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

A $0$-dimensional ring that is not noetherian

2012-04-06T08:21:28Z

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

What is an example of a commutative ring $A$ with $\dim A = 0$ that is not artinian (or equivalently, not noetherian)?

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.

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.

Answer by Mariano Suárez-Alvarez for A $0$-dimensional ring that is not noetherian

2012-04-06T08:23:46Z

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.

Answer by Filippo Alberto Edoardo for A $0$-dimensional ring that is not noetherian

2012-04-06T08:36:31Z

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

Answer by Martin Brandenburg for A $0$-dimensional ring that is not noetherian

2012-04-06T08:43:57Z

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.