$$This question arose while reading the paper [*On the composition of derivations*, by J.Krempa and J.Matczuk](http://w[/code]ww.researchgate.net/publication/226595021_On_the_composition_of_derivations).
Let R denote an associative ring For convenience, we will take $chaR=\infty$,f the additive group of R is torsionfree.Lett $d_{1} ,...,d_{n}$ be derivations of $R$ and $U=\{1,...,n\}$. For any subset of $V$ of $U$ we will put
$d_{v}(x)=d^{\epsilon_{1}}_{1}...d^{\epsilon_{n}}_{n}(x)$, where
$$\epsilon_{i}= \begin{array}{cc}
\Big\{ &
\begin{array}{cc}
1 & i\in V \\
0& i\notin V.
\end{array}
\end{array}$$
In particular, we get $d_{U}(x)=d_{1}...d_{n}(x)$.
Further, by writing $V=V_{1}\sqcup...\sqcup V_{r}$ we will understand that $V$ is a sum of disjoint subset $V_{i}$.
It is easy to compute by induction on cardinality of $V$ contained in U that for any $x,y \in R$,
$$(2) \qquad d_{v}(xy)=\sum\limits_{V_{1}\sqcup V_{2}=V} d_{v_{1}}(x)d_{v_{2}}(y).$$
By substituting $d$ for all $ d_{i}$'s in (2) we easily obtained the Leibniz's formula $d^{r}(xy)=\sum\limits_{i=0}^r \binom{r}{i} \cdot d^{i}(x)d^{ri}(y)$.
For any $k>1$ and $V \subset U, P_{k}(V)$ will stand for the family of all sequences
of sets $V_{1}, ..., V_{k}$ where all $V_{i}$'s are not empty and $V=V_{1}\sqcup...\sqcup V_{k}.$
In the above notation, it is claimed in the paper that (2) takes form of
$$(3) \qquad d_{v}(xy)=d_{v}(x)y+xd_{v}(y)+ \sum\limits_{V_{1},V_{2}
\in P_{2}(V)} d_{v_{1}}(x)d_{v_{2}}(y).$$
My question is: how can we deduce (3) from (2)?$$$$I am trying to understand the concept for (2).If just for x concept easy. if we can take V={1,2,3,} and $V_{1}$={1},$V_{2}$={2,3} then $d_{v}(x)=d_{1}d_{2}d_{3}(x) ,d_{V_{1}}(x)=d_{1}(x),d_{V_{2}}(x)=d_{2}d_{3}(x)$. So, $d_{v}(x)=d_{1}d_{2}d_{3}(x)= d_{V_{1}\sqcup V_{2}=V}(x)=d_{V_{1}} (x).d_{V_{2}}(x)=d_{1}(x)d_{2}d_{3}(x)=d_{1}d_{2}d_{3}(x)$. But if we take x and y,then we can obtain $d_{v}(xy)=d_{1}d_{2}d_{3}(xy)=d_{V_{1}\sqcup V_{2}=V}(xy)=d_{V_{1}}(xy).d_{V_{2}}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$. But how they do reduce (3) from (2)?.Maybe we must understand meaning of $\sum\limits_{V_{1}\sqcup V_{2}=V}$ and $\sum\limits_{V_{1},V_{2}
\in P_{2}(V)}$. this equation $d_{1}d_{2}d_{3}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$ is it true i dont know.]$$
on the composition of derivations

 Posts: 13
 Joined: Tue Oct 06, 2015 12:33 pm
Re: on the composition of derivations
Yes.fbli wrote:$$This question arose while reading the paper [*On the composition of derivations*, by J.Krempa and J.Matczuk](http://w[/code]ww.researchgate.net/publication/226595021_On_the_composition_of_derivations).
Let R denote an associative ring For convenience, we will take $chaR=\infty$,f the additive group of R is torsionfree.Lett $d_{1} ,...,d_{n}$ be derivations of $R$ and $U=\{1,...,n\}$. For any subset of $V$ of $U$ we will put
$d_{v}(x)=d^{\epsilon_{1}}_{1}...d^{\epsilon_{n}}_{n}(x)$, where
$$\epsilon_{i}= \begin{array}{cc}
\Big\{ &
\begin{array}{cc}
1 & i\in V \\
0& i\notin V.
\end{array}
\end{array}$$
In particular, we get $d_{U}(x)=d_{1}...d_{n}(x)$.
Further, by writing $V=V_{1}\sqcup...\sqcup V_{r}$ we will understand that $V$ is a sum of disjoint subset $V_{i}$.
It is easy to compute by induction on cardinality of $V$ contained in U that for any $x,y \in R$,
$$(2) \qquad d_{v}(xy)=\sum\limits_{V_{1}\sqcup V_{2}=V} d_{v_{1}}(x)d_{v_{2}}(y).$$
By substituting $d$ for all $ d_{i}$'s in (2) we easily obtained the Leibniz's formula $d^{r}(xy)=\sum\limits_{i=0}^r \binom{r}{i} \cdot d^{i}(x)d^{ri}(y)$.
For any $k>1$ and $V \subset U, P_{k}(V)$ will stand for the family of all sequences
of sets $V_{1}, ..., V_{k}$ where all $V_{i}$'s are not empty and $V=V_{1}\sqcup...\sqcup V_{k}.$
In the above notation, it is claimed in the paper that (2) takes form of
$$(3) \qquad d_{v}(xy)=d_{v}(x)y+xd_{v}(y)+ \sum\limits_{V_{1},V_{2}
\in P_{2}(V)} d_{v_{1}}(x)d_{v_{2}}(y).$$
My question is: how can we deduce (3) from (2)?$$$$I am trying to understand the concept for (2).If just for x concept easy. if we can take V={1,2,3,} and $V_{1}$={1},$V_{2}$={2,3} then $d_{v}(x)=d_{1}d_{2}d_{3}(x) ,d_{V_{1}}(x)=d_{1}(x),d_{V_{2}}(x)=d_{2}d_{3}(x)$. So, $d_{v}(x)=d_{1}d_{2}d_{3}(x)= d_{V_{1}\sqcup V_{2}=V}(x)=d_{V_{1}} (x).d_{V_{2}}(x)=d_{1}(x)d_{2}d_{3}(x)=d_{1}d_{2}d_{3}(x)$. But if we take x and y,then we can obtain $d_{v}(xy)=d_{1}d_{2}d_{3}(xy)=d_{V_{1}\sqcup V_{2}=V}(xy)=d_{V_{1}}(xy).d_{V_{2}}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$. But how they do reduce (3) from (2)?.Maybe we must understand meaning of $\sum\limits_{V_{1}\sqcup V_{2}=V}$ and $\sum\limits_{V_{1},V_{2}
\in P_{2}(V)}$. this equation $d_{1}d_{2}d_{3}(xy)=d_{1}(xy)d_{2}d_{3}(xy)=d_{1}d_{2}d_{3}(xy)$ is it true i dont know.]$$