# A $0$-dimensional ring that is not noetherian

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

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

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.

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$dim(A)=0$ does not imply that $\mathrm{Spec}(A)$ is discrete. Instead, it implies that $\mathrm{Spec}(A)$ is Hausdorff. –  Martin Brandenburg Apr 6 '12 at 8:40
Martin, ok. Then I made a mistake somewhere. Actually I am very interested in an example of a non-noetherian ring $A$ for which $\Spec A$ is discrete. –  jmc Apr 6 '12 at 8:50
Ok, it appears that my flaw of reasoning was exactly that spectra of $0$-dimensional rings need not be discrete. –  jmc Apr 6 '12 at 9:01
Now, I have also answered Q2 in my answer. –  Martin Brandenburg Apr 6 '12 at 9:32

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.

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Thank you! I'll try to work a proof out myself. –  jmc Apr 6 '12 at 8:52
Thanks for answering Q2. I was looking in a totally different direction, but this is indeed a nice example. –  jmc Apr 6 '12 at 9:48

The quotient of $\mathbb Q[x_1,x_2,\dots]$ by the ideal generated by all products $x_ix_j$ with $1\leq i\leq j<\infty$ is an example.

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Nice example, Mariano. Somehow, I have the feling it is not the last time we are seeing it:) –  Georges Elencwajg Apr 6 '12 at 14:26
Sorry for the dumb question but I am confused - why is this ring zero dimensional? For $i\in \mathbb{N}$ aren't the ideals $(x_j\vert j\neq i)$ prime with quotient $\mathbb{Q}[x_i]$? So this ring would have dimension 1... Even if I made a silly mistake here I don't see how it could be von Neumann regular. –  Greg Stevenson Apr 13 '12 at 5:55
I am modding out by all products $x_ix_j$ with any $i$ and any $j$. In particular, $x_i^2$ is zero in the quotient. –  Mariano Suárez-Alvarez Apr 13 '12 at 6:35
...so there is exactly one prime ideal, the one generated by the variables, so that the Krull dimension is zero. –  Mariano Suárez-Alvarez Apr 13 '12 at 6:41
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Why don't you take infinitely many copies of a field?

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I thought I had a prove that, that will not work. Maybe I made a mistake. Can you prove yours? –  jmc Apr 6 '12 at 8:49
You can find a proof in my answer. –  Martin Brandenburg Apr 6 '12 at 8:49
I might be wrong, but I argue as follows. It is not noetherian because the chain of ideals $I_n=<e_i>_{1\leq i \leq n}$ is ascending and never stabilizes ($e_i$ is the $\infty$-ple having $1$ at the i-th place and $0$ elsewhere). It is of dimension $0$ because every ideal is product of ideals (general in any product of rings) and a quotient of my ring is a domain iff is a field, so every prime is maximal. –  Filippo Alberto Edoardo Apr 6 '12 at 9:07
No, the ideals are more complicated. They correspond to filters on the index set. –  Martin Brandenburg Apr 6 '12 at 9:27
Sequences with finite support. –  Martin Brandenburg Apr 6 '12 at 9:54
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