A module $B$ is flat if Tor $= 0$
$begingroup$
From Weibel's "An Introduction to Homological Algebra":
Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.
It seems to me that the obvious way to do this would be to use the definition:
Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.
We need to show that for an exact sequence:
$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,
$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.
Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.
Any help is appreciated.
homological-algebra exact-sequence
asked Jan 7 at 14:40
DavenDaven
38319
$endgroup$
add a comment |
$begingroup$
From Weibel's "An Introduction to Homological Algebra":
Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.
It seems to me that the obvious way to do this would be to use the definition:
Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.
We need to show that for an exact sequence:
$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,
$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.
Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.
Any help is appreciated.
homological-algebra exact-sequence
asked Jan 7 at 14:40
DavenDaven
38319
$endgroup$
$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53
$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59
1
$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01
$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05
$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08
add a comment |
$begingroup$
From Weibel's "An Introduction to Homological Algebra":
Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.
It seems to me that the obvious way to do this would be to use the definition:
Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.
We need to show that for an exact sequence:
$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,
$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.
Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.
Any help is appreciated.
homological-algebra exact-sequence
asked Jan 7 at 14:40
DavenDaven
38319
$endgroup$
From Weibel's "An Introduction to Homological Algebra":
Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.
It seems to me that the obvious way to do this would be to use the definition:
Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.
We need to show that for an exact sequence:
$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,
$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.
Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.
Any help is appreciated.
homological-algebra exact-sequence
homological-algebra exact-sequence
asked Jan 7 at 14:40
DavenDaven
38319
asked Jan 7 at 14:40
DavenDaven
38319
asked Jan 7 at 14:40
DavenDaven
38319
asked Jan 7 at 14:40
DavenDaven
38319
asked Jan 7 at 14:40
DavenDaven
38319
38319
$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53
$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59
1
$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01
$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05
$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08
add a comment |
$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53
$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59
1
$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01
$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05
$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08
$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53
$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53
$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59
$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59
1
1
$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01
$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01
$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05
$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05
$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08
$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
1
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
add a comment |
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1 Answer
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$begingroup$
We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
1
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
add a comment |
$begingroup$
We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
$endgroup$
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
1
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
add a comment |
$begingroup$
We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
$endgroup$
We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
edited Jan 8 at 8:06
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
answered Jan 7 at 15:48
EnkiduEnkidu
1,36719
1,36719
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
1
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
add a comment |
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
1
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06
1
1
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44
add a comment |
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$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53
$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59
1
$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01
$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
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– Daven
Jan 7 at 15:05
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It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
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– Arthur
Jan 7 at 15:08