17
$\begingroup$

$\DeclareMathOperator{\Spec}{Spec}$ [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.

$\endgroup$
4
  • $\begingroup$ $dim(A)=0$ does not imply that $\mathrm{Spec}(A)$ is discrete. Instead, it implies that $\mathrm{Spec}(A)$ is Hausdorff. $\endgroup$ 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. $\endgroup$ Apr 6, 2012 at 9:32

4 Answers 4

29
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
18
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Nice example, Mariano. Somehow, I have the feling it is not the last time we are seeing it:) $\endgroup$ 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. $\endgroup$ Apr 13, 2012 at 5:55
  • 2
    $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Apr 13, 2012 at 6:41
  • $\begingroup$ Ah, thanks - my bad... I for some reason was reading $i<j$. $\endgroup$ Apr 13, 2012 at 12:00
6
$\begingroup$

Why don't you take infinitely many copies of a field?

$\endgroup$
7
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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. $\endgroup$ Apr 6, 2012 at 9:07
  • 2
    $\begingroup$ No, the ideals are more complicated. They correspond to filters on the index set. $\endgroup$ Apr 6, 2012 at 9:27
  • 1
    $\begingroup$ Sequences with finite support. $\endgroup$ Apr 6, 2012 at 9:54
  • 1
    $\begingroup$ 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
-3
$\begingroup$

A3. A Local ring has one maximal ideal, but it's not artinian in general (for example, $\mathbb{Z}_{(2)}$).

$\endgroup$
1
  • $\begingroup$ $\mathbb Z_{(2)}$ is not $0$-dimensional. $\endgroup$
    – Wojowu
    May 9, 2018 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.