tensoring with flat module factors the kernel
$begingroup$
I want to show that if $F$ is a flat $R$-module, then for any $R$-homomorphism $varphi: M rightarrow N$, we have
$$ker (1_F otimes varphi) cong F otimes ker varphi $$
The $supset$ direction is trivial. For the $subset$ direction, I use the fact that $Fotimes(-)$ preserves the injectivity of the map
$$ M /ker varphi overset{tilde{varphi}}{rightarrow} N $$
Thus if $fotimes m in ker (1_F otimes varphi)$, i.e. $fotimes [m]$ mapped to $0$ under $1_F otimes tilde{varphi}$, we have $fotimes [m] =0$. This suggests me to complete the proof by showing
$$ Fotimes frac{M}{kervarphi} cong frac{Fotimes M}{Fotimes kervarphi} $$
Is this relation true? It seems very plausible for me that $fotimes[m] mapsto [fotimes m]$ gives the isomorphism. But I struggle at the injectivity part of the proof.
abstract-algebra modules homological-algebra flatness
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
$endgroup$
add a comment |
$begingroup$
I want to show that if $F$ is a flat $R$-module, then for any $R$-homomorphism $varphi: M rightarrow N$, we have
$$ker (1_F otimes varphi) cong F otimes ker varphi $$
The $supset$ direction is trivial. For the $subset$ direction, I use the fact that $Fotimes(-)$ preserves the injectivity of the map
$$ M /ker varphi overset{tilde{varphi}}{rightarrow} N $$
Thus if $fotimes m in ker (1_F otimes varphi)$, i.e. $fotimes [m]$ mapped to $0$ under $1_F otimes tilde{varphi}$, we have $fotimes [m] =0$. This suggests me to complete the proof by showing
$$ Fotimes frac{M}{kervarphi} cong frac{Fotimes M}{Fotimes kervarphi} $$
Is this relation true? It seems very plausible for me that $fotimes[m] mapsto [fotimes m]$ gives the isomorphism. But I struggle at the injectivity part of the proof.
abstract-algebra modules homological-algebra flatness
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
$endgroup$
add a comment |
$begingroup$
I want to show that if $F$ is a flat $R$-module, then for any $R$-homomorphism $varphi: M rightarrow N$, we have
$$ker (1_F otimes varphi) cong F otimes ker varphi $$
The $supset$ direction is trivial. For the $subset$ direction, I use the fact that $Fotimes(-)$ preserves the injectivity of the map
$$ M /ker varphi overset{tilde{varphi}}{rightarrow} N $$
Thus if $fotimes m in ker (1_F otimes varphi)$, i.e. $fotimes [m]$ mapped to $0$ under $1_F otimes tilde{varphi}$, we have $fotimes [m] =0$. This suggests me to complete the proof by showing
$$ Fotimes frac{M}{kervarphi} cong frac{Fotimes M}{Fotimes kervarphi} $$
Is this relation true? It seems very plausible for me that $fotimes[m] mapsto [fotimes m]$ gives the isomorphism. But I struggle at the injectivity part of the proof.
abstract-algebra modules homological-algebra flatness
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
$endgroup$
I want to show that if $F$ is a flat $R$-module, then for any $R$-homomorphism $varphi: M rightarrow N$, we have
$$ker (1_F otimes varphi) cong F otimes ker varphi $$
The $supset$ direction is trivial. For the $subset$ direction, I use the fact that $Fotimes(-)$ preserves the injectivity of the map
$$ M /ker varphi overset{tilde{varphi}}{rightarrow} N $$
Thus if $fotimes m in ker (1_F otimes varphi)$, i.e. $fotimes [m]$ mapped to $0$ under $1_F otimes tilde{varphi}$, we have $fotimes [m] =0$. This suggests me to complete the proof by showing
$$ Fotimes frac{M}{kervarphi} cong frac{Fotimes M}{Fotimes kervarphi} $$
Is this relation true? It seems very plausible for me that $fotimes[m] mapsto [fotimes m]$ gives the isomorphism. But I struggle at the injectivity part of the proof.
abstract-algebra modules homological-algebra flatness
abstract-algebra modules homological-algebra flatness
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
asked Dec 25 '18 at 16:54
zudumazicszudumazics
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
Since $F$ is flat, tensoring with it preserves exactness of all exact sequences; from
$$
0toker fto Mxrightarrow{f} N to operatorname{coker}fto 0
$$
you get the exact sequence
$$
0to Fotimesker fto Fotimes Mxrightarrow{1_Fotimes f} Fotimes N to Fotimesoperatorname{coker}fto 0
$$
which is (isomorphic to) the standard kernel-cokernel sequence for $1_Fotimes f$. Hence
$$
ker(1_Fotimes f)cong Fotimesker f
$$
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
$endgroup$
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Since $F$ is flat, tensoring with it preserves exactness of all exact sequences; from
$$
0toker fto Mxrightarrow{f} N to operatorname{coker}fto 0
$$
you get the exact sequence
$$
0to Fotimesker fto Fotimes Mxrightarrow{1_Fotimes f} Fotimes N to Fotimesoperatorname{coker}fto 0
$$
which is (isomorphic to) the standard kernel-cokernel sequence for $1_Fotimes f$. Hence
$$
ker(1_Fotimes f)cong Fotimesker f
$$
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
$endgroup$
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
add a comment |
$begingroup$
Since $F$ is flat, tensoring with it preserves exactness of all exact sequences; from
$$
0toker fto Mxrightarrow{f} N to operatorname{coker}fto 0
$$
you get the exact sequence
$$
0to Fotimesker fto Fotimes Mxrightarrow{1_Fotimes f} Fotimes N to Fotimesoperatorname{coker}fto 0
$$
which is (isomorphic to) the standard kernel-cokernel sequence for $1_Fotimes f$. Hence
$$
ker(1_Fotimes f)cong Fotimesker f
$$
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
$endgroup$
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
add a comment |
$begingroup$
Since $F$ is flat, tensoring with it preserves exactness of all exact sequences; from
$$
0toker fto Mxrightarrow{f} N to operatorname{coker}fto 0
$$
you get the exact sequence
$$
0to Fotimesker fto Fotimes Mxrightarrow{1_Fotimes f} Fotimes N to Fotimesoperatorname{coker}fto 0
$$
which is (isomorphic to) the standard kernel-cokernel sequence for $1_Fotimes f$. Hence
$$
ker(1_Fotimes f)cong Fotimesker f
$$
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
$endgroup$
Since $F$ is flat, tensoring with it preserves exactness of all exact sequences; from
$$
0toker fto Mxrightarrow{f} N to operatorname{coker}fto 0
$$
you get the exact sequence
$$
0to Fotimesker fto Fotimes Mxrightarrow{1_Fotimes f} Fotimes N to Fotimesoperatorname{coker}fto 0
$$
which is (isomorphic to) the standard kernel-cokernel sequence for $1_Fotimes f$. Hence
$$
ker(1_Fotimes f)cong Fotimesker f
$$
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
answered Dec 25 '18 at 17:00
egregegreg
182k1486204
182k1486204
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
add a comment |
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
$endgroup$
– zudumazics
Dec 25 '18 at 17:28
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
$begingroup$
@zudumazics You're basically trying to prove it in the same fashion, considering instead the two exact sequences $0toker fto Mto M/ker fto 0$ and $0to M/ker fto Ntooperatorname{coker}fto0$ (where $operatorname{coker}f=N/operatorname{im}f$).
$endgroup$
– egreg
Dec 25 '18 at 17:33
add a comment |
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