= 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 chain
$$
P_1\supsetneq P_1P_2\supsetneq P_1P_2P_3\supsetneq\cdots,
$$
contradicting 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 $I\ne0$, choose an ideal $J$ minimal subject to $IJ\ne0$, then choose $x\in J$ with $Ix\ne0$. 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 $P_1,\ldots,P_s$, the <Chinese remainder theorem> gives
$$
R/N\cong\prod_{i=1}^sR/P_i,
$$
a 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 $y=x^ru$; then $u\notin P$, so $u$ is a unit. Hence
$$
J\subseteq(x^r)=(y)\subseteq J,
$$
and \b[every ideal of $R$ is principal].
Finally, the <Artin–Wedderburn theorem> says that a finite-dimensional semisimple $k$-algebra has the form
$$
\boxed{T\cong\prod_{i=1}^sM_{n_i}(D_i),}
$$
where 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 as
$$
T_T\cong\bigoplus_{i=1}^sS_i^{n_i}
$$
with pairwise nonisomorphic simple right modules $S_i$. The <Schur lemma> says that $D_i=\operatorname{End}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the <endomorphism ring> of the displayed direct sum therefore give
$$
T\cong\operatorname{End}_T(T_T)
\cong\prod_iM_{n_i}(D_i).
$$
Conversely, 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.
Back to article page
