Mixing
How to prove that for $K$ $mathbb{Z}$-free & $Q$ projective, $Kotimes_{mathbb{Z}}Q$ is a projective?
0
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Let $K$ be $mathbb{Z}$-free and $Q$ a projective $mathbb{Z}G$ module then $Kotimes_{mathbb{Z}}Q$ is a projective $mathbb{Z}G$-module. I believe that this follows from the adjoint isomorphism theorem:
$Hom_{mathbb{Z}G}(Kotimes Q, N)=Hom_{mathbb{Z}}(K,Hom_{mathbb{Z}G}(Q,N))$
But I don't understand why? Any ideas would be helpful, I just don't know where to start.
homological-algebra projective-module free-modules
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
$endgroup$
|
show 3 more comments
0
$begingroup$
Let $K$ be $mathbb{Z}$-free and $Q$ a projective $mathbb{Z}G$ module then $Kotimes_{mathbb{Z}}Q$ is a projective $mathbb{Z}G$-module. I believe that this follows from the adjoint isomorphism theorem:
$Hom_{mathbb{Z}G}(Kotimes Q, N)=Hom_{mathbb{Z}}(K,Hom_{mathbb{Z}G}(Q,N))$
But I don't understand why? Any ideas would be helpful, I just don't know where to start.
homological-algebra projective-module free-modules
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
$endgroup$
$begingroup$
Hint : using the equality you wrote as an isomorphism of functors in $N$, show that the functors $Hom_{mathbb{Z}G}(Kotimes Q,.)$ is exact.
$endgroup$
– Roland
Jan 24 at 13:20
$begingroup$
Well $Hom_{mathbb{Z}}(K,-)$ is exact as it is over an abelian category so $Hom_{mathbb{Z}G}(Kotimes Q,-)$ must also be exact?
$endgroup$
– Rhoswyn
Jan 24 at 13:33
$begingroup$
Remember a module $M$ is projective iff $Hom(M,.)$ is exact. Then $Hom_{mathbb{Z}G}(Kotimes Q,.)$ can be written as the composite of two exact functors...
$endgroup$
– Roland
Jan 24 at 13:35
2
$begingroup$
You don't know that the LHS is exact, this is what you are trying to prove !! The RHS is because this is the composite of two exact functors and thus the LHS is.
$endgroup$
– Roland
Jan 24 at 13:43
1
$begingroup$
Sorry I meant LHS and RHS the other way around facepalm I think I get it now though, thanks.
$endgroup$
– Rhoswyn
Jan 24 at 13:45
|
show 3 more comments
0
0
0
$begingroup$
Let $K$ be $mathbb{Z}$-free and $Q$ a projective $mathbb{Z}G$ module then $Kotimes_{mathbb{Z}}Q$ is a projective $mathbb{Z}G$-module. I believe that this follows from the adjoint isomorphism theorem:
$Hom_{mathbb{Z}G}(Kotimes Q, N)=Hom_{mathbb{Z}}(K,Hom_{mathbb{Z}G}(Q,N))$
But I don't understand why? Any ideas would be helpful, I just don't know where to start.
homological-algebra projective-module free-modules
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
$endgroup$
Let $K$ be $mathbb{Z}$-free and $Q$ a projective $mathbb{Z}G$ module then $Kotimes_{mathbb{Z}}Q$ is a projective $mathbb{Z}G$-module. I believe that this follows from the adjoint isomorphism theorem:
$Hom_{mathbb{Z}G}(Kotimes Q, N)=Hom_{mathbb{Z}}(K,Hom_{mathbb{Z}G}(Q,N))$
But I don't understand why? Any ideas would be helpful, I just don't know where to start.
homological-algebra projective-module free-modules
homological-algebra projective-module free-modules
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
asked Jan 24 at 13:07
RhoswynRhoswyn
388210
388210
$begingroup$
Hint : using the equality you wrote as an isomorphism of functors in $N$, show that the functors $Hom_{mathbb{Z}G}(Kotimes Q,.)$ is exact.
$endgroup$
– Roland
Jan 24 at 13:20
$begingroup$
Well $Hom_{mathbb{Z}}(K,-)$ is exact as it is over an abelian category so $Hom_{mathbb{Z}G}(Kotimes Q,-)$ must also be exact?
$endgroup$
– Rhoswyn
Jan 24 at 13:33
$begingroup$
Remember a module $M$ is projective iff $Hom(M,.)$ is exact. Then $Hom_{mathbb{Z}G}(Kotimes Q,.)$ can be written as the composite of two exact functors...
$endgroup$
– Roland
Jan 24 at 13:35
2
$begingroup$
You don't know that the LHS is exact, this is what you are trying to prove !! The RHS is because this is the composite of two exact functors and thus the LHS is.
$endgroup$
– Roland
Jan 24 at 13:43
1
$begingroup$
Sorry I meant LHS and RHS the other way around facepalm I think I get it now though, thanks.
$endgroup$
– Rhoswyn
Jan 24 at 13:45
|
show 3 more comments
$begingroup$
Hint : using the equality you wrote as an isomorphism of functors in $N$, show that the functors $Hom_{mathbb{Z}G}(Kotimes Q,.)$ is exact.
$endgroup$
– Roland
Jan 24 at 13:20
$begingroup$
Well $Hom_{mathbb{Z}}(K,-)$ is exact as it is over an abelian category so $Hom_{mathbb{Z}G}(Kotimes Q,-)$ must also be exact?
$endgroup$
– Rhoswyn
Jan 24 at 13:33
$begingroup$
Remember a module $M$ is projective iff $Hom(M,.)$ is exact. Then $Hom_{mathbb{Z}G}(Kotimes Q,.)$ can be written as the composite of two exact functors...
$endgroup$
– Roland
Jan 24 at 13:35
2
$begingroup$
You don't know that the LHS is exact, this is what you are trying to prove !! The RHS is because this is the composite of two exact functors and thus the LHS is.
$endgroup$
– Roland
Jan 24 at 13:43
1
$begingroup$
Sorry I meant LHS and RHS the other way around facepalm I think I get it now though, thanks.
$endgroup$
– Rhoswyn
Jan 24 at 13:45
$begingroup$
Hint : using the equality you wrote as an isomorphism of functors in $N$, show that the functors $Hom_{mathbb{Z}G}(Kotimes Q,.)$ is exact.
$endgroup$
– Roland
Jan 24 at 13:20
$begingroup$
Hint : using the equality you wrote as an isomorphism of functors in $N$, show that the functors $Hom_{mathbb{Z}G}(Kotimes Q,.)$ is exact.
$endgroup$
– Roland
Jan 24 at 13:20
$begingroup$
Well $Hom_{mathbb{Z}}(K,-)$ is exact as it is over an abelian category so $Hom_{mathbb{Z}G}(Kotimes Q,-)$ must also be exact?
$endgroup$
– Rhoswyn
Jan 24 at 13:33
$begingroup$
Well $Hom_{mathbb{Z}}(K,-)$ is exact as it is over an abelian category so $Hom_{mathbb{Z}G}(Kotimes Q,-)$ must also be exact?
$endgroup$
– Rhoswyn
Jan 24 at 13:33
$begingroup$
Remember a module $M$ is projective iff $Hom(M,.)$ is exact. Then $Hom_{mathbb{Z}G}(Kotimes Q,.)$ can be written as the composite of two exact functors...
$endgroup$
– Roland
Jan 24 at 13:35
$begingroup$
Remember a module $M$ is projective iff $Hom(M,.)$ is exact. Then $Hom_{mathbb{Z}G}(Kotimes Q,.)$ can be written as the composite of two exact functors...
$endgroup$
– Roland
Jan 24 at 13:35
2
2
$begingroup$
You don't know that the LHS is exact, this is what you are trying to prove !! The RHS is because this is the composite of two exact functors and thus the LHS is.
$endgroup$
– Roland
Jan 24 at 13:43
$begingroup$
You don't know that the LHS is exact, this is what you are trying to prove !! The RHS is because this is the composite of two exact functors and thus the LHS is.
$endgroup$
– Roland
Jan 24 at 13:43
1
1
$begingroup$
Sorry I meant LHS and RHS the other way around facepalm I think I get it now though, thanks.
$endgroup$
– Rhoswyn
Jan 24 at 13:45
$begingroup$
Sorry I meant LHS and RHS the other way around facepalm I think I get it now though, thanks.
$endgroup$
– Rhoswyn
Jan 24 at 13:45
|
show 3 more comments
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$begingroup$
Hint : using the equality you wrote as an isomorphism of functors in $N$, show that the functors $Hom_{mathbb{Z}G}(Kotimes Q,.)$ is exact.
$endgroup$
– Roland
Jan 24 at 13:20
$begingroup$
Well $Hom_{mathbb{Z}}(K,-)$ is exact as it is over an abelian category so $Hom_{mathbb{Z}G}(Kotimes Q,-)$ must also be exact?
$endgroup$
– Rhoswyn
Jan 24 at 13:33
$begingroup$
Remember a module $M$ is projective iff $Hom(M,.)$ is exact. Then $Hom_{mathbb{Z}G}(Kotimes Q,.)$ can be written as the composite of two exact functors...
$endgroup$
– Roland
Jan 24 at 13:35
2
$begingroup$
You don't know that the LHS is exact, this is what you are trying to prove !! The RHS is because this is the composite of two exact functors and thus the LHS is.
$endgroup$
– Roland
Jan 24 at 13:43
1
$begingroup$
Sorry I meant LHS and RHS the other way around facepalm I think I get it now though, thanks.
$endgroup$
– Rhoswyn
Jan 24 at 13:45