## on the composition of derivations

Forum for the GRE subject test in mathematics.
fbli
Posts: 1
Joined: Wed Oct 28, 2015 7:21 am

### on the composition of derivations

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

GeraltofRivia
Posts: 13
Joined: Tue Oct 06, 2015 12:33 pm

### Re: on the composition of derivations

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