Group cohomology over ring
$begingroup$
For any group $G$ and $G$-module ($mathbb{Z}[G]$-module) $M$, we can define a group cohomology $H^{n}(G, M)$ as
$$
H^{n}(G, M):=mathrm{Ext}_{mathbb{Z}[G]}^{n}(mathbb{Z}, M).
$$
However, I think one can replace $mathbb{Z}$ with other rings $R$, if $M$ is $R$-module and $G$ acts on it (i.e. $M$ is $R[G]$-module). We can define
$$
H^{n}_{R}(G, M):=mathrm{Ext}_{R[G]}^{n}(R, M).
$$
Is there any reference about this cohomology group? Actually, there is an exercise about group cohomology of finite dimensional $mathbb{F}_{p}$-vector space in Dummit-Foote Algebra (Exercise 20, 21 of chapter 17.2). In this case, it seems that we are computing cohomology group when $R=mathbb{F}_{p}$, not $mathbb{Z}$.
Also, is this group is useful for number theory? Thanks in advance.
group-cohomology
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
$endgroup$
add a comment |
$begingroup$
For any group $G$ and $G$-module ($mathbb{Z}[G]$-module) $M$, we can define a group cohomology $H^{n}(G, M)$ as
$$
H^{n}(G, M):=mathrm{Ext}_{mathbb{Z}[G]}^{n}(mathbb{Z}, M).
$$
However, I think one can replace $mathbb{Z}$ with other rings $R$, if $M$ is $R$-module and $G$ acts on it (i.e. $M$ is $R[G]$-module). We can define
$$
H^{n}_{R}(G, M):=mathrm{Ext}_{R[G]}^{n}(R, M).
$$
Is there any reference about this cohomology group? Actually, there is an exercise about group cohomology of finite dimensional $mathbb{F}_{p}$-vector space in Dummit-Foote Algebra (Exercise 20, 21 of chapter 17.2). In this case, it seems that we are computing cohomology group when $R=mathbb{F}_{p}$, not $mathbb{Z}$.
Also, is this group is useful for number theory? Thanks in advance.
group-cohomology
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
$endgroup$
add a comment |
$begingroup$
For any group $G$ and $G$-module ($mathbb{Z}[G]$-module) $M$, we can define a group cohomology $H^{n}(G, M)$ as
$$
H^{n}(G, M):=mathrm{Ext}_{mathbb{Z}[G]}^{n}(mathbb{Z}, M).
$$
However, I think one can replace $mathbb{Z}$ with other rings $R$, if $M$ is $R$-module and $G$ acts on it (i.e. $M$ is $R[G]$-module). We can define
$$
H^{n}_{R}(G, M):=mathrm{Ext}_{R[G]}^{n}(R, M).
$$
Is there any reference about this cohomology group? Actually, there is an exercise about group cohomology of finite dimensional $mathbb{F}_{p}$-vector space in Dummit-Foote Algebra (Exercise 20, 21 of chapter 17.2). In this case, it seems that we are computing cohomology group when $R=mathbb{F}_{p}$, not $mathbb{Z}$.
Also, is this group is useful for number theory? Thanks in advance.
group-cohomology
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
$endgroup$
For any group $G$ and $G$-module ($mathbb{Z}[G]$-module) $M$, we can define a group cohomology $H^{n}(G, M)$ as
$$
H^{n}(G, M):=mathrm{Ext}_{mathbb{Z}[G]}^{n}(mathbb{Z}, M).
$$
However, I think one can replace $mathbb{Z}$ with other rings $R$, if $M$ is $R$-module and $G$ acts on it (i.e. $M$ is $R[G]$-module). We can define
$$
H^{n}_{R}(G, M):=mathrm{Ext}_{R[G]}^{n}(R, M).
$$
Is there any reference about this cohomology group? Actually, there is an exercise about group cohomology of finite dimensional $mathbb{F}_{p}$-vector space in Dummit-Foote Algebra (Exercise 20, 21 of chapter 17.2). In this case, it seems that we are computing cohomology group when $R=mathbb{F}_{p}$, not $mathbb{Z}$.
Also, is this group is useful for number theory? Thanks in advance.
group-cohomology
group-cohomology
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
asked Nov 8 '17 at 5:21
Seewoo LeeSeewoo Lee
6,895927
6,895927
add a comment |
add a comment |
1 Answer
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$begingroup$
Actually, for every (commutative) ring $R$ we have an isomorphism of $mathbb Z$-modules:
$DeclareMathOperator{Ext}{Ext} Ext^n_{R[G]}(R, M) cong Ext^n_{mathbb Z[G]}(mathbb Z, M),$
see https://stacks.math.columbia.edu/tag/0DVD (they discuss there topological groups, but you can always take the discrete topology).
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
$endgroup$
add a comment |
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$begingroup$
Actually, for every (commutative) ring $R$ we have an isomorphism of $mathbb Z$-modules:
$DeclareMathOperator{Ext}{Ext} Ext^n_{R[G]}(R, M) cong Ext^n_{mathbb Z[G]}(mathbb Z, M),$
see https://stacks.math.columbia.edu/tag/0DVD (they discuss there topological groups, but you can always take the discrete topology).
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
$endgroup$
add a comment |
$begingroup$
Actually, for every (commutative) ring $R$ we have an isomorphism of $mathbb Z$-modules:
$DeclareMathOperator{Ext}{Ext} Ext^n_{R[G]}(R, M) cong Ext^n_{mathbb Z[G]}(mathbb Z, M),$
see https://stacks.math.columbia.edu/tag/0DVD (they discuss there topological groups, but you can always take the discrete topology).
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
$endgroup$
add a comment |
$begingroup$
Actually, for every (commutative) ring $R$ we have an isomorphism of $mathbb Z$-modules:
$DeclareMathOperator{Ext}{Ext} Ext^n_{R[G]}(R, M) cong Ext^n_{mathbb Z[G]}(mathbb Z, M),$
see https://stacks.math.columbia.edu/tag/0DVD (they discuss there topological groups, but you can always take the discrete topology).
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
$endgroup$
Actually, for every (commutative) ring $R$ we have an isomorphism of $mathbb Z$-modules:
$DeclareMathOperator{Ext}{Ext} Ext^n_{R[G]}(R, M) cong Ext^n_{mathbb Z[G]}(mathbb Z, M),$
see https://stacks.math.columbia.edu/tag/0DVD (they discuss there topological groups, but you can always take the discrete topology).
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
answered Jan 8 at 13:21
Jędrzej GarnekJędrzej Garnek
416
416
add a comment |
add a comment |
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