= Solution
The assertion is true. This is the <Artinian commutative ring is Noetherian theorem>. One proof uses the nilpotent <nilradical> $N$ of an <Artinian ring> $A$. The quotient $A/N$ is a finite product of <field>[fields]. Each quotient $N^j/N^{j+1}$ is an Artinian module over the <semisimple ring> $A/N$, hence has <length of a module>[finite length] and is <Noetherian module>[Noetherian]. The finite filtration
$$
A\supseteq N\supseteq\cdots\supseteq N^r=0
$$
then makes $A$ a Noetherian module over itself, which is exactly the <ascending chain condition> on its ideals.
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