Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 85 / 4

Use the descending chain condition on the finite intersections of maximal right ideals. It gives a minimal such intersection $K=M_1\cap \cdots \cap M_r$. Intersecting $K$ with any other maximal right ideal cannot decrease it, by minimality. Thus $K$ lies in every maximal right ideal, and $K=J(R)$. When $J(R)=0$, the diagonal mapis injective, and each summand on the right is a simple module.

For completeness, a submodule $N$ of a finite direct sum of simple modules is semisimple. Induct on the number of summands. For $S\oplus T$ with $S$ simple, $N\cap S$ is either $0$ or $S$. In the first case projection into $T$ identifies $N$ with a submodule of $T$, so induction applies. In the second case subtracting the $S$ coordinate shows $N=S\oplus (N\cap T)$, and induction again applies. This also shows finite module length. Apply it to the image of the diagonal map to obtainThis is the semisimplicity of a right Artinian ring with zero radical. It does not assume right Noetherianity in order to prove it.

The descending powers $J\supseteq J^2\supseteq \cdots$ stabilize because they are right ideals. Choose $m\ge 1$ with $J^m=J^{m+1}$ and put $I=J^m$. Then $I=J^{2m}=I^2$.

Suppose $I\ne 0$. Among the right ideals $K$ such that $KI\ne 0$, choose a minimal one; the set is nonempty since $RI=I\ne 0$. The right ideal $KI$ lies in $K$, and $(KI)I=K{I}^2=KI\ne 0$, so minimality forces $KI=K$. Choose $k\in K$ with $kI\ne 0$. Then $(kR)I=kI\ne 0$, and $kR\subseteq K$, so the same minimality gives $K=kR$.

Since $K=KI$ and $I\subseteq J$, its generator satisfies $k={\sum }_{\nu }k{r}_{\nu }{i}_{\nu }=kj$ for an element $j={\sum }_{\nu }{r}_{\nu }{i}_{\nu }\in J$. Thus $k(1-j)=0$. The unit criterion for the Jacobson radical makes $1-j$ invertible, implying $k=0$, contrary to $kI\ne 0$. Therefore $I=0$, andThe minimal-right-ideal proof of radical nilpotence is important here: one cannot apply the finite-generator form of Nakayama lemma directly to $J^m$ before proving it finitely generated. We made a suitable auxiliary right ideal cyclic before using the unit argument.

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.

Let $K=k(z)$, where $z$ is an indeterminate, and form the right Artinian triangular ring that is not left NoetherianMultiplication uses the natural $k$–$K$ bimodule structure on the upper-right entry. With the two diagonal idempotents $e_1,e_2$, the right regular module is $e_{1}T\oplus e_{2}T$. The off-diagonal submodule of $e_{1}T$ is one-dimensional over the bottom-right field $K$, hence simple. Its quotient is one-dimensional over the top-left field $k$, hence simple. Also $e_{2}T$ is a simple right module over $K$. Thus $T_T$ has a composition series of length three, and in particular $T$ is right Artinian.

On the left, multiplication of an off-diagonal matrix givesEvery $k$-subspace $U\subseteq K$ therefore yields a left ideal of off-diagonal matrices with entries in $U$. The subspaces $U_m={\mathrm{span}}_k\{1,z,\dots ,z^m\}$ are strictly increasing because $z$ is transcendental. Their associated left ideals violate the ascending chain condition. Hence $T$ is not left Noetherian, even though it is right Artinian and right Noetherian.