$\DeclareMathOperator{\Spec}{Spec}$
[Edit] Martin pointed out that $\dim A = 0$ does not imply that $\Spec A$ is discrete. Therefore I changed the wording of question 2.[/Edit]




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With dimension of a ring I mean the Krull-dimension.




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It is well-known that for a commutative ring $A$ the following are equivelent


* $A$ is noetherian and $\dim A = 0$;
* $A$ is artinian.


It is easy to think of noetherian rings that are not artinian ($\mathbb{Z}$). However I cannot find an example of a $0$-dimensional ring that is not artinian.



> 
> Questions
> ---------
> 
> 
> 1. What is an example of a commutative ring $A$ with $\dim A = 0$ that is not artinian (or equivalently, not noetherian)?
> 2. A related question is: Give an example of an affine scheme $X$, such that $X$ is discrete as topological space, but $\mathcal{O}\_X(X)$ is not noetherian/artinian.
> 3. Yet another question: Why does the converse of proposition 8.3 in Atiyah-MacDonald fail for a ring $A$ with $\dim A = 0$? (The proposition says that artinian rings have finitely many maximal ideals.)
> 
> 
> 


I have tried various constructions, but they all fail somehow.



A $0$-dimensional ring that is not noetherian
Take any compact totally disconnected Hausdorff space $X$ (for example the Cantor set, or the one-point compactification of $\mathbb{N}$). Then $\mathcal{C}(X,\mathbb{F}\_2)$ is a ring whose spectrum is homeomorphic to $X$. In particular, this ring is zero-dimensional, but this ring is noetherian iff $X$ is finite.


More generally, a commutative ring is called von Neumann regular when for every $x$ we have $x^2 | x$ (in particular, boolean rings qualify). Equivalently, every localization at a prime ideal is a field. In particular, they are zero-dimensional (in fact, they are precisely the reduced zero-dimensional rings). It is easy to check that these rings are closed under infinite products.


In particular, an infinite product of fields is a zero-dimensional ring, which is not noetherian. If the index set is $I$, the spectrum is the space of ultrafilters on $I$.


EDIT: It is even more trivial to give non-reduced examples. If $V$ is any $k$-module, then $A=k \oplus V$ is a $k$-algebra (with $V^2=0$). Then $A\_{\mathrm{red}}=k$ is a field, in particular $\mathrm{Spec}(A)$ is just a single point. If $V$ is not noetherian as a module, it is clear that $A$ won't be noetherian as a ring.


