A relation between the Jacobson radicals of a ring and those of a certain quotient ring

Question

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 $a\in J(R)$

Answer 1

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 $\phi (u) (1-sx)=1$.

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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 $x\in m$.

Answer 2

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 $R/I$.