If $R$ is a ring without identity element then one can embed $R$ in a ring $S$ with identity element by Dorroh extension theorem. In fact, $S=\mathbb Z × R$ with element-wise addition, and the multiplication law $(n,a)(m,b)=(mn, nb+ma+ab)$ would be a candidate for the extension ring of $R$ with identity $(1,0)$.
Is the Jacobson radical of $S$ equal to that of $R$?
I mean, could we apply the rule $J(R_1 × R_2)=J(R_1)× J(R_2)$, which holds for the usual direct product of rings $R_1$ and $R_2$, as $J(S)=J(\mathbb Z)× J(R)=0× J(R)=J(R)$?
Thanks for any suggestion!