Jacobson radical of extension ring 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 $(1,0)$.
My question: 

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! 
 A: After realizing that $J(S)\subseteq R$ as you have, it is easy to see that $J(R)=J(S)$ using the definition of the Jacobson radical via quasi-regularity. Specifically:

$x\in J(R)$ iff $xr$ is right quasi-regular for every $r\in R$.

"right quasi-regular" means there exists $y\in R$ such that $x + y -xy=0$.
Suppose $x\in J(R)$. It's easy to show that if $I\lhd R$, then $(0,I)\lhd S$, and in particular $J(R)=(0, J(R))\lhd S$. Then for any $s\in S$, $xs\in J(R)$ and is r.q.r. in $R$, hence also in $S$. This shows $J(R)\subseteq J(S)$.
Now let $x\in J(S)$. Then for every $s\in S$, there exists $y\in S$ such that $xs+y -xsy=0$. But because $x\in R\lhd S$ (your discovery that $J(S)\subseteq R$ comes in here), $y=xsy-xs\in R$. In particular, $xr$ is r.q.r. for every $r\in R$. Thus $x\in J(R)$.
A: Let $g : S \rightarrow \mathbb Z$ be the map sending $(m,r)$ to $m$. It is definitely an epimorphism of rings. Since the kernel of $g$ would be the set $\{ (0,r) \mid r\in R\}$, and it is identical with $R$, we deduce that $S/R$ is isomorphic to the ring $\mathbb Z$. So, $J(S/R)=0$. Regarding $R$ as an ideal of $S$ we get $J(S)\subseteq R$. Now, if $R$ is, for example, a nil ring   then $J(S)=J(R)$.
