord Shark the Unknown
            Aug 27 at 16:14










          • Ahh yes, of course! Thank you.
            – Rhoswyn
            Aug 27 at 16:16















          So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
          – Rhoswyn
          Aug 27 at 16:14




          So would $A$ and $B$ be replaced by $G$ and $H$? But if so where does $R$ come in?
          – Rhoswyn
          Aug 27 at 16:14












          $A=RH$, $B=RG$.
          – Lord Shark the Unknown
          Aug 27 at 16:14




          $A=RH$, $B=RG$.
          – Lord Shark the Unknown
          Aug 27 at 16:14












          Ahh yes, of course! Thank you.
          – Rhoswyn
          Aug 27 at 16:16




          Ahh yes, of course! Thank you.
          – Rhoswyn
          Aug 27 at 16:16

















           

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