Mixing
about the cohomology of a tensor module.
$begingroup$
Let $F$ be a field and let $C_*$ be a graded vector space over $F$ such that $C_i = 0$ for $i<0$ and $i>N$ for some integer $N$. Consider $R=F[t]$ as a graded ring (with the usual grading) and $M = C_* otimes_F R$ be a graded $R$-module with also the usual tensor product grading (the total complex)
Suppose that $(M,d)$ is a cochain complex : $d: M_* rightarrow M_{*+1}$ is a $R$-linear map such that $d^2 = 0$ and such that no element of the form $x otimes 1$ belongs to $im(d)$. Then $H^*(M)$ is a free $R$-module if and only if $d = 0$.
The reverse direction is easy: if $d = 0$, then $H^*(M) = M$, as $C_*$ has a $F$-basis it induces an $R$-basis on $M$.
Now I am trying to show that if $d neq 0$, $H^*(M)$ can't be free. As $d$ is a $R$-linear map, we may assume that $d(c otimes 1) neq 0$ for some $c in C_*$. If $d(c otimes 1) = x otimes p$ for $x in C_*$ and $p in R$, then $d(x otimes 1) = 0$ (otherwise $0 neq pcdot d(x otimes 1) = d(x otimes p) = d^2(c otimes 1) = 0$). Therefore, the cohomology class $ [x otimes 1] neq 0$ is a torsion element of $H^*(M)$. For the general case we assume $d(c otimes 1) = c_1 otimes p_1 + x_2 otimes p_2$. I am trying to rule a linear dependence of a chosen basis of $H^*(M)$ using that $d(c otimes 1) neq 0$ but I am stuck here.
homology-cohomology homological-algebra
asked Jan 27 at 4:17
VitoloVitolo
454
$endgroup$
add a comment |
$begingroup$
Let $F$ be a field and let $C_*$ be a graded vector space over $F$ such that $C_i = 0$ for $i<0$ and $i>N$ for some integer $N$. Consider $R=F[t]$ as a graded ring (with the usual grading) and $M = C_* otimes_F R$ be a graded $R$-module with also the usual tensor product grading (the total complex)
Suppose that $(M,d)$ is a cochain complex : $d: M_* rightarrow M_{*+1}$ is a $R$-linear map such that $d^2 = 0$ and such that no element of the form $x otimes 1$ belongs to $im(d)$. Then $H^*(M)$ is a free $R$-module if and only if $d = 0$.
The reverse direction is easy: if $d = 0$, then $H^*(M) = M$, as $C_*$ has a $F$-basis it induces an $R$-basis on $M$.
Now I am trying to show that if $d neq 0$, $H^*(M)$ can't be free. As $d$ is a $R$-linear map, we may assume that $d(c otimes 1) neq 0$ for some $c in C_*$. If $d(c otimes 1) = x otimes p$ for $x in C_*$ and $p in R$, then $d(x otimes 1) = 0$ (otherwise $0 neq pcdot d(x otimes 1) = d(x otimes p) = d^2(c otimes 1) = 0$). Therefore, the cohomology class $ [x otimes 1] neq 0$ is a torsion element of $H^*(M)$. For the general case we assume $d(c otimes 1) = c_1 otimes p_1 + x_2 otimes p_2$. I am trying to rule a linear dependence of a chosen basis of $H^*(M)$ using that $d(c otimes 1) neq 0$ but I am stuck here.
homology-cohomology homological-algebra
asked Jan 27 at 4:17
VitoloVitolo
454
$endgroup$
$begingroup$
You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(cotimes 1)neq 0$, and this maybe helps into getting a contradiction.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 11:30
$begingroup$
It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 14:02
$begingroup$
@PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables.
$endgroup$
– Vitolo
Feb 2 at 16:32
$begingroup$
If you found an answer then please do post it here!
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 17:25
add a comment |
$begingroup$
Let $F$ be a field and let $C_*$ be a graded vector space over $F$ such that $C_i = 0$ for $i<0$ and $i>N$ for some integer $N$. Consider $R=F[t]$ as a graded ring (with the usual grading) and $M = C_* otimes_F R$ be a graded $R$-module with also the usual tensor product grading (the total complex)
Suppose that $(M,d)$ is a cochain complex : $d: M_* rightarrow M_{*+1}$ is a $R$-linear map such that $d^2 = 0$ and such that no element of the form $x otimes 1$ belongs to $im(d)$. Then $H^*(M)$ is a free $R$-module if and only if $d = 0$.
The reverse direction is easy: if $d = 0$, then $H^*(M) = M$, as $C_*$ has a $F$-basis it induces an $R$-basis on $M$.
Now I am trying to show that if $d neq 0$, $H^*(M)$ can't be free. As $d$ is a $R$-linear map, we may assume that $d(c otimes 1) neq 0$ for some $c in C_*$. If $d(c otimes 1) = x otimes p$ for $x in C_*$ and $p in R$, then $d(x otimes 1) = 0$ (otherwise $0 neq pcdot d(x otimes 1) = d(x otimes p) = d^2(c otimes 1) = 0$). Therefore, the cohomology class $ [x otimes 1] neq 0$ is a torsion element of $H^*(M)$. For the general case we assume $d(c otimes 1) = c_1 otimes p_1 + x_2 otimes p_2$. I am trying to rule a linear dependence of a chosen basis of $H^*(M)$ using that $d(c otimes 1) neq 0$ but I am stuck here.
homology-cohomology homological-algebra
asked Jan 27 at 4:17
VitoloVitolo
454
$endgroup$
Let $F$ be a field and let $C_*$ be a graded vector space over $F$ such that $C_i = 0$ for $i<0$ and $i>N$ for some integer $N$. Consider $R=F[t]$ as a graded ring (with the usual grading) and $M = C_* otimes_F R$ be a graded $R$-module with also the usual tensor product grading (the total complex)
Suppose that $(M,d)$ is a cochain complex : $d: M_* rightarrow M_{*+1}$ is a $R$-linear map such that $d^2 = 0$ and such that no element of the form $x otimes 1$ belongs to $im(d)$. Then $H^*(M)$ is a free $R$-module if and only if $d = 0$.
The reverse direction is easy: if $d = 0$, then $H^*(M) = M$, as $C_*$ has a $F$-basis it induces an $R$-basis on $M$.
Now I am trying to show that if $d neq 0$, $H^*(M)$ can't be free. As $d$ is a $R$-linear map, we may assume that $d(c otimes 1) neq 0$ for some $c in C_*$. If $d(c otimes 1) = x otimes p$ for $x in C_*$ and $p in R$, then $d(x otimes 1) = 0$ (otherwise $0 neq pcdot d(x otimes 1) = d(x otimes p) = d^2(c otimes 1) = 0$). Therefore, the cohomology class $ [x otimes 1] neq 0$ is a torsion element of $H^*(M)$. For the general case we assume $d(c otimes 1) = c_1 otimes p_1 + x_2 otimes p_2$. I am trying to rule a linear dependence of a chosen basis of $H^*(M)$ using that $d(c otimes 1) neq 0$ but I am stuck here.
homology-cohomology homological-algebra
homology-cohomology homological-algebra
asked Jan 27 at 4:17
VitoloVitolo
454
asked Jan 27 at 4:17
VitoloVitolo
454
asked Jan 27 at 4:17
VitoloVitolo
454
asked Jan 27 at 4:17
VitoloVitolo
454
asked Jan 27 at 4:17
VitoloVitolo
454
454
$begingroup$
You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(cotimes 1)neq 0$, and this maybe helps into getting a contradiction.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 11:30
$begingroup$
It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 14:02
$begingroup$
@PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables.
$endgroup$
– Vitolo
Feb 2 at 16:32
$begingroup$
If you found an answer then please do post it here!
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 17:25
add a comment |
$begingroup$
You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(cotimes 1)neq 0$, and this maybe helps into getting a contradiction.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 11:30
$begingroup$
It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 14:02
$begingroup$
@PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables.
$endgroup$
– Vitolo
Feb 2 at 16:32
$begingroup$
If you found an answer then please do post it here!
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 17:25
$begingroup$
You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(cotimes 1)neq 0$, and this maybe helps into getting a contradiction.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 11:30
$begingroup$
You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(cotimes 1)neq 0$, and this maybe helps into getting a contradiction.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 11:30
$begingroup$
It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 14:02
$begingroup$
It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 14:02
$begingroup$
@PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables.
$endgroup$
– Vitolo
Feb 2 at 16:32
$begingroup$
@PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables.
$endgroup$
– Vitolo
Feb 2 at 16:32
$begingroup$
If you found an answer then please do post it here!
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 17:25
$begingroup$
If you found an answer then please do post it here!
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 17:25
add a comment |
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$begingroup$
You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(cotimes 1)neq 0$, and this maybe helps into getting a contradiction.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 11:30
$begingroup$
It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false.
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 14:02
$begingroup$
@PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables.
$endgroup$
– Vitolo
Feb 2 at 16:32
$begingroup$
If you found an answer then please do post it here!
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 17:25