Do I need commutativity of $A$ to say that $Der_mathbb{K}(A)$ is a left $A$-module?
$begingroup$
Let $A$ be an associative $mathbb{K}$-algebra, with $mathbb{K}$ a field.
A derivation of $A$ is a $mathbb{K}$- linear map $D:Ato A$ such that for each $a,bin A$ we have $D(ab)=D(a)b+a D(b)$ $quad[1]$
For each $ain A$ and for each derivation $D$ define $aD:Ato A$ with $(aD)(b):=a cdot D(b)$.
I want to show that $aD$ is still a derivation of $A$.
In particular let's prove the property $[1]$.
Let $b,cin A$, we have $(aD)(bc)=acdot D(bc)=a (D(b)c+bD(c))=a cdot (D(b)cdot c)+abD(c)=$ (associativity) $=(aD)(b) cdot c + abD(c)$.
Do I nedd commutativity of the multiplication in $A$ to continue? I would like to say that $abD(c)=b (aD)(c)$ but without commutativity I can't "exchange" the position of $a$ and $b$ rigth?
abstract-algebra differential-geometry algebraic-geometry
asked Jan 7 at 11:34
MinatoMinato
600314
$endgroup$
add a comment |
$begingroup$
Let $A$ be an associative $mathbb{K}$-algebra, with $mathbb{K}$ a field.
A derivation of $A$ is a $mathbb{K}$- linear map $D:Ato A$ such that for each $a,bin A$ we have $D(ab)=D(a)b+a D(b)$ $quad[1]$
For each $ain A$ and for each derivation $D$ define $aD:Ato A$ with $(aD)(b):=a cdot D(b)$.
I want to show that $aD$ is still a derivation of $A$.
In particular let's prove the property $[1]$.
Let $b,cin A$, we have $(aD)(bc)=acdot D(bc)=a (D(b)c+bD(c))=a cdot (D(b)cdot c)+abD(c)=$ (associativity) $=(aD)(b) cdot c + abD(c)$.
Do I nedd commutativity of the multiplication in $A$ to continue? I would like to say that $abD(c)=b (aD)(c)$ but without commutativity I can't "exchange" the position of $a$ and $b$ rigth?
abstract-algebra differential-geometry algebraic-geometry
asked Jan 7 at 11:34
MinatoMinato
600314
$endgroup$
$begingroup$
You appear to be correct. Are you asked to prove this without commutativity, or is this your own initiative?
$endgroup$
– Matt Samuel
Jan 7 at 11:59
$begingroup$
I were asked to prove this without commutativity, maybe a typo in my notes
$endgroup$
– Minato
Jan 8 at 16:41
add a comment |
$begingroup$
Let $A$ be an associative $mathbb{K}$-algebra, with $mathbb{K}$ a field.
A derivation of $A$ is a $mathbb{K}$- linear map $D:Ato A$ such that for each $a,bin A$ we have $D(ab)=D(a)b+a D(b)$ $quad[1]$
For each $ain A$ and for each derivation $D$ define $aD:Ato A$ with $(aD)(b):=a cdot D(b)$.
I want to show that $aD$ is still a derivation of $A$.
In particular let's prove the property $[1]$.
Let $b,cin A$, we have $(aD)(bc)=acdot D(bc)=a (D(b)c+bD(c))=a cdot (D(b)cdot c)+abD(c)=$ (associativity) $=(aD)(b) cdot c + abD(c)$.
Do I nedd commutativity of the multiplication in $A$ to continue? I would like to say that $abD(c)=b (aD)(c)$ but without commutativity I can't "exchange" the position of $a$ and $b$ rigth?
abstract-algebra differential-geometry algebraic-geometry
asked Jan 7 at 11:34
MinatoMinato
600314
$endgroup$
Let $A$ be an associative $mathbb{K}$-algebra, with $mathbb{K}$ a field.
A derivation of $A$ is a $mathbb{K}$- linear map $D:Ato A$ such that for each $a,bin A$ we have $D(ab)=D(a)b+a D(b)$ $quad[1]$
For each $ain A$ and for each derivation $D$ define $aD:Ato A$ with $(aD)(b):=a cdot D(b)$.
I want to show that $aD$ is still a derivation of $A$.
In particular let's prove the property $[1]$.
Let $b,cin A$, we have $(aD)(bc)=acdot D(bc)=a (D(b)c+bD(c))=a cdot (D(b)cdot c)+abD(c)=$ (associativity) $=(aD)(b) cdot c + abD(c)$.
Do I nedd commutativity of the multiplication in $A$ to continue? I would like to say that $abD(c)=b (aD)(c)$ but without commutativity I can't "exchange" the position of $a$ and $b$ rigth?
abstract-algebra differential-geometry algebraic-geometry
abstract-algebra differential-geometry algebraic-geometry
asked Jan 7 at 11:34
MinatoMinato
600314
asked Jan 7 at 11:34
MinatoMinato
600314
asked Jan 7 at 11:34
MinatoMinato
600314
asked Jan 7 at 11:34
MinatoMinato
600314
asked Jan 7 at 11:34
MinatoMinato
600314
600314
$begingroup$
You appear to be correct. Are you asked to prove this without commutativity, or is this your own initiative?
$endgroup$
– Matt Samuel
Jan 7 at 11:59
$begingroup$
I were asked to prove this without commutativity, maybe a typo in my notes
$endgroup$
– Minato
Jan 8 at 16:41
add a comment |
$begingroup$
You appear to be correct. Are you asked to prove this without commutativity, or is this your own initiative?
$endgroup$
– Matt Samuel
Jan 7 at 11:59
$begingroup$
I were asked to prove this without commutativity, maybe a typo in my notes
$endgroup$
– Minato
Jan 8 at 16:41
$begingroup$
You appear to be correct. Are you asked to prove this without commutativity, or is this your own initiative?
$endgroup$
– Matt Samuel
Jan 7 at 11:59
$begingroup$
You appear to be correct. Are you asked to prove this without commutativity, or is this your own initiative?
$endgroup$
– Matt Samuel
Jan 7 at 11:59
$begingroup$
I were asked to prove this without commutativity, maybe a typo in my notes
$endgroup$
– Minato
Jan 8 at 16:41
$begingroup$
I were asked to prove this without commutativity, maybe a typo in my notes
$endgroup$
– Minato
Jan 8 at 16:41
add a comment |
1 Answer
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You are correct that $Der(A)$ will not be an $A$ module, for example take $A = klangle a,b,crangle$ the free algebra and the derivation determined by $$D(a) = a, D(b) = b, D(c) = 1.$$ Then $abD(c) neq baD(c)$ because $ab neq ba$ in the free algebra.
In general the natural structure on $Der(A)$ is a $Z(A)$-Lie algebra and in particular a bimodule over $Z(A)$.
More generally you can consider derivations in a bimodule $M$, and ask when $Der(A,M)$ is an $A$ module. The same problem arises that $abD(c) neq baD(c)$. One condition that suffices to make $Der(A,M)$ an $A$-module is that $M$ is a central bimodule. This means that the left and right actions are the same, i.e. for all $a in A, m in M$ then $a.m = m.a$. This is equivalent to saying that $[a,b].m = 0$ for all $a,bin A$. The data of a central bimodule $M$ is the same as that of a module over the abelianization of $A$, the commutative ring $A_{ab} = A/F^1A$ where $F^1A = A[A,A]A$ is the two-sided ideal generated by commutators of elements of $A$.
(There is another notion of central bimodule which says instead only that $z.m = m.z$ for $z in Z(A)$. If $M$ is this kind of central bimodule, then $Der(A,M)$ is a central $Z(A)$-module.)
answered Jan 7 at 14:16
BenBen
4,373617
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$begingroup$
You are correct that $Der(A)$ will not be an $A$ module, for example take $A = klangle a,b,crangle$ the free algebra and the derivation determined by $$D(a) = a, D(b) = b, D(c) = 1.$$ Then $abD(c) neq baD(c)$ because $ab neq ba$ in the free algebra.
In general the natural structure on $Der(A)$ is a $Z(A)$-Lie algebra and in particular a bimodule over $Z(A)$.
More generally you can consider derivations in a bimodule $M$, and ask when $Der(A,M)$ is an $A$ module. The same problem arises that $abD(c) neq baD(c)$. One condition that suffices to make $Der(A,M)$ an $A$-module is that $M$ is a central bimodule. This means that the left and right actions are the same, i.e. for all $a in A, m in M$ then $a.m = m.a$. This is equivalent to saying that $[a,b].m = 0$ for all $a,bin A$. The data of a central bimodule $M$ is the same as that of a module over the abelianization of $A$, the commutative ring $A_{ab} = A/F^1A$ where $F^1A = A[A,A]A$ is the two-sided ideal generated by commutators of elements of $A$.
(There is another notion of central bimodule which says instead only that $z.m = m.z$ for $z in Z(A)$. If $M$ is this kind of central bimodule, then $Der(A,M)$ is a central $Z(A)$-module.)
answered Jan 7 at 14:16
BenBen
4,373617
$endgroup$
add a comment |
$begingroup$
You are correct that $Der(A)$ will not be an $A$ module, for example take $A = klangle a,b,crangle$ the free algebra and the derivation determined by $$D(a) = a, D(b) = b, D(c) = 1.$$ Then $abD(c) neq baD(c)$ because $ab neq ba$ in the free algebra.
In general the natural structure on $Der(A)$ is a $Z(A)$-Lie algebra and in particular a bimodule over $Z(A)$.
More generally you can consider derivations in a bimodule $M$, and ask when $Der(A,M)$ is an $A$ module. The same problem arises that $abD(c) neq baD(c)$. One condition that suffices to make $Der(A,M)$ an $A$-module is that $M$ is a central bimodule. This means that the left and right actions are the same, i.e. for all $a in A, m in M$ then $a.m = m.a$. This is equivalent to saying that $[a,b].m = 0$ for all $a,bin A$. The data of a central bimodule $M$ is the same as that of a module over the abelianization of $A$, the commutative ring $A_{ab} = A/F^1A$ where $F^1A = A[A,A]A$ is the two-sided ideal generated by commutators of elements of $A$.
(There is another notion of central bimodule which says instead only that $z.m = m.z$ for $z in Z(A)$. If $M$ is this kind of central bimodule, then $Der(A,M)$ is a central $Z(A)$-module.)
answered Jan 7 at 14:16
BenBen
4,373617
$endgroup$
add a comment |
$begingroup$
You are correct that $Der(A)$ will not be an $A$ module, for example take $A = klangle a,b,crangle$ the free algebra and the derivation determined by $$D(a) = a, D(b) = b, D(c) = 1.$$ Then $abD(c) neq baD(c)$ because $ab neq ba$ in the free algebra.
In general the natural structure on $Der(A)$ is a $Z(A)$-Lie algebra and in particular a bimodule over $Z(A)$.
More generally you can consider derivations in a bimodule $M$, and ask when $Der(A,M)$ is an $A$ module. The same problem arises that $abD(c) neq baD(c)$. One condition that suffices to make $Der(A,M)$ an $A$-module is that $M$ is a central bimodule. This means that the left and right actions are the same, i.e. for all $a in A, m in M$ then $a.m = m.a$. This is equivalent to saying that $[a,b].m = 0$ for all $a,bin A$. The data of a central bimodule $M$ is the same as that of a module over the abelianization of $A$, the commutative ring $A_{ab} = A/F^1A$ where $F^1A = A[A,A]A$ is the two-sided ideal generated by commutators of elements of $A$.
(There is another notion of central bimodule which says instead only that $z.m = m.z$ for $z in Z(A)$. If $M$ is this kind of central bimodule, then $Der(A,M)$ is a central $Z(A)$-module.)
answered Jan 7 at 14:16
BenBen
4,373617
$endgroup$
You are correct that $Der(A)$ will not be an $A$ module, for example take $A = klangle a,b,crangle$ the free algebra and the derivation determined by $$D(a) = a, D(b) = b, D(c) = 1.$$ Then $abD(c) neq baD(c)$ because $ab neq ba$ in the free algebra.
In general the natural structure on $Der(A)$ is a $Z(A)$-Lie algebra and in particular a bimodule over $Z(A)$.
More generally you can consider derivations in a bimodule $M$, and ask when $Der(A,M)$ is an $A$ module. The same problem arises that $abD(c) neq baD(c)$. One condition that suffices to make $Der(A,M)$ an $A$-module is that $M$ is a central bimodule. This means that the left and right actions are the same, i.e. for all $a in A, m in M$ then $a.m = m.a$. This is equivalent to saying that $[a,b].m = 0$ for all $a,bin A$. The data of a central bimodule $M$ is the same as that of a module over the abelianization of $A$, the commutative ring $A_{ab} = A/F^1A$ where $F^1A = A[A,A]A$ is the two-sided ideal generated by commutators of elements of $A$.
(There is another notion of central bimodule which says instead only that $z.m = m.z$ for $z in Z(A)$. If $M$ is this kind of central bimodule, then $Der(A,M)$ is a central $Z(A)$-module.)
answered Jan 7 at 14:16
BenBen
4,373617
edited Jan 7 at 14:22
answered Jan 7 at 14:16
BenBen
4,373617
answered Jan 7 at 14:16
BenBen
4,373617
answered Jan 7 at 14:16
BenBen
4,373617
4,373617
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$begingroup$
You appear to be correct. Are you asked to prove this without commutativity, or is this your own initiative?
$endgroup$
– Matt Samuel
Jan 7 at 11:59
$begingroup$
I were asked to prove this without commutativity, maybe a typo in my notes
$endgroup$
– Minato
Jan 8 at 16:41