Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 85 / 4 / c / 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 filtrationEach 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 $R$ is right Noetherian. This is the Hopkins-Levitzki theorem, obtained from the two preceding results rather than invoked in their proofs.