Radical/Prime/Maximal ideals under quotient maps

Solution 1:

You actually already proved the difficult part.

If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.

Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.

Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.