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$\DeclareMathOperator{\Spec}{Spec}$ [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]

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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  • $\begingroup$ $dim(A)=0$ does not imply that $\mathrm{Spec}(A)$ is discrete. Instead, it implies that $\mathrm{Spec}(A)$ is Hausdorff. $\endgroup$
    – Martin Brandenburg
    Apr 6, 2012 at 8:40
  • $\begingroup$ 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. $\endgroup$
    – jmc
    Apr 6, 2012 at 8:50
  • $\begingroup$ Ok, it appears that my flaw of reasoning was exactly that spectra of $0$-dimensional rings need not be discrete. $\endgroup$
    – jmc
    Apr 6, 2012 at 9:01
  • $\begingroup$ Now, I have also answered Q2 in my answer. $\endgroup$
    – Martin Brandenburg
    Apr 6, 2012 at 9:32

4 Answers 4

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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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  • $\begingroup$ Thanks for answering Q2. I was looking in a totally different direction, but this is indeed a nice example. $\endgroup$
    – jmc
    Apr 6, 2012 at 9:48
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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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  • $\begingroup$ Nice example, Mariano. Somehow, I have the feling it is not the last time we are seeing it:) $\endgroup$
    – Georges Elencwajg
    Apr 6, 2012 at 14:26
  • $\begingroup$ 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. $\endgroup$
    – Greg Stevenson
    Apr 13, 2012 at 5:55
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    $\begingroup$ 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. $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 13, 2012 at 6:35
  • $\begingroup$ ...so there is exactly one prime ideal, the one generated by the variables, so that the Krull dimension is zero. $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 13, 2012 at 6:41
  • $\begingroup$ Ah, thanks - my bad... I for some reason was reading $i<j$. $\endgroup$
    – Greg Stevenson
    Apr 13, 2012 at 12:00
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Why don't you take infinitely many copies of a field?

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  • $\begingroup$ I thought I had a prove that, that will not work. Maybe I made a mistake. Can you prove yours? $\endgroup$
    – jmc
    Apr 6, 2012 at 8:49
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    $\begingroup$ 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. $\endgroup$
    – Filippo Alberto Edoardo
    Apr 6, 2012 at 9:07
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    $\begingroup$ No, the ideals are more complicated. They correspond to filters on the index set. $\endgroup$
    – Martin Brandenburg
    Apr 6, 2012 at 9:27
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    $\begingroup$ Sequences with finite support. $\endgroup$
    – Martin Brandenburg
    Apr 6, 2012 at 9:54
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    $\begingroup$ This is a very simple answer. To prove the prime ideals are maximal, just note that this ring is Von Neumann regular. The quotient by a prime ideal is a Von Neumann regular domain, end it is easy to show this is a field. So prime ideals are all maximal. $\endgroup$
    – rschwieb
    Dec 9, 2015 at 1:44
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A3. A Local ring has one maximal ideal, but it's not artinian in general (for example, $\mathbb{Z}_{(2)}$).

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  • $\begingroup$ $\mathbb Z_{(2)}$ is not $0$-dimensional. $\endgroup$
    – Wojowu
    May 9, 2018 at 15:17

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