Finitely generated free module is projective.
$begingroup$
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:
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 $C$?
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
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
$endgroup$
add a comment |
$begingroup$
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:
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 $C$?
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
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
$endgroup$
add a comment |
$begingroup$
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:
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 $C$?
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
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
$endgroup$
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:
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 $C$?
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
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
edited Nov 25 '18 at 6:13
Lord Shark the Unknown
102k1059132
102k1059132
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
asked Nov 25 '18 at 5:10
tangentbundletangentbundle
410211
410211
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
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 $hcirc g=text{id}_C$.
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $hcirc g=text{id}_C$.
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
add a comment |
$begingroup$
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 $hcirc g=text{id}_C$.
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
add a comment |
$begingroup$
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 $hcirc g=text{id}_C$.
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
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 $hcirc g=text{id}_C$.
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Nov 25 '18 at 6:20
Lord Shark the UnknownLord Shark the Unknown
102k1059132
102k1059132
add a comment |
add a comment |
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