Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 148 / 4 / Solution

If $P$ is a prime ideal of the commutative Artinian ring $R$, then $R/P$ is an Artinian domain. For $0\ne x\in R/P$, the descending chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes, so $x^n=x^{n+1}a$ for some $n$; cancellation gives $xa=1$. Thus $R/P$ is a field and $P$ is maximal.

There are only finitely many maximal ideals. Otherwise, distinct maximal ideals $P_i$ are pairwise comaximal and their finite products give the strictly descending chaincontradicting the Artinian condition. This proves the statement about the prime ideals of a commutative Artinian ring.

The nilradical $N$ is the ideal of all nilpotent elements, equivalently the intersection of all prime ideals. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, then choose $x\in J$ with $Ix\ne 0$. Since $I^2=I$, minimality gives $Rx=Ix$, so $x=ax$ for some $a\in I$. But $a$ is nilpotent, hence $1-a$ is a unit, contradicting $(1-a)x=0$. Therefore the nilradical of a commutative Artinian ring is nilpotent.

If the maximal ideals are $P_1,\dots ,P_s$, the Chinese remainder theorem givesa finite product of fields. Each layer $N^j/N^{j+1}$ is an Artinian module over this product and therefore finite-dimensional. Since $N$ is nilpotent, these finitely many layers show that every ideal of $R$ is finitely generated. This proves the Artinian commutative ring is Noetherian theorem.

Now assume $R$ is local with maximal ideal $P$ and ${\dim }_{R/P}P/P^2=1$. The Nakayama lemma gives $P=(x)$. Since $P$ is nilpotent, its powers form a finite chain ending in zero. For a nonzero ideal $J$, choose the largest $r$ with $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Write $y=x^{r}u$; then $u\notin P$, so $u$ is a unit. Henceand every ideal of $R$ is principal.

Finally, the Artin–Wedderburn theorem says that a finite-dimensional semisimple $k$-algebra has the formwhere each $D_i$ is a finite-dimensional division ring over $k$; the factors are unique up to permutation and isomorphism. By the assumed complete reducibility, write the right regular module aswith pairwise nonisomorphic simple right modules $S_i$. The Schur lemma says that $D_i={\mathrm{End}}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the endomorphism ring of the displayed direct sum therefore giveConversely, the regular module of $M_n(D)$ is a direct sum of $n$ simple column modules, proving that every algebra on the right is semisimple. Uniqueness follows from uniqueness of the simple summands and their multiplicities.