Exercise A0.V.1.10

Proposition A1

Let $R$ be an Artinian ring and let $I$ be an ideal of $R.$ Then $R/I$ is Artinian.

Proof

Let $\pi : R \to R/I$ be the quotient map. Take a descending chain $U_1 \supseteq U_2 \supseteq \ldots$ of ideals of $R/I.$ Since preimages preserve inclusions, this gives us a descending chain $\pi^{-1}(U_1) \supseteq \pi^{-1}(U_2) \supseteq \ldots$ of ideals of $R.$ Since $R$ is Artinian, this chain has to stabilize at some point, say $n.$ So $\pi^{-1}(U_n) = \pi^{-1}(U_{n+k}),$ and hence $U_n = U_{n + k},$ showing that the chain in $R/I$ stabilizes as well. Thus we conclude that $R/I$ is Artinian.

$\blacksquare$

Proposition A2

Let $R$ be an Artinian integral domain. Then $R$ is a field.

Proof

Let $r \in R$ be non-zero. We have a descending chain of ideals $(r^1) \supseteq (r^2) \supseteq (r^3) \supseteq \ldots$ which must stabilize, say at $(r^n).$ So $(r^n) = (r^{n + 1}),$ making $r^n$ and $r^{n + 1}$ associates. By lemma 1.5 this gives us that $r^{n + 1} = u r^n$ where $u \in R$ is a unit. Hence $r r^n = u r^n$ and since multiplication by a non-zero is cancellative in an integral domain, $r = u,$ making $r$ a unit. Since $r$ is an arbitrary non-zero element of $R,$ this holds for every non-zero element of $R,$ making it into a field.

$\blacksquare$

Proposition A3

Let $R$ be an Artinian ring. Then $R$ has Krull dimension $0.$

Proof

Let $I$ be a prime ideal of $R.$ By A1 we know that $R/I$ is Artinian. Since $I$ is a prime ideal, $R/I$ is an integral domain. By A2 we know that an Artinian integral domain is a field. Since $R/I$ is a field if and only if $I$ is a maximal ideal, we can conclude that $I$ is maximal. Since $I$ was arbitrary, this tells us that all prime ideals of $R$ are maximal, and hence the Krull dimension must be $0.$

$\blacksquare$