Prove: $N$ is a submodule of $M iff N$ is a kernel of some homomorphism.Torsion submodule equal to kernel of canonical mapHomomorphism and kernel proof help$N$ is the Kernel of some homomorphismImage of homomorphism with kernel $N$ isomophic to $M/N$Proving that particular ideal is the kernel of a homomorphism of polynomial rings$ker(alpha)$ is a direct summand of the pullbackIf $f:Grightarrow H$ is a homomorphism of group and $Ntriangleleft G$…kernel, group, and subgroup…Kernel of induced mapKernel is a submodule
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Prove: $N$ is a submodule of $M iff N$ is a kernel of some homomorphism.
Torsion submodule equal to kernel of canonical mapHomomorphism and kernel proof help$N$ is the Kernel of some homomorphismImage of homomorphism with kernel $N$ isomophic to $M/N$Proving that particular ideal is the kernel of a homomorphism of polynomial rings$ker(alpha)$ is a direct summand of the pullbackIf $f:Grightarrow H$ is a homomorphism of group and $Ntriangleleft G$…kernel, group, and subgroup…Kernel of induced mapKernel is a submodule
$begingroup$
Prove: $N$ is a submodule of $M iff N$ is a kernel of some homomorphism.
$(implies)$ Fix $N<M$, $kappa:M rightarrow M/N. kappa$ is the canonical epimorphism. Then $N=kerkappa.$
How to prove the other direction?
abstract-algebra modules
edited Mar 16 at 13:23
Andrews
1,2762422
asked Mar 16 at 11:55
ToidiToidi
254
$endgroup$
add a comment |
$begingroup$
Prove: $N$ is a submodule of $M iff N$ is a kernel of some homomorphism.
$(implies)$ Fix $N<M$, $kappa:M rightarrow M/N. kappa$ is the canonical epimorphism. Then $N=kerkappa.$
How to prove the other direction?
abstract-algebra modules
edited Mar 16 at 13:23
Andrews
1,2762422
asked Mar 16 at 11:55
ToidiToidi
254
$endgroup$
add a comment |
$begingroup$
Prove: $N$ is a submodule of $M iff N$ is a kernel of some homomorphism.
$(implies)$ Fix $N<M$, $kappa:M rightarrow M/N. kappa$ is the canonical epimorphism. Then $N=kerkappa.$
How to prove the other direction?
abstract-algebra modules
edited Mar 16 at 13:23
Andrews
1,2762422
asked Mar 16 at 11:55
ToidiToidi
254
$endgroup$
Prove: $N$ is a submodule of $M iff N$ is a kernel of some homomorphism.
$(implies)$ Fix $N<M$, $kappa:M rightarrow M/N. kappa$ is the canonical epimorphism. Then $N=kerkappa.$
How to prove the other direction?
abstract-algebra modules
abstract-algebra modules
edited Mar 16 at 13:23
Andrews
1,2762422
asked Mar 16 at 11:55
ToidiToidi
254
edited Mar 16 at 13:23
Andrews
1,2762422
asked Mar 16 at 11:55
ToidiToidi
254
edited Mar 16 at 13:23
Andrews
1,2762422
edited Mar 16 at 13:23
Andrews
1,2762422
edited Mar 16 at 13:23
Andrews
1,2762422
1,2762422
asked Mar 16 at 11:55
ToidiToidi
254
asked Mar 16 at 11:55
ToidiToidi
254
asked Mar 16 at 11:55
ToidiToidi
254
254
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $n,n'inkerkappa$, then $kappa(n+n')=kappa(n)+kappa(n')=0$, and therefore $n+n'inkerkappa$. And if $r$ belongs to the ring that you are working with, then $kappa(rn)=rkappa(n)=0$, and therefore $rninkerkappa$.
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If $n,n'inkerkappa$, then $kappa(n+n')=kappa(n)+kappa(n')=0$, and therefore $n+n'inkerkappa$. And if $r$ belongs to the ring that you are working with, then $kappa(rn)=rkappa(n)=0$, and therefore $rninkerkappa$.
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
If $n,n'inkerkappa$, then $kappa(n+n')=kappa(n)+kappa(n')=0$, and therefore $n+n'inkerkappa$. And if $r$ belongs to the ring that you are working with, then $kappa(rn)=rkappa(n)=0$, and therefore $rninkerkappa$.
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
If $n,n'inkerkappa$, then $kappa(n+n')=kappa(n)+kappa(n')=0$, and therefore $n+n'inkerkappa$. And if $r$ belongs to the ring that you are working with, then $kappa(rn)=rkappa(n)=0$, and therefore $rninkerkappa$.
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
If $n,n'inkerkappa$, then $kappa(n+n')=kappa(n)+kappa(n')=0$, and therefore $n+n'inkerkappa$. And if $r$ belongs to the ring that you are working with, then $kappa(rn)=rkappa(n)=0$, and therefore $rninkerkappa$.
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 16 at 12:00
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
add a comment |
add a comment |
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