A relation between the Jacobson radicals of a ring and those of a certain quotient ring Let $R$ be a ring $J(R)$ the Jacobson radical of $R$ which we define for this problem to be all the maximal left ideals of $R.$  I'm trying to prove the following proposition with only the definition (i.e. not using anything about simple modules ect)

Let $I\subset J(R)$ be an ideal of $R.$  Then $J(R/I)\cong J(R)/I.$ 

Attempt:  Since $I\subset J(R)$ it is contained in every maximal ideal and hence the maximal ideals of $R$ are in bijection with the maximal ideals of $R/I.$  Then the map $\phi:J(R)\to J(R/I)$ given by $x\mapsto x+I$ is a $R$-module hom with kernel $I.$  So $\phi(J(R))\cong J(R)/I.$  But I am having a hard time showing this map is onto.  I don't see a good reason why every $x\in J(R/I) = \bigcap( M_{k}/I)$ must look like $a + I$ where $a\in J(R)$ 

Lemma. Let $\phi : R\to S$ be an epimorphism. Then $\phi (J(R))\subset J(S)$:  
proof. Let $x\in \phi (J(R))$. by Atiyah-Macdonald, Proposition 1.9, we need to prove that for all $s\in S$, $1-sx$ is a unit in $S$. let $x= \phi (t)$ for some $t\in J(R)$, and $s=\phi (r)$. we have $1-sx=\phi (1-rt)$. note that $1-rt$ has an inverse, say $u$. So $\phi (u) (1-sx)=1$.  

In this case $S =R/I$ and $J(R)+I/I\subset J(R/I)$.
If $I\subset J(R)$ we have equality:
let $x+I\in J(R/I)$ and $m$ an arbitrary maximal ideal of $R$. we should prove that $x\in m$. note that $I\subset m$. so $m/I$ is a maximal ideal of $R/I$. by assumption $x+I\in m/I$. therefor $x\in m$.
