A Jordan derivation which is not a derivation

Posted: May 8, 2011 in Examples & Counter-Examples, Noncommutative Ring Theory Notes
Tags: ,

Let $R$ be a ring. Recall that an additive map $\delta : R \longrightarrow R$ is called a derivation if $\delta(r_1r_2)=\delta(r_1)r_2 + r_1 \delta(r_2)$ for all $r_1,r_2 \in R.$ Thus if $\delta$ is a derivation of $R,$ then $\delta(r^2)=\delta(r)r + r \delta(r)$ for all $r \in R.$

Definition. Let $R$ be a ring. An additive map $\delta : R \longrightarrow R$ is called a Jordan derivation if $\delta(r^2)=\delta(r)r+r\delta(r)$ for all $r \in R.$

So every derivation is a Jordan derivation . But the converse is not true and this is not surprising:

Example. Let $S=\mathbb{C}[x]$ with the relation $x^2=0.$ Let $I=\mathbb{C}x,$ which is an ideal of $S$ because $x^2=0.$ Let

$R = \begin{pmatrix} S & S \\ I & S \end{pmatrix}.$

See that $R,$ with matrix addition and multiplication, is a ring because $I$ is an ideal of $S.$ For any $r = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \in R$ define $\delta(r)=\begin{pmatrix}0 & c \\ 0 & 0 \end{pmatrix}.$ Then $\delta : R \longrightarrow R$ is a Jordan derivation but not a derivation.

Proof. It is obvious that $\delta$ is additive. Let $r = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \in R.$ Then, since $c^2=0,$ we have

$\delta(r^2)=\delta(r)r+r \delta(r) = \begin{pmatrix}0 & (a+d)c \\ 0 & 0 \end{pmatrix}$

and so $\delta$ is a Jordan derivation. To see why $\delta$ is not a derivation, let $r_1=\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}$ and $r_2=\begin{pmatrix}0 & 0 \\ x & 0 \end{pmatrix}.$ Then $\delta(r_1r_2)=0_R$ but $\delta(r_1)r_2 + r_1 \delta(r_2)=\begin{pmatrix}0 & x \\ 0 & 0 \end{pmatrix}. \ \Box$