Proposition

Let $A$ be a Noetherian ring. Prove that the following are equivalent:

$A$ is Artinian;
$\Spec(A)$ is discrete and finite;
$\Spec(A)$ is discrete.

Solution

$1 \rightarrow 2$

By Proposition 8.1[Atiyah], $\Spec(A)$ only contains maximal ideals. By Proposition 8.3[Atiyah], $\Spec(A)$ only contains finitely many elements. As shown previously, a singleton of $\Spec(A)$ is closed if and only if the element is a maximal ideal. Since a finite union of closed sets is closed, $\Spec(A)$ is closed.

$2 \rightarrow 3$

Obvious.

$3 \rightarrow 2$

Since $\Spec(A)$ is discrete, every element in $\Spec(A)$ is maximal as shown previously, In other words, every prime ideal in $\Spec(A)$ is maximal. Suppose there exists a chain of length $\geq 1$. Since a subsequence of a chain is also a chain, there exists a chain of length 1, $p_0 \subsetneq p_1$. However, this is impossible because $p_0$ is a maximal ideal, so there cannot be a prime ideal containing it properly. Hence, the dimension of $A$ is 0. Since $A$ is Noetherian and $\dim A = 0$, $A$ is Artinian by Theorem 8.5.