let $R$ be a ring , Show that $J(R/J(R))=\{0\}$. 
$(a)$ let $R$ be a ring , Show that $J(R/J(R))=\{0\}$.
$(b)$ Let $(R_i)_{i\in I}$ be a family of rings. Show that $J(\prod _{i\in I} R_i)= \prod _{i\in I} J(R_i). $

Definition: The Jacobson radical of a ring $R$ is
$$J(R) = \bigcap \{I\mid I\text{ primitive ideal of }R\} $$
$$=\bigcap \{\operatorname{Ann}_R M\mid M\text{ simple }R\text{-module}\}$$
also $J(R)$ is an ideal of R.

I found the ans to $(a)$ HERE
can I use $(a)$ to ans $(b)??$
 A: Let $j = (j_i)_{i \in I}$ be an element of $J(\prod_{i \in I} R_i)$. Since a simple $R_k$-module $M_k$ can be thought of a simple $\prod_{i \in I} R_i$-module by
$$ mr := mr_k $$
for all $m \in M_k$ and $r = (r_i)_{i \in I}$ and hence
$$ mj_k = mj = 0 $$
from assumption.
So $j_k \in \bigcap \{\, \mathrm{ann}\,M_k \mid \text{$M_k$ is a simple $R_k$-module} \,\} = J(R_k)$ and this shows the desired inclusion.

Let $j = (j_i)_{i \in I}$ be an element of $\prod_{i \in I} J(R_i)$ and $M$ a simple $\prod_{i \in I} R_i$-module. Then $M$ can be thought of a simple $R_k$-module by
$$ mr_k := m\overline{r_k} $$
where $\overline{r_k} := (r_i)_{i \in I}$ and $r_h = \delta_{hk}r_k$.
Hence if $I$ is a finite index set,
$$ mj = m\sum_{i \in I}\overline{\jmath_i} = \sum_{i \in I}m\overline{\jmath_i} = 0 $$
from assumption. This gives the desired inclusion.

%% I'm afraid I'm not sure the finiteness assumption can be omitted or not.
A: One can prove (b) by Proposition 1 [N.Jacobson, Structure of Rings, Chapt.1, Sec.6]:
$$J(R) = \{z | bza \ {\rm is\  quasi-regular \ for\ all\ } a,b\in R\}$$
(an element $z$ is quasi-regular if there exists $z'\in R$ such that $z-z'+zz'=0, z'z=zz'$).
Evidently, $(x,y)\in R_1\times R_2$  is quasi-regular iff $x,y$ are quasi-regular (in $R_1, R_2$ resp.)
From here it is easy to prove (b).
