Mixing
A natural isomomorphism of $G$-modules
$begingroup$
Let $A$ be a $G$ module, $Bbb Z$ regarded as a trivial $Bbb ZG$ module . Then
$$ Bbb Z otimes _G A rightarrow A/mathcal{G}A, quad motimes a mapsto ma+ mathcal{G}A $$
is an isomorphism. $mathcal{G}$ is the defined as the kernel of the augmentation map,
$$ 0 rightarrow mathcal{G} rightarrow Bbb Z G xrightarrow{epsilon} Bbb Z rightarrow 0 $$
$epsilon (sum m_g g): = sum m_g $
So my strategy to this problem is $-otimes_G A$ the exact sequence. So that $Bbb Z G otimes A/mathcal{G} otimes A cong Bbb Z otimes A$
So I guess the general question is: if $I$ is an ideal of ring $R$, $A$ is an $R$ module $I otimes_R A cong IA$? But this does not seem to be true in general.
abstract-algebra group-theory modules homology-cohomology homological-algebra
asked Jan 18 at 7:58
CL.CL.
2,2402925
$endgroup$
add a comment |
$begingroup$
Let $A$ be a $G$ module, $Bbb Z$ regarded as a trivial $Bbb ZG$ module . Then
$$ Bbb Z otimes _G A rightarrow A/mathcal{G}A, quad motimes a mapsto ma+ mathcal{G}A $$
is an isomorphism. $mathcal{G}$ is the defined as the kernel of the augmentation map,
$$ 0 rightarrow mathcal{G} rightarrow Bbb Z G xrightarrow{epsilon} Bbb Z rightarrow 0 $$
$epsilon (sum m_g g): = sum m_g $
So my strategy to this problem is $-otimes_G A$ the exact sequence. So that $Bbb Z G otimes A/mathcal{G} otimes A cong Bbb Z otimes A$
So I guess the general question is: if $I$ is an ideal of ring $R$, $A$ is an $R$ module $I otimes_R A cong IA$? But this does not seem to be true in general.
abstract-algebra group-theory modules homology-cohomology homological-algebra
asked Jan 18 at 7:58
CL.CL.
2,2402925
$endgroup$
add a comment |
$begingroup$
Let $A$ be a $G$ module, $Bbb Z$ regarded as a trivial $Bbb ZG$ module . Then
$$ Bbb Z otimes _G A rightarrow A/mathcal{G}A, quad motimes a mapsto ma+ mathcal{G}A $$
is an isomorphism. $mathcal{G}$ is the defined as the kernel of the augmentation map,
$$ 0 rightarrow mathcal{G} rightarrow Bbb Z G xrightarrow{epsilon} Bbb Z rightarrow 0 $$
$epsilon (sum m_g g): = sum m_g $
So my strategy to this problem is $-otimes_G A$ the exact sequence. So that $Bbb Z G otimes A/mathcal{G} otimes A cong Bbb Z otimes A$
So I guess the general question is: if $I$ is an ideal of ring $R$, $A$ is an $R$ module $I otimes_R A cong IA$? But this does not seem to be true in general.
abstract-algebra group-theory modules homology-cohomology homological-algebra
asked Jan 18 at 7:58
CL.CL.
2,2402925
$endgroup$
Let $A$ be a $G$ module, $Bbb Z$ regarded as a trivial $Bbb ZG$ module . Then
$$ Bbb Z otimes _G A rightarrow A/mathcal{G}A, quad motimes a mapsto ma+ mathcal{G}A $$
is an isomorphism. $mathcal{G}$ is the defined as the kernel of the augmentation map,
$$ 0 rightarrow mathcal{G} rightarrow Bbb Z G xrightarrow{epsilon} Bbb Z rightarrow 0 $$
$epsilon (sum m_g g): = sum m_g $
So my strategy to this problem is $-otimes_G A$ the exact sequence. So that $Bbb Z G otimes A/mathcal{G} otimes A cong Bbb Z otimes A$
So I guess the general question is: if $I$ is an ideal of ring $R$, $A$ is an $R$ module $I otimes_R A cong IA$? But this does not seem to be true in general.
abstract-algebra group-theory modules homology-cohomology homological-algebra
abstract-algebra group-theory modules homology-cohomology homological-algebra
asked Jan 18 at 7:58
CL.CL.
2,2402925
asked Jan 18 at 7:58
CL.CL.
2,2402925
edited Jan 18 at 8:19
CL.
asked Jan 18 at 7:58
CL.CL.
2,2402925
asked Jan 18 at 7:58
CL.CL.
2,2402925
asked Jan 18 at 7:58
CL.CL.
2,2402925
2,2402925
add a comment |
add a comment |
2 Answers
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$begingroup$
It's not necessary to have $Iotimes Asimeq IA$ for it to work.
Just note that the image of $Iotimes A$ under $iotimes amapsto ia$ is $IA$ - this is easy.
Then you have an exact sequence $mathcal{G}otimes Ato Ato mathbb{Z}otimes Ato 0$ and the image of the first map is $mathcal{G}A$; therefore the cokernel of this first map is $A/mathcal{G}A$
Note that in general for an ideal $I$, you have a short exact sequence $0to Ito Rto R/Ito 0$, so that tensoring with an $R$-module $A$ yields an exact sequence $mathrm{Tor}^R_1(R/I,A) to Iotimes Ato Ato A/IAto 0$; so the kernel of the map $Iotimes Ato IA$ is $mathrm{Tor}^R_1(R/I,A)$ : it may not be trivial.
answered Jan 18 at 9:08
MaxMax
14.7k11143
$endgroup$
add a comment |
$begingroup$
This is not a special property of $G$-modules.
Let $R$ be a ring (not necessarily commutative), $I$ a right ideal of $R$ and $M$ a left $R$-module. Then
$$
R/Iotimes_R Mcong M/IM
$$
as abelian groups.
Just show that $M/IM$ has the required universal property of the tensor product.
answered Jan 18 at 9:01
egregegreg
183k1486205
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
It's not necessary to have $Iotimes Asimeq IA$ for it to work.
Just note that the image of $Iotimes A$ under $iotimes amapsto ia$ is $IA$ - this is easy.
Then you have an exact sequence $mathcal{G}otimes Ato Ato mathbb{Z}otimes Ato 0$ and the image of the first map is $mathcal{G}A$; therefore the cokernel of this first map is $A/mathcal{G}A$
Note that in general for an ideal $I$, you have a short exact sequence $0to Ito Rto R/Ito 0$, so that tensoring with an $R$-module $A$ yields an exact sequence $mathrm{Tor}^R_1(R/I,A) to Iotimes Ato Ato A/IAto 0$; so the kernel of the map $Iotimes Ato IA$ is $mathrm{Tor}^R_1(R/I,A)$ : it may not be trivial.
answered Jan 18 at 9:08
MaxMax
14.7k11143
$endgroup$
add a comment |
$begingroup$
It's not necessary to have $Iotimes Asimeq IA$ for it to work.
Just note that the image of $Iotimes A$ under $iotimes amapsto ia$ is $IA$ - this is easy.
Then you have an exact sequence $mathcal{G}otimes Ato Ato mathbb{Z}otimes Ato 0$ and the image of the first map is $mathcal{G}A$; therefore the cokernel of this first map is $A/mathcal{G}A$
Note that in general for an ideal $I$, you have a short exact sequence $0to Ito Rto R/Ito 0$, so that tensoring with an $R$-module $A$ yields an exact sequence $mathrm{Tor}^R_1(R/I,A) to Iotimes Ato Ato A/IAto 0$; so the kernel of the map $Iotimes Ato IA$ is $mathrm{Tor}^R_1(R/I,A)$ : it may not be trivial.
answered Jan 18 at 9:08
MaxMax
14.7k11143
$endgroup$
add a comment |
$begingroup$
It's not necessary to have $Iotimes Asimeq IA$ for it to work.
Just note that the image of $Iotimes A$ under $iotimes amapsto ia$ is $IA$ - this is easy.
Then you have an exact sequence $mathcal{G}otimes Ato Ato mathbb{Z}otimes Ato 0$ and the image of the first map is $mathcal{G}A$; therefore the cokernel of this first map is $A/mathcal{G}A$
Note that in general for an ideal $I$, you have a short exact sequence $0to Ito Rto R/Ito 0$, so that tensoring with an $R$-module $A$ yields an exact sequence $mathrm{Tor}^R_1(R/I,A) to Iotimes Ato Ato A/IAto 0$; so the kernel of the map $Iotimes Ato IA$ is $mathrm{Tor}^R_1(R/I,A)$ : it may not be trivial.
answered Jan 18 at 9:08
MaxMax
14.7k11143
$endgroup$
It's not necessary to have $Iotimes Asimeq IA$ for it to work.
Just note that the image of $Iotimes A$ under $iotimes amapsto ia$ is $IA$ - this is easy.
Then you have an exact sequence $mathcal{G}otimes Ato Ato mathbb{Z}otimes Ato 0$ and the image of the first map is $mathcal{G}A$; therefore the cokernel of this first map is $A/mathcal{G}A$
Note that in general for an ideal $I$, you have a short exact sequence $0to Ito Rto R/Ito 0$, so that tensoring with an $R$-module $A$ yields an exact sequence $mathrm{Tor}^R_1(R/I,A) to Iotimes Ato Ato A/IAto 0$; so the kernel of the map $Iotimes Ato IA$ is $mathrm{Tor}^R_1(R/I,A)$ : it may not be trivial.
answered Jan 18 at 9:08
MaxMax
14.7k11143
answered Jan 18 at 9:08
MaxMax
14.7k11143
answered Jan 18 at 9:08
MaxMax
14.7k11143
answered Jan 18 at 9:08
MaxMax
14.7k11143
14.7k11143
add a comment |
add a comment |
$begingroup$
This is not a special property of $G$-modules.
Let $R$ be a ring (not necessarily commutative), $I$ a right ideal of $R$ and $M$ a left $R$-module. Then
$$
R/Iotimes_R Mcong M/IM
$$
as abelian groups.
Just show that $M/IM$ has the required universal property of the tensor product.
answered Jan 18 at 9:01
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
This is not a special property of $G$-modules.
Let $R$ be a ring (not necessarily commutative), $I$ a right ideal of $R$ and $M$ a left $R$-module. Then
$$
R/Iotimes_R Mcong M/IM
$$
as abelian groups.
Just show that $M/IM$ has the required universal property of the tensor product.
answered Jan 18 at 9:01
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
This is not a special property of $G$-modules.
Let $R$ be a ring (not necessarily commutative), $I$ a right ideal of $R$ and $M$ a left $R$-module. Then
$$
R/Iotimes_R Mcong M/IM
$$
as abelian groups.
Just show that $M/IM$ has the required universal property of the tensor product.
answered Jan 18 at 9:01
egregegreg
183k1486205
$endgroup$
This is not a special property of $G$-modules.
Let $R$ be a ring (not necessarily commutative), $I$ a right ideal of $R$ and $M$ a left $R$-module. Then
$$
R/Iotimes_R Mcong M/IM
$$
as abelian groups.
Just show that $M/IM$ has the required universal property of the tensor product.
answered Jan 18 at 9:01
egregegreg
183k1486205
answered Jan 18 at 9:01
egregegreg
183k1486205
answered Jan 18 at 9:01
egregegreg
183k1486205
answered Jan 18 at 9:01
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
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