Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 2 / 1 / Solution

Let $R$ be a commutative Artinian ring. We prove finite composition length first, without assuming the conclusion that $R$ is Noetherian.

Every prime ideal is maximal. Indeed, an Artinian integral domain is a field: for $a\ne 0$, stabilization of $(a^m)$ gives $a^m=a^{m+1}b$, and cancellation gives $ab=1$. There are only finitely many maximal ideals. To see this, choose a minimal finite intersection $L={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_s$ using the descending chain condition. Minimality gives $L\subseteq \mathfrak{m}$ for any other maximal ideal $\mathfrak{m}$. Since ${\mathfrak{m}}_1\cdots {\mathfrak{m}}_s\subseteq L$, primeness forces one ${\mathfrak{m}}_i\subseteq \mathfrak{m}$, and maximality makes them equal. Thus the Jacobson radical $J$ is this finite intersection, and the Chinese remainder theorem gives

We also need nilpotence of $J$. Its powers stabilize; put $I=J^m$ at a stabilization stage, so $I^2=I$. If $I\ne 0$, choose an ideal $L$ minimal subject to $IL\ne 0$. Since $I(IL)=IL\ne 0$, minimality gives $IL=L$. Choose $a\in L$ with $Ia\ne 0$. Minimality applied to the subideal $Ra$ gives $L=Ra$. The equality $IL=L$ then yields $a=xa$ for some $x\in I\subseteq J$. But $1-x$ is a unit by the unit criterion for the Jacobson radical; hence $a=0$, a contradiction. Therefore $J^m=0$.

Each layer $J^i/J^{i+1}$ is an Artinian module over the finite product of fields $R/J$. Its components over those fields are finite-dimensional: an infinite-dimensional vector space has a strictly descending chain of subspaces. Each layer consequently has finite composition length. The finite filtrationproves that $R$ itself has finite composition length, and therefore satisfies the ascending chain condition on ideals. Every commutative Artinian ring is Noetherian. The zero ring satisfies the conclusion trivially.

The relevant Artin–Wedderburn theorem says that a semisimple ring which is a right Artinian ring has the formwhere the $D_i$ are division rings and the positive integers $n_i$ and division rings are unique up to permutation and isomorphism. The division rings need not be commutative or finite-dimensional over a field. Conversely every such finite product is semisimple and a right Artinian ring.

Here is a proof which also works if semisimplicity is initially expressed as $J(R)=0$. A nonzero right ideal contains a minimal nonzero right ideal $L$ by the descending chain condition; $L$ is a simple module. If $L^2=0$, then for every $l\in L$ and $r\in R$, $(lr{)}^2=0$, so $1-lr$ has inverse $1+lr$. The unit criterion for the Jacobson radical would give $L\subseteq J(R)=0$, a contradiction. Thus some $a\in L$ induces a nonzero map $L\to L$ by left multiplication. This map is an isomorphism, since $L$ is simple. Choose $e\in L$ with $ae=a$. Then $a(e^2-e)=0$ and injectivity on $L$ gives $e^2=e$. We obtain $L=eR$ and the splittingFor a right ideal $K$ containing $e$, the same argument splits $K=eR\oplus (K\cap (1-e)R)$. Repeatedly split a minimal right ideal out of the remaining summand. The remaining summands form a strictly descending chain of right ideals, so the process terminates. Thusfor pairwise nonisomorphic simple modules $S_i$.

A nonzero map between simple modules is an isomorphism: its kernel and image leave no alternatives. Thus $D_i={\mathrm{End}}_R(S_i)$ is a division ring, and ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. This is the needed Schur lemma. Taking endomorphisms of the direct sum gives ${\prod }_i{M}_{n_i}(D_i)$. Finally, left multiplication identifies $R$ with ${\mathrm{End}}_R(R_R)$: a right-module endomorphism sends $r$ to $f(1)r$, and composition corresponds to multiplication in $R$, not its opposite. This proves the displayed ring isomorphism.

For the converse, the right regular module of $M_n(D)$ is the direct sum of its $n$ row modules. Each row module is simple, since elementary matrices and nonzero scalar inverses send any nonzero row to any prescribed row. A finite product therefore has a finite semisimple regular module and is a right Artinian ring. The simple-module multiplicities and their endomorphism division rings recover the factors, establishing uniqueness.