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Is pullback of non-commutative rings well defined?
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I know that pullback is defined for commutative ring, but what about non-commutative case?
Let's consider the following diagram, where $R_i,barR$ are rings and $R$ is the pullback:
Then $1in R$ and $(R,+)$ is an additive group since it'a a pullback of additive groups. Also $(R-0,cdot)$ is closed since if $x^1,x^2in R$, $x^i=(x_1^i,x_2^i)$ with $nu_1(x_1^i)=nu_2(x_2^i)$, then $x^1x^2in R$ too.
This does not means that $R$ is a ring too? Or maybe it's a ring but it's not the pullback of the diagram?
ring-theory noncommutative-algebra pullback
asked Jul 19 at 10:07
user84976
433213
add a comment |Â
up vote
0
down vote
favorite
I know that pullback is defined for commutative ring, but what about non-commutative case?
Let's consider the following diagram, where $R_i,barR$ are rings and $R$ is the pullback:
Then $1in R$ and $(R,+)$ is an additive group since it'a a pullback of additive groups. Also $(R-0,cdot)$ is closed since if $x^1,x^2in R$, $x^i=(x_1^i,x_2^i)$ with $nu_1(x_1^i)=nu_2(x_2^i)$, then $x^1x^2in R$ too.
This does not means that $R$ is a ring too? Or maybe it's a ring but it's not the pullback of the diagram?
ring-theory noncommutative-algebra pullback
asked Jul 19 at 10:07
user84976
433213
5
The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't?
– Mees de Vries
Jul 19 at 10:27
@MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more
– user84976
Jul 19 at 10:32
2
It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures
– Max
Jul 19 at 11:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know that pullback is defined for commutative ring, but what about non-commutative case?
Let's consider the following diagram, where $R_i,barR$ are rings and $R$ is the pullback:
Then $1in R$ and $(R,+)$ is an additive group since it'a a pullback of additive groups. Also $(R-0,cdot)$ is closed since if $x^1,x^2in R$, $x^i=(x_1^i,x_2^i)$ with $nu_1(x_1^i)=nu_2(x_2^i)$, then $x^1x^2in R$ too.
This does not means that $R$ is a ring too? Or maybe it's a ring but it's not the pullback of the diagram?
ring-theory noncommutative-algebra pullback
asked Jul 19 at 10:07
user84976
433213
I know that pullback is defined for commutative ring, but what about non-commutative case?
Let's consider the following diagram, where $R_i,barR$ are rings and $R$ is the pullback:
Then $1in R$ and $(R,+)$ is an additive group since it'a a pullback of additive groups. Also $(R-0,cdot)$ is closed since if $x^1,x^2in R$, $x^i=(x_1^i,x_2^i)$ with $nu_1(x_1^i)=nu_2(x_2^i)$, then $x^1x^2in R$ too.
This does not means that $R$ is a ring too? Or maybe it's a ring but it's not the pullback of the diagram?
ring-theory noncommutative-algebra pullback
asked Jul 19 at 10:07
user84976
433213
edited Jul 19 at 16:50
asked Jul 19 at 10:07
user84976
433213
asked Jul 19 at 10:07
user84976
433213
asked Jul 19 at 10:07
user84976
433213
433213
5
The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't?
– Mees de Vries
Jul 19 at 10:27
@MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more
– user84976
Jul 19 at 10:32
2
It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures
– Max
Jul 19 at 11:56
add a comment |Â
5
The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't?
– Mees de Vries
Jul 19 at 10:27
@MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more
– user84976
Jul 19 at 10:32
2
It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures
– Max
Jul 19 at 11:56
5
5
The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't?
– Mees de Vries
Jul 19 at 10:27
The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't?
– Mees de Vries
Jul 19 at 10:27
@MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more
– user84976
Jul 19 at 10:32
@MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more
– user84976
Jul 19 at 10:32
2
2
It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures
– Max
Jul 19 at 11:56
It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures
– Max
Jul 19 at 11:56
add a comment |Â
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5
The pullback of (possibly) non-commutative rings is indeed perfectly well-defined. What made you think it wasn't?
– Mees de Vries
Jul 19 at 10:27
@MeesdeVries: Every time I find "pullback of rings" just the commutative case is cited. Anyway I could not find a good reference for the commutative case no more
– user84976
Jul 19 at 10:32
2
It's probably because the rererences you read are specifically about commutative algebra. In fact the pullback of two morphisms always exists for algebraic structures
– Max
Jul 19 at 11:56