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Thread: Derivation and Jordan derivation

  1. #1
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    Derivation and Jordan derivation

    Let R be ring, and d:R$\displaystyle \to$R is aditif mapping then
    d is called derivation if
    d(ab)=d(a)b+ad(b) $\displaystyle \forall $a,b$\displaystyle \in $ R
    d is called Jordan derivation if
    d(a^2)=d(a)a+ad(a)$\displaystyle \forall $a$\displaystyle \in $ R
    Obviously any derivation is Jordan derivation.

    But the converse is not true. (is there any example to show this statement?)
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  2. #2
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    Quote Originally Posted by Shurelia View Post
    Let R be ring, and d:R$\displaystyle \to$R is aditif mapping then
    d is called derivation if
    d(ab)=d(a)b+ad(b) $\displaystyle \forall $a,b$\displaystyle \in $ R
    d is called Jordan derivation if
    d(a^2)=d(a)a+ad(a)$\displaystyle \forall $a$\displaystyle \in $ R
    Obviously any derivation is Jordan derivation.

    But the converse is not true. (is there any example to show this statement?)
    it's a little bit tricky! here is an example: let $\displaystyle S$ be the algebra $\displaystyle \mathbb{C}[x]$ with the relation $\displaystyle x^2=0$. let $\displaystyle I = \mathbb{C}x$. note that $\displaystyle I$ is an ideal of $\displaystyle S$ because $\displaystyle x^2=0$. now let $\displaystyle R$ be the ring of all $\displaystyle 2 \times 2$ matrices in the form $\displaystyle r = \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ where $\displaystyle a,b,d \in S$ and$\displaystyle c \in I$. define the map $\displaystyle \delta : R \longrightarrow R $ by $\displaystyle \delta(r)=\begin{pmatrix} 0 & c \\ 0 & 0 \end{pmatrix}$. then

    1) $\displaystyle \delta(r_1+r_2)=\delta(r_1)+\delta(r_2),$ for all $\displaystyle r_1,r_2 \in R.$
    2) $\displaystyle \delta(r^2)=r \delta(r) + \delta(r)r,$ for all $\displaystyle r \in R.$
    3) there exist $\displaystyle r_1,r_2 \in R$ such that $\displaystyle \delta(r_1r_2) \neq \delta(r_1)r_2 + r_1 \delta(r_2).$

    i'll leave it to you to check that 1), 2) and 3) hold.
    Last edited by NonCommAlg; May 8th 2011 at 09:41 PM.
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