Does $M otimes mathbb{Z} = M$ for any module M?
Does $M otimes mathbb{Z} = M$ for any $R$-module $M$, where we have some $R$-module structure on $mathbb{Z}$?
My thinking: We have a map $phi:M otimes mathbb{Z} rightarrow M$, defined as follows on simple tensors:
$phi(motimes z) = z cdot m$,
which is clearly surjective, and I am fairly sure it is injective, as for a sum $sum_{i} m_i otimes z_i$, we have:
$phi(sum_{i} m_i otimes z_i) = sum_{i}z_i cdot m_i$, which is $0$ only if $sum_{i} m_i otimes z_i = sum_{i} z_i cdot m_i otimes 1_i = 0$
If the first part of the question holds true, then is there a monoid structure on $R$-modules with $mathbb{Z}$ as the identity?
Thanks.
modules tensor-products
asked 2 days ago
DavenDaven
36619
add a comment |
Does $M otimes mathbb{Z} = M$ for any $R$-module $M$, where we have some $R$-module structure on $mathbb{Z}$?
My thinking: We have a map $phi:M otimes mathbb{Z} rightarrow M$, defined as follows on simple tensors:
$phi(motimes z) = z cdot m$,
which is clearly surjective, and I am fairly sure it is injective, as for a sum $sum_{i} m_i otimes z_i$, we have:
$phi(sum_{i} m_i otimes z_i) = sum_{i}z_i cdot m_i$, which is $0$ only if $sum_{i} m_i otimes z_i = sum_{i} z_i cdot m_i otimes 1_i = 0$
If the first part of the question holds true, then is there a monoid structure on $R$-modules with $mathbb{Z}$ as the identity?
Thanks.
modules tensor-products
asked 2 days ago
DavenDaven
36619
add a comment |
Does $M otimes mathbb{Z} = M$ for any $R$-module $M$, where we have some $R$-module structure on $mathbb{Z}$?
My thinking: We have a map $phi:M otimes mathbb{Z} rightarrow M$, defined as follows on simple tensors:
$phi(motimes z) = z cdot m$,
which is clearly surjective, and I am fairly sure it is injective, as for a sum $sum_{i} m_i otimes z_i$, we have:
$phi(sum_{i} m_i otimes z_i) = sum_{i}z_i cdot m_i$, which is $0$ only if $sum_{i} m_i otimes z_i = sum_{i} z_i cdot m_i otimes 1_i = 0$
If the first part of the question holds true, then is there a monoid structure on $R$-modules with $mathbb{Z}$ as the identity?
Thanks.
modules tensor-products
asked 2 days ago
DavenDaven
36619
Does $M otimes mathbb{Z} = M$ for any $R$-module $M$, where we have some $R$-module structure on $mathbb{Z}$?
My thinking: We have a map $phi:M otimes mathbb{Z} rightarrow M$, defined as follows on simple tensors:
$phi(motimes z) = z cdot m$,
which is clearly surjective, and I am fairly sure it is injective, as for a sum $sum_{i} m_i otimes z_i$, we have:
$phi(sum_{i} m_i otimes z_i) = sum_{i}z_i cdot m_i$, which is $0$ only if $sum_{i} m_i otimes z_i = sum_{i} z_i cdot m_i otimes 1_i = 0$
If the first part of the question holds true, then is there a monoid structure on $R$-modules with $mathbb{Z}$ as the identity?
Thanks.
modules tensor-products
modules tensor-products
asked 2 days ago
DavenDaven
36619
asked 2 days ago
DavenDaven
36619
asked 2 days ago
DavenDaven
36619
asked 2 days ago
DavenDaven
36619
asked 2 days ago
DavenDaven
36619
36619
add a comment |
add a comment |
1 Answer
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If you view $M$ as $mathbb{Z}$-Module (i.e. as abelian group) then indeed $M otimes_mathbb{Z} mathbb{Z} = M$. However, if you view $M$ as $R$-Module for some ring $R$ other than $mathbb{Z}$ you would need to specify how $mathbb{Z}$ is a $R$-Module in order to construct $M otimes_R mathbb{Z}$. I'm not sure this is possible for $R neq mathbb{Z}$.
answered 2 days ago
0x5390x539
1,047316
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
add a comment |
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1 Answer
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If you view $M$ as $mathbb{Z}$-Module (i.e. as abelian group) then indeed $M otimes_mathbb{Z} mathbb{Z} = M$. However, if you view $M$ as $R$-Module for some ring $R$ other than $mathbb{Z}$ you would need to specify how $mathbb{Z}$ is a $R$-Module in order to construct $M otimes_R mathbb{Z}$. I'm not sure this is possible for $R neq mathbb{Z}$.
answered 2 days ago
0x5390x539
1,047316
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
add a comment |
If you view $M$ as $mathbb{Z}$-Module (i.e. as abelian group) then indeed $M otimes_mathbb{Z} mathbb{Z} = M$. However, if you view $M$ as $R$-Module for some ring $R$ other than $mathbb{Z}$ you would need to specify how $mathbb{Z}$ is a $R$-Module in order to construct $M otimes_R mathbb{Z}$. I'm not sure this is possible for $R neq mathbb{Z}$.
answered 2 days ago
0x5390x539
1,047316
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
add a comment |
If you view $M$ as $mathbb{Z}$-Module (i.e. as abelian group) then indeed $M otimes_mathbb{Z} mathbb{Z} = M$. However, if you view $M$ as $R$-Module for some ring $R$ other than $mathbb{Z}$ you would need to specify how $mathbb{Z}$ is a $R$-Module in order to construct $M otimes_R mathbb{Z}$. I'm not sure this is possible for $R neq mathbb{Z}$.
answered 2 days ago
0x5390x539
1,047316
If you view $M$ as $mathbb{Z}$-Module (i.e. as abelian group) then indeed $M otimes_mathbb{Z} mathbb{Z} = M$. However, if you view $M$ as $R$-Module for some ring $R$ other than $mathbb{Z}$ you would need to specify how $mathbb{Z}$ is a $R$-Module in order to construct $M otimes_R mathbb{Z}$. I'm not sure this is possible for $R neq mathbb{Z}$.
answered 2 days ago
0x5390x539
1,047316
answered 2 days ago
0x5390x539
1,047316
answered 2 days ago
0x5390x539
1,047316
answered 2 days ago
0x5390x539
1,047316
1,047316
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
add a comment |
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism
– Daven
2 days ago
add a comment |
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