Mixing
Problem on exact sequences
$begingroup$
Let for all $A$-modules $M$, $0rightarrow operatorname{Hom}(M,N^{prime})xrightarrow{overline{f}}operatorname{Hom}(M,N)xrightarrow{overline{g}}operatorname{Hom}(M,N^{primeprime})$ be exact. Thus $overline{f}$ is injective and $ker(overline{g})=operatorname{im}(overline{f})$.
We want to show that $0rightarrow N^{prime}xrightarrow{f}Nxrightarrow{g}N^{primeprime}$ is an exact sequence. Let $M=N^{prime}/ker(f)$. If I can prove $N^{prime}cong M$, then $f$ will be one-one. But how to show that? Any hint will be appreciated.
commutative-algebra modules exact-sequence
edited Jan 1 at 12:34
Bernard
119k639112
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
$endgroup$
add a comment |
$begingroup$
Let for all $A$-modules $M$, $0rightarrow operatorname{Hom}(M,N^{prime})xrightarrow{overline{f}}operatorname{Hom}(M,N)xrightarrow{overline{g}}operatorname{Hom}(M,N^{primeprime})$ be exact. Thus $overline{f}$ is injective and $ker(overline{g})=operatorname{im}(overline{f})$.
We want to show that $0rightarrow N^{prime}xrightarrow{f}Nxrightarrow{g}N^{primeprime}$ is an exact sequence. Let $M=N^{prime}/ker(f)$. If I can prove $N^{prime}cong M$, then $f$ will be one-one. But how to show that? Any hint will be appreciated.
commutative-algebra modules exact-sequence
edited Jan 1 at 12:34
Bernard
119k639112
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
$endgroup$
$begingroup$
Shall I choose $M=A$, the commutative ring?
$endgroup$
– Anupam
Jan 1 at 12:11
$begingroup$
Yes, exactly$ $.
$endgroup$
– Bernard
Jan 1 at 12:38
add a comment |
$begingroup$
Let for all $A$-modules $M$, $0rightarrow operatorname{Hom}(M,N^{prime})xrightarrow{overline{f}}operatorname{Hom}(M,N)xrightarrow{overline{g}}operatorname{Hom}(M,N^{primeprime})$ be exact. Thus $overline{f}$ is injective and $ker(overline{g})=operatorname{im}(overline{f})$.
We want to show that $0rightarrow N^{prime}xrightarrow{f}Nxrightarrow{g}N^{primeprime}$ is an exact sequence. Let $M=N^{prime}/ker(f)$. If I can prove $N^{prime}cong M$, then $f$ will be one-one. But how to show that? Any hint will be appreciated.
commutative-algebra modules exact-sequence
edited Jan 1 at 12:34
Bernard
119k639112
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
$endgroup$
Let for all $A$-modules $M$, $0rightarrow operatorname{Hom}(M,N^{prime})xrightarrow{overline{f}}operatorname{Hom}(M,N)xrightarrow{overline{g}}operatorname{Hom}(M,N^{primeprime})$ be exact. Thus $overline{f}$ is injective and $ker(overline{g})=operatorname{im}(overline{f})$.
We want to show that $0rightarrow N^{prime}xrightarrow{f}Nxrightarrow{g}N^{primeprime}$ is an exact sequence. Let $M=N^{prime}/ker(f)$. If I can prove $N^{prime}cong M$, then $f$ will be one-one. But how to show that? Any hint will be appreciated.
commutative-algebra modules exact-sequence
commutative-algebra modules exact-sequence
edited Jan 1 at 12:34
Bernard
119k639112
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
edited Jan 1 at 12:34
Bernard
119k639112
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
edited Jan 1 at 12:34
Bernard
119k639112
edited Jan 1 at 12:34
Bernard
119k639112
edited Jan 1 at 12:34
Bernard
119k639112
119k639112
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
asked Jan 1 at 12:02
AnupamAnupam
2,3841824
2,3841824
$begingroup$
Shall I choose $M=A$, the commutative ring?
$endgroup$
– Anupam
Jan 1 at 12:11
$begingroup$
Yes, exactly$ $.
$endgroup$
– Bernard
Jan 1 at 12:38
add a comment |
$begingroup$
Shall I choose $M=A$, the commutative ring?
$endgroup$
– Anupam
Jan 1 at 12:11
$begingroup$
Yes, exactly$ $.
$endgroup$
– Bernard
Jan 1 at 12:38
$begingroup$
Shall I choose $M=A$, the commutative ring?
$endgroup$
– Anupam
Jan 1 at 12:11
$begingroup$
Shall I choose $M=A$, the commutative ring?
$endgroup$
– Anupam
Jan 1 at 12:11
$begingroup$
Yes, exactly$ $.
$endgroup$
– Bernard
Jan 1 at 12:38
$begingroup$
Yes, exactly$ $.
$endgroup$
– Bernard
Jan 1 at 12:38
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Take $M=A$. Now, $mbox{Hom}left( A,Nright)simeq N$, $mbox{Hom}left( A,N'right)simeq N'$, $mbox{Hom}left( A,N''right)simeq N''$. Now, by diagram chasing, you get that $0to N'to Nto N''$ is exact.
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $M=A$. Now, $mbox{Hom}left( A,Nright)simeq N$, $mbox{Hom}left( A,N'right)simeq N'$, $mbox{Hom}left( A,N''right)simeq N''$. Now, by diagram chasing, you get that $0to N'to Nto N''$ is exact.
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
Take $M=A$. Now, $mbox{Hom}left( A,Nright)simeq N$, $mbox{Hom}left( A,N'right)simeq N'$, $mbox{Hom}left( A,N''right)simeq N''$. Now, by diagram chasing, you get that $0to N'to Nto N''$ is exact.
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
Take $M=A$. Now, $mbox{Hom}left( A,Nright)simeq N$, $mbox{Hom}left( A,N'right)simeq N'$, $mbox{Hom}left( A,N''right)simeq N''$. Now, by diagram chasing, you get that $0to N'to Nto N''$ is exact.
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
Take $M=A$. Now, $mbox{Hom}left( A,Nright)simeq N$, $mbox{Hom}left( A,N'right)simeq N'$, $mbox{Hom}left( A,N''right)simeq N''$. Now, by diagram chasing, you get that $0to N'to Nto N''$ is exact.
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 15:37
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
825110
add a comment |
add a comment |
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$begingroup$
Shall I choose $M=A$, the commutative ring?
$endgroup$
– Anupam
Jan 1 at 12:11
$begingroup$
Yes, exactly$ $.
$endgroup$
– Bernard
Jan 1 at 12:38