= Solution
Maximal <right ideals> of $R/J$ correspond to <maximal right ideals> of $R$, all of which contain $J$. Their intersection in the quotient is zero. Hence $R/J$ is a <right Artinian ring> with zero <Jacobson radical>, and part (a) makes it semisimple.
Part (b) gives the finite radical filtration
$$
R\supseteq J\supseteq J^2\supseteq\cdots\supseteq J^m=0.
$$
Each factor $J^i/J^{i+1}$ is annihilated by $J$, so it is a right <module> over $R/J$. Every <module> over a semisimple <ring> is semisimple: a free <module> is a direct sum of simple summands of copies of the regular <module>, and a quotient of a semisimple <module> is again semisimple, because the images of its simple summands are either zero or simple and span the quotient. A maximal independent family of these simple images spans: any simple image not contained in the sum intersects it trivially and could be added. Each factor is also <Artinian>, as a subquotient of $R_R$. A semisimple <Artinian module> has only finitely many simple summands, since an infinite direct sum would give a descending chain by successively deleting summands. Consequently every factor has finite <module length>.
Concatenating composition series of these finitely many factors makes $R_R$ a <module> of finite <module length>. Such a <module> satisfies both chain conditions, since every strict change consumes a positive amount of length. Therefore \b[$R$ is right Noetherian]. This is the <Hopkins-Levitzki theorem>, obtained from the two preceding results rather than invoked in their proofs.
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