tensoring with flat module factors the kernel
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 2 days ago
zudumazics
82
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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 2 days ago
zudumazics
82
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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 2 days ago
zudumazics
82
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
zudumazics
82
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
zudumazics
82
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
zudumazics
82
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
zudumazics
82
asked 2 days ago
zudumazics
82
82
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
zudumazics is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
add a comment |
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1 Answer
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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 2 days ago
egreg
177k1484200
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
add a comment |
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1 Answer
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1 Answer
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votes
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 2 days ago
egreg
177k1484200
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
add a comment |
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 2 days ago
egreg
177k1484200
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
add a comment |
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 2 days ago
egreg
177k1484200
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 2 days ago
egreg
177k1484200
answered 2 days ago
egreg
177k1484200
answered 2 days ago
egreg
177k1484200
answered 2 days ago
egreg
177k1484200
177k1484200
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
add a comment |
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
very neat! thanks! what about the isomorphism of tensor products I tried to prove? Is it true?
– zudumazics
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
@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 $operatorname{coker}f=N/operatorname{im}f$).
– egreg
2 days ago
add a comment |
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