Tensor product terminology in category theory?
$begingroup$
Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.
category-theory terminology tensor-products
edited Nov 25 '18 at 6:51
Shaun
9,366113684
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
$endgroup$
add a comment |
$begingroup$
Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.
category-theory terminology tensor-products
edited Nov 25 '18 at 6:51
Shaun
9,366113684
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
$endgroup$
add a comment |
$begingroup$
Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.
category-theory terminology tensor-products
edited Nov 25 '18 at 6:51
Shaun
9,366113684
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
$endgroup$
Say that I have any homomorphisms of commutative rings, $A rightarrow B, A rightarrow C.$ I recently read that $B otimes_A C$ is the pushout of the morphisms in the category of commutative rings. However, I understood tensor products as defined for modules over a commutative ring. Can we somehow realize $B, C$ as modules over $A$ via the homomorphisms or is $B otimes_A C$ as defined by the pushout a generalization of the 'normal' definition of tensors? It seems unlikely that it is a generalization as it seems to depend on these morphisms whereas the tensors of modules doesn't depend on any sort of morphisms.
category-theory terminology tensor-products
category-theory terminology tensor-products
edited Nov 25 '18 at 6:51
Shaun
9,366113684
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
edited Nov 25 '18 at 6:51
Shaun
9,366113684
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
edited Nov 25 '18 at 6:51
Shaun
9,366113684
edited Nov 25 '18 at 6:51
Shaun
9,366113684
edited Nov 25 '18 at 6:51
Shaun
9,366113684
9,366113684
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
asked Nov 25 '18 at 6:47
伽罗瓦伽罗瓦
1,214615
1,214615
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Take all rings here to be commutative. A ring homomorphism $f:Ato B$
makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
and can form the tensor product $Botimes_A C$.
At first $Botimes_A C$ is just a module. But it has a multiplication,
defined as the composition
$$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$
The middle map is just permuting the factors, and the last map is induced
by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
In terms of elements:
$$(botimes c)(b'otimes c')=bb'otimes cc'.$$
Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
is the pushout of $Ato B$ and $Ato C$.
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
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$begingroup$
Take all rings here to be commutative. A ring homomorphism $f:Ato B$
makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
and can form the tensor product $Botimes_A C$.
At first $Botimes_A C$ is just a module. But it has a multiplication,
defined as the composition
$$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$
The middle map is just permuting the factors, and the last map is induced
by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
In terms of elements:
$$(botimes c)(b'otimes c')=bb'otimes cc'.$$
Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
is the pushout of $Ato B$ and $Ato C$.
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
Take all rings here to be commutative. A ring homomorphism $f:Ato B$
makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
and can form the tensor product $Botimes_A C$.
At first $Botimes_A C$ is just a module. But it has a multiplication,
defined as the composition
$$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$
The middle map is just permuting the factors, and the last map is induced
by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
In terms of elements:
$$(botimes c)(b'otimes c')=bb'otimes cc'.$$
Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
is the pushout of $Ato B$ and $Ato C$.
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
Take all rings here to be commutative. A ring homomorphism $f:Ato B$
makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
and can form the tensor product $Botimes_A C$.
At first $Botimes_A C$ is just a module. But it has a multiplication,
defined as the composition
$$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$
The middle map is just permuting the factors, and the last map is induced
by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
In terms of elements:
$$(botimes c)(b'otimes c')=bb'otimes cc'.$$
Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
is the pushout of $Ato B$ and $Ato C$.
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
Take all rings here to be commutative. A ring homomorphism $f:Ato B$
makes $B$ into an $A$-module. In detail, the module action is $acdot b=f(a)b$.
With another ring homomorphism $g:Ato C$ then we have two $A$-modules,
and can form the tensor product $Botimes_A C$.
At first $Botimes_A C$ is just a module. But it has a multiplication,
defined as the composition
$$(Botimes_A C)times(Botimes_A C)to (Botimes_A C)otimes_A (Botimes_A C)
to (Botimes_A B)otimes_A (Cotimes_A C)to Botimes_A C.$$
The middle map is just permuting the factors, and the last map is induced
by $(botimes b')otimes(cotimes c')mapsto bb'otimes cc'$.
In terms of elements:
$$(botimes c)(b'otimes c')=bb'otimes cc'.$$
Then $Botimes_A C$ is a ring. There are ring homomorphisms from $B$
and $C$ to it, the first given by $amapsto aotimes 1_C$. Now one sits
down with a large sheet of paper, and proves that the map $Ato Botimes_A C$
is the pushout of $Ato B$ and $Ato C$.
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Nov 25 '18 at 7:00
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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