Orthogonal traces on semi-prime gamma rings (Q2747289)

Let \(M\) be a \(\Gamma\)-ring. A symmetric bi-derivation on \(M\) is a map \(D\colon M\times M\to M\) satisfying (i) \(D(x+y,z)=D(x,z)+D(y,z)\); (ii) \(D(x,y)=D(y,x)\); (iii) \(D(x\gamma y)=D(x)\gamma y+x\gamma D(y)\) for all \(x,y,z\in M\) and \(\gamma\in\Gamma\). The trace of \(D\) is the mapping \(d\colon M\to M\) defined by \(d(x)=D(x,x)\) for all \(x\in M\). The traces \(d_1\) and \(d_2\) of two derivations \(D_1\) and \(D_2\), respectively, are orthogonal if \(d_1(x)\Gamma M\Gamma d_2(y)=0=d_2(y)\Gamma M\Gamma d_1(x)\) for all \(x,y\in M\). Let \(M\) be a 2-, 3-torsion free semiprime \(\Gamma\)-ring and let \(U\) be a nonzero ideal of \(M\) with zero left annihilator, and let \(D_1\) and \(D_2\) be symmetric bi-derivations with traces \(d_1\) and \(d_2\), respectively. ThenNEWLINENEWLINENEWLINE\(\bullet\) \(d_1\) and \(d_2\) are orthogonal if and only if \(d_1(u)\Gamma d_2(v)+d_2(u)\Gamma d_1(v)=0\) for all \(u,v\in U\).NEWLINENEWLINENEWLINE\(\bullet\) The following are equivalent if \(d_2(U)\subset U\): (a) \(d_1\) and \(d_2\) are orthogonal; (b) \(d_1d_2=0\); (c) There exist \(a,b\in M\) and \(\gamma,\beta\in\Gamma\) such that \(d_1d_2(u)=a\beta u+u\gamma b\) for all \(u\in U\); (d) \(d_1d_2=f\) where \(f\) is the trace of a symmetric bi-additive mapping \(F\) of \(M\).