A Jordan derivation which is not a derivation

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

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