Mixing
Different definitions of an algebra over a commutative ring
$begingroup$
Let $R$ be a commutative ring. Here are two definitions of an $R$-algebra:
- An $R$-algebra is a ring $A$ together with a ring homomorphism $f: Rto A$ (Atiyah)
- An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear mapping $cdot: Atimes A to A$ (https://proofwiki.org/wiki/Definition:Algebra_over_Ring)
Are these two equivalent?
I can see that there is an $R$-module structure on $A$ in the first definition; addition is given by addition in $A$ and multiplication by elts of $R$ is given by $(r,a)mapsto f(r)a$ (the product in $A$). But definition 2 says that there is also an $R$-bilinear mapping. What is it in the first definition?
linear-algebra abstract-algebra modules
asked Jan 17 at 0:28
user437309user437309
695313
$endgroup$
|
show 6 more comments
$begingroup$
Let $R$ be a commutative ring. Here are two definitions of an $R$-algebra:
- An $R$-algebra is a ring $A$ together with a ring homomorphism $f: Rto A$ (Atiyah)
- An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear mapping $cdot: Atimes A to A$ (https://proofwiki.org/wiki/Definition:Algebra_over_Ring)
Are these two equivalent?
I can see that there is an $R$-module structure on $A$ in the first definition; addition is given by addition in $A$ and multiplication by elts of $R$ is given by $(r,a)mapsto f(r)a$ (the product in $A$). But definition 2 says that there is also an $R$-bilinear mapping. What is it in the first definition?
linear-algebra abstract-algebra modules
asked Jan 17 at 0:28
user437309user437309
695313
$endgroup$
$begingroup$
Is the Bilinear map associative?
$endgroup$
– Charlie Frohman
Jan 17 at 0:30
$begingroup$
The bilinear mapping in the second is the ring multiplication in the first.
$endgroup$
– user3482749
Jan 17 at 0:31
1
$begingroup$
An extra question: here math.stackexchange.com/questions/349968/… there is the requirement that $f(R)$ is contained in the center of $A$. Is that a different kind of algebra? I believe it doesn't follow automatically from either of the definitions above.
$endgroup$
– user437309
Jan 17 at 0:42
2
$begingroup$
@user348274 If the image of the homomorphism is not in the center, it's not usually called an $R$-algebra. It would instead be an $R$-ring. This came up in my dissertation.
$endgroup$
– Matt Samuel
Jan 17 at 1:40
1
$begingroup$
@user437 And Herstein calls an injective homomorphism that isn't necessarily surjective an isomorphism. The standard, modern definition requires that it be in the center. If you show people a paper where $R$ is not in the center and call it an $R$-algebra, they will be puzzled.
$endgroup$
– Matt Samuel
Jan 17 at 1:57
|
show 6 more comments
$begingroup$
Let $R$ be a commutative ring. Here are two definitions of an $R$-algebra:
- An $R$-algebra is a ring $A$ together with a ring homomorphism $f: Rto A$ (Atiyah)
- An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear mapping $cdot: Atimes A to A$ (https://proofwiki.org/wiki/Definition:Algebra_over_Ring)
Are these two equivalent?
I can see that there is an $R$-module structure on $A$ in the first definition; addition is given by addition in $A$ and multiplication by elts of $R$ is given by $(r,a)mapsto f(r)a$ (the product in $A$). But definition 2 says that there is also an $R$-bilinear mapping. What is it in the first definition?
linear-algebra abstract-algebra modules
asked Jan 17 at 0:28
user437309user437309
695313
$endgroup$
Let $R$ be a commutative ring. Here are two definitions of an $R$-algebra:
- An $R$-algebra is a ring $A$ together with a ring homomorphism $f: Rto A$ (Atiyah)
- An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear mapping $cdot: Atimes A to A$ (https://proofwiki.org/wiki/Definition:Algebra_over_Ring)
Are these two equivalent?
I can see that there is an $R$-module structure on $A$ in the first definition; addition is given by addition in $A$ and multiplication by elts of $R$ is given by $(r,a)mapsto f(r)a$ (the product in $A$). But definition 2 says that there is also an $R$-bilinear mapping. What is it in the first definition?
linear-algebra abstract-algebra modules
linear-algebra abstract-algebra modules
asked Jan 17 at 0:28
user437309user437309
695313
asked Jan 17 at 0:28
user437309user437309
695313
edited Jan 17 at 0:30
user437309
asked Jan 17 at 0:28
user437309user437309
695313
asked Jan 17 at 0:28
user437309user437309
695313
asked Jan 17 at 0:28
user437309user437309
695313
695313
$begingroup$
Is the Bilinear map associative?
$endgroup$
– Charlie Frohman
Jan 17 at 0:30
$begingroup$
The bilinear mapping in the second is the ring multiplication in the first.
$endgroup$
– user3482749
Jan 17 at 0:31
1
$begingroup$
An extra question: here math.stackexchange.com/questions/349968/… there is the requirement that $f(R)$ is contained in the center of $A$. Is that a different kind of algebra? I believe it doesn't follow automatically from either of the definitions above.
$endgroup$
– user437309
Jan 17 at 0:42
2
$begingroup$
@user348274 If the image of the homomorphism is not in the center, it's not usually called an $R$-algebra. It would instead be an $R$-ring. This came up in my dissertation.
$endgroup$
– Matt Samuel
Jan 17 at 1:40
1
$begingroup$
@user437 And Herstein calls an injective homomorphism that isn't necessarily surjective an isomorphism. The standard, modern definition requires that it be in the center. If you show people a paper where $R$ is not in the center and call it an $R$-algebra, they will be puzzled.
$endgroup$
– Matt Samuel
Jan 17 at 1:57
|
show 6 more comments
$begingroup$
Is the Bilinear map associative?
$endgroup$
– Charlie Frohman
Jan 17 at 0:30
$begingroup$
The bilinear mapping in the second is the ring multiplication in the first.
$endgroup$
– user3482749
Jan 17 at 0:31
1
$begingroup$
An extra question: here math.stackexchange.com/questions/349968/… there is the requirement that $f(R)$ is contained in the center of $A$. Is that a different kind of algebra? I believe it doesn't follow automatically from either of the definitions above.
$endgroup$
– user437309
Jan 17 at 0:42
2
$begingroup$
@user348274 If the image of the homomorphism is not in the center, it's not usually called an $R$-algebra. It would instead be an $R$-ring. This came up in my dissertation.
$endgroup$
– Matt Samuel
Jan 17 at 1:40
1
$begingroup$
@user437 And Herstein calls an injective homomorphism that isn't necessarily surjective an isomorphism. The standard, modern definition requires that it be in the center. If you show people a paper where $R$ is not in the center and call it an $R$-algebra, they will be puzzled.
$endgroup$
– Matt Samuel
Jan 17 at 1:57
$begingroup$
Is the Bilinear map associative?
$endgroup$
– Charlie Frohman
Jan 17 at 0:30
$begingroup$
Is the Bilinear map associative?
$endgroup$
– Charlie Frohman
Jan 17 at 0:30
$begingroup$
The bilinear mapping in the second is the ring multiplication in the first.
$endgroup$
– user3482749
Jan 17 at 0:31
$begingroup$
The bilinear mapping in the second is the ring multiplication in the first.
$endgroup$
– user3482749
Jan 17 at 0:31
1
1
$begingroup$
An extra question: here math.stackexchange.com/questions/349968/… there is the requirement that $f(R)$ is contained in the center of $A$. Is that a different kind of algebra? I believe it doesn't follow automatically from either of the definitions above.
$endgroup$
– user437309
Jan 17 at 0:42
$begingroup$
An extra question: here math.stackexchange.com/questions/349968/… there is the requirement that $f(R)$ is contained in the center of $A$. Is that a different kind of algebra? I believe it doesn't follow automatically from either of the definitions above.
$endgroup$
– user437309
Jan 17 at 0:42
2
2
$begingroup$
@user348274 If the image of the homomorphism is not in the center, it's not usually called an $R$-algebra. It would instead be an $R$-ring. This came up in my dissertation.
$endgroup$
– Matt Samuel
Jan 17 at 1:40
$begingroup$
@user348274 If the image of the homomorphism is not in the center, it's not usually called an $R$-algebra. It would instead be an $R$-ring. This came up in my dissertation.
$endgroup$
– Matt Samuel
Jan 17 at 1:40
1
1
$begingroup$
@user437 And Herstein calls an injective homomorphism that isn't necessarily surjective an isomorphism. The standard, modern definition requires that it be in the center. If you show people a paper where $R$ is not in the center and call it an $R$-algebra, they will be puzzled.
$endgroup$
– Matt Samuel
Jan 17 at 1:57
$begingroup$
@user437 And Herstein calls an injective homomorphism that isn't necessarily surjective an isomorphism. The standard, modern definition requires that it be in the center. If you show people a paper where $R$ is not in the center and call it an $R$-algebra, they will be puzzled.
$endgroup$
– Matt Samuel
Jan 17 at 1:57
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
No, they are not. To back up a bit, there are two different (NOT equivalent) standard definitions of an algebra over a commutative ring $R$. (Here rings always have unit; if you allow non-unital rings there are some modifications.)
Definition 1: An $R$-algebra is a ring $A$ together with a homomorphism $f:Rto A$ such that the image of $f$ is contained in the center of $A$.
Definition 2: An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear map $mu:Atimes Ato A$.
How are these definitions related? Well, if you require $mu$ in Definition 2 to be associative and have a unit, then it makes $A$ a ring. Moreover, there is then a ring-homomorphism $f:Rto A$, given by $f(r)=rcdot 1$ (here the multiplication is scalar multiplication and $1$ is the unit of $mu$), and the image of $f$ is contained in the center of $A$. (The proof that $f$ is a ring-homomorphism and that its image is contained in the center uses the fact that $mu$ is $R$-bilinear.)
So, a unital associative $R$-algebra by Definition 2 is also an $R$-algebra by Definition 1. Conversely, if $f:Rto A$ makes $A$ an $R$-algebra by Definition 1, then $A$ is an $R$-module by defining $rcdot a=f(r)a$ (the multiplication on the right being the ring structure of $R$. Moreover, the multiplication map of $A$ is $R$-bilinear (the proof of this uses the fact that the image of $f$ is contained in the center of $A$).
It is easy to see that these two constructions are inverse, so an $R$-algebra in the sense of Definition 1 is the same as a unital associative $R$-algebra in the sense of Definition 2. In contexts where Definition 2 is used, an $R$-algebra by Definition 1 is typically referred to as a (unital) associative $R$-algebra. In contexts where Definition 1 is used, an $R$-algebra by Definition 2 is typically referred to as a nonassociative $R$-algebra.
As a final note, the first definition you quoted is like Definition 1, but without the requirement that the image of $f$ is contained in the center of $A$. This is not a standard definition of an $R$-algebra. However, I suspect that the context in which you encountered the definition was one where all rings are assumed to be commutative (as is typical in commutative algebra), and so the image of $f$ is automatically contained in the center of $A$. Note that in commutative algebra, it is common for "$R$-algebra" to mean "commutative $R$-algebra according to Definition 1", just like how "ring" is often taken to mean "commutative ring".
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
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$begingroup$
No, they are not. To back up a bit, there are two different (NOT equivalent) standard definitions of an algebra over a commutative ring $R$. (Here rings always have unit; if you allow non-unital rings there are some modifications.)
Definition 1: An $R$-algebra is a ring $A$ together with a homomorphism $f:Rto A$ such that the image of $f$ is contained in the center of $A$.
Definition 2: An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear map $mu:Atimes Ato A$.
How are these definitions related? Well, if you require $mu$ in Definition 2 to be associative and have a unit, then it makes $A$ a ring. Moreover, there is then a ring-homomorphism $f:Rto A$, given by $f(r)=rcdot 1$ (here the multiplication is scalar multiplication and $1$ is the unit of $mu$), and the image of $f$ is contained in the center of $A$. (The proof that $f$ is a ring-homomorphism and that its image is contained in the center uses the fact that $mu$ is $R$-bilinear.)
So, a unital associative $R$-algebra by Definition 2 is also an $R$-algebra by Definition 1. Conversely, if $f:Rto A$ makes $A$ an $R$-algebra by Definition 1, then $A$ is an $R$-module by defining $rcdot a=f(r)a$ (the multiplication on the right being the ring structure of $R$. Moreover, the multiplication map of $A$ is $R$-bilinear (the proof of this uses the fact that the image of $f$ is contained in the center of $A$).
It is easy to see that these two constructions are inverse, so an $R$-algebra in the sense of Definition 1 is the same as a unital associative $R$-algebra in the sense of Definition 2. In contexts where Definition 2 is used, an $R$-algebra by Definition 1 is typically referred to as a (unital) associative $R$-algebra. In contexts where Definition 1 is used, an $R$-algebra by Definition 2 is typically referred to as a nonassociative $R$-algebra.
As a final note, the first definition you quoted is like Definition 1, but without the requirement that the image of $f$ is contained in the center of $A$. This is not a standard definition of an $R$-algebra. However, I suspect that the context in which you encountered the definition was one where all rings are assumed to be commutative (as is typical in commutative algebra), and so the image of $f$ is automatically contained in the center of $A$. Note that in commutative algebra, it is common for "$R$-algebra" to mean "commutative $R$-algebra according to Definition 1", just like how "ring" is often taken to mean "commutative ring".
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
No, they are not. To back up a bit, there are two different (NOT equivalent) standard definitions of an algebra over a commutative ring $R$. (Here rings always have unit; if you allow non-unital rings there are some modifications.)
Definition 1: An $R$-algebra is a ring $A$ together with a homomorphism $f:Rto A$ such that the image of $f$ is contained in the center of $A$.
Definition 2: An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear map $mu:Atimes Ato A$.
How are these definitions related? Well, if you require $mu$ in Definition 2 to be associative and have a unit, then it makes $A$ a ring. Moreover, there is then a ring-homomorphism $f:Rto A$, given by $f(r)=rcdot 1$ (here the multiplication is scalar multiplication and $1$ is the unit of $mu$), and the image of $f$ is contained in the center of $A$. (The proof that $f$ is a ring-homomorphism and that its image is contained in the center uses the fact that $mu$ is $R$-bilinear.)
So, a unital associative $R$-algebra by Definition 2 is also an $R$-algebra by Definition 1. Conversely, if $f:Rto A$ makes $A$ an $R$-algebra by Definition 1, then $A$ is an $R$-module by defining $rcdot a=f(r)a$ (the multiplication on the right being the ring structure of $R$. Moreover, the multiplication map of $A$ is $R$-bilinear (the proof of this uses the fact that the image of $f$ is contained in the center of $A$).
It is easy to see that these two constructions are inverse, so an $R$-algebra in the sense of Definition 1 is the same as a unital associative $R$-algebra in the sense of Definition 2. In contexts where Definition 2 is used, an $R$-algebra by Definition 1 is typically referred to as a (unital) associative $R$-algebra. In contexts where Definition 1 is used, an $R$-algebra by Definition 2 is typically referred to as a nonassociative $R$-algebra.
As a final note, the first definition you quoted is like Definition 1, but without the requirement that the image of $f$ is contained in the center of $A$. This is not a standard definition of an $R$-algebra. However, I suspect that the context in which you encountered the definition was one where all rings are assumed to be commutative (as is typical in commutative algebra), and so the image of $f$ is automatically contained in the center of $A$. Note that in commutative algebra, it is common for "$R$-algebra" to mean "commutative $R$-algebra according to Definition 1", just like how "ring" is often taken to mean "commutative ring".
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
No, they are not. To back up a bit, there are two different (NOT equivalent) standard definitions of an algebra over a commutative ring $R$. (Here rings always have unit; if you allow non-unital rings there are some modifications.)
Definition 1: An $R$-algebra is a ring $A$ together with a homomorphism $f:Rto A$ such that the image of $f$ is contained in the center of $A$.
Definition 2: An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear map $mu:Atimes Ato A$.
How are these definitions related? Well, if you require $mu$ in Definition 2 to be associative and have a unit, then it makes $A$ a ring. Moreover, there is then a ring-homomorphism $f:Rto A$, given by $f(r)=rcdot 1$ (here the multiplication is scalar multiplication and $1$ is the unit of $mu$), and the image of $f$ is contained in the center of $A$. (The proof that $f$ is a ring-homomorphism and that its image is contained in the center uses the fact that $mu$ is $R$-bilinear.)
So, a unital associative $R$-algebra by Definition 2 is also an $R$-algebra by Definition 1. Conversely, if $f:Rto A$ makes $A$ an $R$-algebra by Definition 1, then $A$ is an $R$-module by defining $rcdot a=f(r)a$ (the multiplication on the right being the ring structure of $R$. Moreover, the multiplication map of $A$ is $R$-bilinear (the proof of this uses the fact that the image of $f$ is contained in the center of $A$).
It is easy to see that these two constructions are inverse, so an $R$-algebra in the sense of Definition 1 is the same as a unital associative $R$-algebra in the sense of Definition 2. In contexts where Definition 2 is used, an $R$-algebra by Definition 1 is typically referred to as a (unital) associative $R$-algebra. In contexts where Definition 1 is used, an $R$-algebra by Definition 2 is typically referred to as a nonassociative $R$-algebra.
As a final note, the first definition you quoted is like Definition 1, but without the requirement that the image of $f$ is contained in the center of $A$. This is not a standard definition of an $R$-algebra. However, I suspect that the context in which you encountered the definition was one where all rings are assumed to be commutative (as is typical in commutative algebra), and so the image of $f$ is automatically contained in the center of $A$. Note that in commutative algebra, it is common for "$R$-algebra" to mean "commutative $R$-algebra according to Definition 1", just like how "ring" is often taken to mean "commutative ring".
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
$endgroup$
No, they are not. To back up a bit, there are two different (NOT equivalent) standard definitions of an algebra over a commutative ring $R$. (Here rings always have unit; if you allow non-unital rings there are some modifications.)
Definition 1: An $R$-algebra is a ring $A$ together with a homomorphism $f:Rto A$ such that the image of $f$ is contained in the center of $A$.
Definition 2: An $R$-algebra is an $R$-module $A$ together with an $R$-bilinear map $mu:Atimes Ato A$.
How are these definitions related? Well, if you require $mu$ in Definition 2 to be associative and have a unit, then it makes $A$ a ring. Moreover, there is then a ring-homomorphism $f:Rto A$, given by $f(r)=rcdot 1$ (here the multiplication is scalar multiplication and $1$ is the unit of $mu$), and the image of $f$ is contained in the center of $A$. (The proof that $f$ is a ring-homomorphism and that its image is contained in the center uses the fact that $mu$ is $R$-bilinear.)
So, a unital associative $R$-algebra by Definition 2 is also an $R$-algebra by Definition 1. Conversely, if $f:Rto A$ makes $A$ an $R$-algebra by Definition 1, then $A$ is an $R$-module by defining $rcdot a=f(r)a$ (the multiplication on the right being the ring structure of $R$. Moreover, the multiplication map of $A$ is $R$-bilinear (the proof of this uses the fact that the image of $f$ is contained in the center of $A$).
It is easy to see that these two constructions are inverse, so an $R$-algebra in the sense of Definition 1 is the same as a unital associative $R$-algebra in the sense of Definition 2. In contexts where Definition 2 is used, an $R$-algebra by Definition 1 is typically referred to as a (unital) associative $R$-algebra. In contexts where Definition 1 is used, an $R$-algebra by Definition 2 is typically referred to as a nonassociative $R$-algebra.
As a final note, the first definition you quoted is like Definition 1, but without the requirement that the image of $f$ is contained in the center of $A$. This is not a standard definition of an $R$-algebra. However, I suspect that the context in which you encountered the definition was one where all rings are assumed to be commutative (as is typical in commutative algebra), and so the image of $f$ is automatically contained in the center of $A$. Note that in commutative algebra, it is common for "$R$-algebra" to mean "commutative $R$-algebra according to Definition 1", just like how "ring" is often taken to mean "commutative ring".
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 3:43
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
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$begingroup$
Is the Bilinear map associative?
$endgroup$
– Charlie Frohman
Jan 17 at 0:30
$begingroup$
The bilinear mapping in the second is the ring multiplication in the first.
$endgroup$
– user3482749
Jan 17 at 0:31
1
$begingroup$
An extra question: here math.stackexchange.com/questions/349968/… there is the requirement that $f(R)$ is contained in the center of $A$. Is that a different kind of algebra? I believe it doesn't follow automatically from either of the definitions above.
$endgroup$
– user437309
Jan 17 at 0:42
2
$begingroup$
@user348274 If the image of the homomorphism is not in the center, it's not usually called an $R$-algebra. It would instead be an $R$-ring. This came up in my dissertation.
$endgroup$
– Matt Samuel
Jan 17 at 1:40
1
$begingroup$
@user437 And Herstein calls an injective homomorphism that isn't necessarily surjective an isomorphism. The standard, modern definition requires that it be in the center. If you show people a paper where $R$ is not in the center and call it an $R$-algebra, they will be puzzled.
$endgroup$
– Matt Samuel
Jan 17 at 1:57