TITLE: A relation between the Jacobson radicals of a ring and those of a certain quotient ring
QUESTION [6 upvotes]: 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 $a\in J(R)$

REPLY [2 votes]: The Jacobson radical of $R$ is the intersection of the maximal left ideals of $R$.
The maximal left ideals of $R/I$, if $I\subseteq J(R)$, are of the form $K/I$, where $K$ is a maximal left ideal of $R$.
The intersection of the left ideals $K/I$ is exactly $J(R)/I$ by the correspondence theorem that gives an order preserving bijection between the submodules of $R$ containing $I$ and the submodules of $R/I$.