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Does Distributivity Imply Power Associativity?
What is this 2D division algebra?Showing that a Boolean algebra is a Boolean ringIf $R = langle n mathbbZ, +, cdotrangle$, is it a ring? Is it commutative, does it have a unity, is it a field?Is multiplication the only operation that satisfies the associative, commutative and distributive law?Axioms of associative algebra?Proving the uniqueness of the additive inverse in a field without the commutative propertyHow to “establish” the basic properties of a commutative ring?Prove that Gl(n, F) is a groupProving Quaternions are a ringA question about rings, true or false statementsWhen does commutativity imply associativity in a set of operators?
$begingroup$
Say we have an algebra $(A, +, cdot)$, where $(A, +)$ is an Abelian Group. All we know about $cdot$ is that it is both left and right distributive over addition. So, $forall a,b,c in A, a cdot (b+c) = (a cdot b) + (a cdot c)$ and $(b+c) cdot a = (b cdot a) + (c cdot a)$.
We can't assume $ cdot $ is associative, commutative, or anything else besides distributive.
Do we know whether multiplication is power associative or not? That is, for all $a$, powers of $a$ are associative (e.g $a cdot (a cdot a) = (a cdot a) cdot a$).
If so, what would the proof look like? If not, is there a counterexample?
I attempted this myself, but I couldn't find any hints of a proof, so I tried to produce a counterexample, similarly with no luck.
abstract-algebra nonassociative-algebras
asked May 1 at 16:12
RothXRothX
690713
$endgroup$
|
show 1 more comment
$begingroup$
Say we have an algebra $(A, +, cdot)$, where $(A, +)$ is an Abelian Group. All we know about $cdot$ is that it is both left and right distributive over addition. So, $forall a,b,c in A, a cdot (b+c) = (a cdot b) + (a cdot c)$ and $(b+c) cdot a = (b cdot a) + (c cdot a)$.
We can't assume $ cdot $ is associative, commutative, or anything else besides distributive.
Do we know whether multiplication is power associative or not? That is, for all $a$, powers of $a$ are associative (e.g $a cdot (a cdot a) = (a cdot a) cdot a$).
If so, what would the proof look like? If not, is there a counterexample?
I attempted this myself, but I couldn't find any hints of a proof, so I tried to produce a counterexample, similarly with no luck.
abstract-algebra nonassociative-algebras
asked May 1 at 16:12
RothXRothX
690713
$endgroup$
$begingroup$
@AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what?
$endgroup$
– RothX
May 1 at 16:15
$begingroup$
he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way
$endgroup$
– gt6989b
May 1 at 16:16
$begingroup$
Would you be able to argue something from definition of multiplication via addition, like $$a cdot (a cdot a) = a cdot (a + a + ldots + a) = acdot a + ldots + acdot a = (a cdot a) cdot a?$$
$endgroup$
– gt6989b
May 1 at 16:18
2
$begingroup$
Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard.
$endgroup$
– coffeemath
May 1 at 16:19
1
$begingroup$
For another example of a distributive algebra that isn't power-associative see this question.
$endgroup$
– pregunton
May 1 at 16:46
|
show 1 more comment
$begingroup$
Say we have an algebra $(A, +, cdot)$, where $(A, +)$ is an Abelian Group. All we know about $cdot$ is that it is both left and right distributive over addition. So, $forall a,b,c in A, a cdot (b+c) = (a cdot b) + (a cdot c)$ and $(b+c) cdot a = (b cdot a) + (c cdot a)$.
We can't assume $ cdot $ is associative, commutative, or anything else besides distributive.
Do we know whether multiplication is power associative or not? That is, for all $a$, powers of $a$ are associative (e.g $a cdot (a cdot a) = (a cdot a) cdot a$).
If so, what would the proof look like? If not, is there a counterexample?
I attempted this myself, but I couldn't find any hints of a proof, so I tried to produce a counterexample, similarly with no luck.
abstract-algebra nonassociative-algebras
asked May 1 at 16:12
RothXRothX
690713
$endgroup$
Say we have an algebra $(A, +, cdot)$, where $(A, +)$ is an Abelian Group. All we know about $cdot$ is that it is both left and right distributive over addition. So, $forall a,b,c in A, a cdot (b+c) = (a cdot b) + (a cdot c)$ and $(b+c) cdot a = (b cdot a) + (c cdot a)$.
We can't assume $ cdot $ is associative, commutative, or anything else besides distributive.
Do we know whether multiplication is power associative or not? That is, for all $a$, powers of $a$ are associative (e.g $a cdot (a cdot a) = (a cdot a) cdot a$).
If so, what would the proof look like? If not, is there a counterexample?
I attempted this myself, but I couldn't find any hints of a proof, so I tried to produce a counterexample, similarly with no luck.
abstract-algebra nonassociative-algebras
abstract-algebra nonassociative-algebras
asked May 1 at 16:12
RothXRothX
690713
asked May 1 at 16:12
RothXRothX
690713
edited May 1 at 19:36
RothX
asked May 1 at 16:12
RothXRothX
690713
asked May 1 at 16:12
RothXRothX
690713
asked May 1 at 16:12
RothXRothX
690713
690713
$begingroup$
@AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what?
$endgroup$
– RothX
May 1 at 16:15
$begingroup$
he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way
$endgroup$
– gt6989b
May 1 at 16:16
$begingroup$
Would you be able to argue something from definition of multiplication via addition, like $$a cdot (a cdot a) = a cdot (a + a + ldots + a) = acdot a + ldots + acdot a = (a cdot a) cdot a?$$
$endgroup$
– gt6989b
May 1 at 16:18
2
$begingroup$
Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard.
$endgroup$
– coffeemath
May 1 at 16:19
1
$begingroup$
For another example of a distributive algebra that isn't power-associative see this question.
$endgroup$
– pregunton
May 1 at 16:46
|
show 1 more comment
$begingroup$
@AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what?
$endgroup$
– RothX
May 1 at 16:15
$begingroup$
he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way
$endgroup$
– gt6989b
May 1 at 16:16
$begingroup$
Would you be able to argue something from definition of multiplication via addition, like $$a cdot (a cdot a) = a cdot (a + a + ldots + a) = acdot a + ldots + acdot a = (a cdot a) cdot a?$$
$endgroup$
– gt6989b
May 1 at 16:18
2
$begingroup$
Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard.
$endgroup$
– coffeemath
May 1 at 16:19
1
$begingroup$
For another example of a distributive algebra that isn't power-associative see this question.
$endgroup$
– pregunton
May 1 at 16:46
$begingroup$
@AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what?
$endgroup$
– RothX
May 1 at 16:15
$begingroup$
@AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what?
$endgroup$
– RothX
May 1 at 16:15
$begingroup$
he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way
$endgroup$
– gt6989b
May 1 at 16:16
$begingroup$
he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way
$endgroup$
– gt6989b
May 1 at 16:16
$begingroup$
Would you be able to argue something from definition of multiplication via addition, like $$a cdot (a cdot a) = a cdot (a + a + ldots + a) = acdot a + ldots + acdot a = (a cdot a) cdot a?$$
$endgroup$
– gt6989b
May 1 at 16:18
$begingroup$
Would you be able to argue something from definition of multiplication via addition, like $$a cdot (a cdot a) = a cdot (a + a + ldots + a) = acdot a + ldots + acdot a = (a cdot a) cdot a?$$
$endgroup$
– gt6989b
May 1 at 16:18
2
2
$begingroup$
Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard.
$endgroup$
– coffeemath
May 1 at 16:19
$begingroup$
Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard.
$endgroup$
– coffeemath
May 1 at 16:19
1
1
$begingroup$
For another example of a distributive algebra that isn't power-associative see this question.
$endgroup$
– pregunton
May 1 at 16:46
$begingroup$
For another example of a distributive algebra that isn't power-associative see this question.
$endgroup$
– pregunton
May 1 at 16:46
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.
Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $x, xx, x(xx), (xx)x, ldots$.
Then, form formal combinations using integers as coefficients $sum_min M z_mm$ which are stipulated to be finitely supported.
Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)xneq x(xx)$.
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
$endgroup$
add a comment |
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$begingroup$
It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.
Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $x, xx, x(xx), (xx)x, ldots$.
Then, form formal combinations using integers as coefficients $sum_min M z_mm$ which are stipulated to be finitely supported.
Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)xneq x(xx)$.
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
$endgroup$
add a comment |
$begingroup$
It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.
Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $x, xx, x(xx), (xx)x, ldots$.
Then, form formal combinations using integers as coefficients $sum_min M z_mm$ which are stipulated to be finitely supported.
Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)xneq x(xx)$.
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
$endgroup$
add a comment |
$begingroup$
It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.
Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $x, xx, x(xx), (xx)x, ldots$.
Then, form formal combinations using integers as coefficients $sum_min M z_mm$ which are stipulated to be finitely supported.
Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)xneq x(xx)$.
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
$endgroup$
It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.
Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $x, xx, x(xx), (xx)x, ldots$.
Then, form formal combinations using integers as coefficients $sum_min M z_mm$ which are stipulated to be finitely supported.
Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)xneq x(xx)$.
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
answered May 1 at 16:25
rschwiebrschwieb
108k12105254
108k12105254
add a comment |
add a comment |
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$begingroup$
@AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what?
$endgroup$
– RothX
May 1 at 16:15
$begingroup$
he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way
$endgroup$
– gt6989b
May 1 at 16:16
$begingroup$
Would you be able to argue something from definition of multiplication via addition, like $$a cdot (a cdot a) = a cdot (a + a + ldots + a) = acdot a + ldots + acdot a = (a cdot a) cdot a?$$
$endgroup$
– gt6989b
May 1 at 16:18
2
$begingroup$
Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard.
$endgroup$
– coffeemath
May 1 at 16:19
1
$begingroup$
For another example of a distributive algebra that isn't power-associative see this question.
$endgroup$
– pregunton
May 1 at 16:46