The ring $(\mathbb{Z}\mathrm{/}2{)}^{\mathbb{N}}$ is zero-dimensional but not Noetherian

Every element $x\in R$ is idempotent, because $x^2=x$ coordinatewise in $\mathbb{Z}\mathrm{/}2$. Let $\mathfrak{p}$ be a prime ideal of $R$. Then $R\mathrm{/}P$ is an integral domain if and only if $P$ is a prime ideal makes $R\mathrm{/}\mathfrak{p}$ an integral domain, and every class $\bar{x}\in R\mathrm{/}\mathfrak{p}$ still satisfies ${\bar{x}}^2=\bar{x}$. So $\bar{x}(\bar{x}-1)=0$ forces $\bar{x}=0$ or $\bar{x}=1$. Thus $R\mathrm{/}\mathfrak{p}$ has exactly two elements and is a field, so $\mathfrak{p}$ is maximal.

For each $n\ge 0$, let $e_n$ be the sequence with $1$ in coordinate $n$ and $0$ elsewhere, and let $I_n:=R{e}_0+\cdots +R{e}_n$. Then $I_0⊊I_1⊊I_2⊊\cdots$ is a strict ascending chain of ideals, because $e_{n+1}\in I_{n+1}\setminus I_n$ for every $n$. Therefore $R$ is not Noetherian.

Let ${\mathfrak{p}}_0=\{x\in R:x_0=0\}\mathrm{.}$ The first-coordinate projection $R\to \mathbb{Z}\mathrm{/}2$ is a surjective ring homomorphism with kernel ${\mathfrak{p}}_0$, so $R\mathrm{/}{\mathfrak{p}}_0\cong \mathbb{Z}\mathrm{/}2$ is a field and therefore an integral domain. Thus $R\mathrm{/}P$ is an integral domain if and only if $P$ is a prime ideal makes ${\mathfrak{p}}_0$ a prime ideal. By step 1.1 every prime ideal of $R$ is maximal, so no strict chain of prime ideals can have length greater than $0$. Since ${\mathfrak{p}}_0$ provides a prime ideal, chains of length $0$ do occur. Therefore Krull dimension of a nonzero ring gives $\dim R=0$. Together with step 2.1, this ring is zero-dimensional but not Noetherian, so it is a concrete witness that the Noetherian hypothesis in the prime-maximal Artinian criterion cannot be dropped.