Let $\mathfrak m$ be the <maximal ideal> and $k=A/\mathfrak m$ the finite <residue field>. The maximal ideal of an <Artinian local ring> is <nilpotent ideal>[nilpotent], so $\mathfrak m^r=0$ for some $r$. Every quotient
$$
\mathfrak m^j/\mathfrak m^{j+1}
$$
is both an Artinian $A$-module and a <vector space> over $k$. An Artinian vector space is finite-dimensional, hence each quotient is a <finite set>. The finite filtration
$$
A\supset\mathfrak m\supset\cdots\supset\mathfrak m^r=0
$$
therefore proves that the underlying set of $A$ is finite. This is the <Finiteness criterion for an Artinian local ring>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 1 / b / Solution

{c}

An Artinian local ring is a finite set exactly when its residue field is finite. If $\mathfrak m^r=0$, each layer $\mathfrak m^i/\mathfrak m^{i+1}$ is a finite-dimensional vector space over that residue field.


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