A $0$-dimensional ring that is not noetherian - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:21:58Z http://mathoverflow.net/feeds/question/93289 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93289/a-0-dimensional-ring-that-is-not-noetherian A $0$-dimensional ring that is not noetherian Johan Commelin 2012-04-06T08:21:28Z 2012-04-06T09:31:39Z <p>$\DeclareMathOperator{\Spec}{Spec}$  Martin pointed out that $\dim A = 0$ does not imply that $\Spec A$ is discrete. Therefore I changed the wording of question 2.[/Edit]</p> <hr> <p>With dimension of a ring I mean the Krull-dimension.</p> <hr> <p>It is well-known that for a commutative ring $A$ the following are equivelent</p> <ul> <li>$A$ is noetherian and $\dim A = 0$;</li> <li>$A$ is artinian.</li> </ul> <p>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.</p> <blockquote> <h2>Questions</h2> <ol> <li><p>What is an example of a commutative ring $A$ with $\dim A = 0$ that is not artinian (or equivalently, not noetherian)?</p></li> <li><p>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.</p></li> <li><p>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.)</p></li> </ol> </blockquote> <p>I have tried various constructions, but they all fail somehow.</p> http://mathoverflow.net/questions/93289/a-0-dimensional-ring-that-is-not-noetherian/93290#93290 Answer by Mariano Suárez-Alvarez for A $0$-dimensional ring that is not noetherian Mariano Suárez-Alvarez 2012-04-06T08:23:46Z 2012-04-06T08:23:46Z <p>The quotient of $\mathbb Q[x_1,x_2,\dots]$ by the ideal generated by all products $x_ix_j$ with $1\leq i\leq j&lt;\infty$ is an example.</p> http://mathoverflow.net/questions/93289/a-0-dimensional-ring-that-is-not-noetherian/93291#93291 Answer by Filippo Alberto Edoardo for A $0$-dimensional ring that is not noetherian Filippo Alberto Edoardo 2012-04-06T08:36:31Z 2012-04-06T08:36:31Z <p>Why don't you take infinitely many copies of a field?</p> http://mathoverflow.net/questions/93289/a-0-dimensional-ring-that-is-not-noetherian/93292#93292 Answer by Martin Brandenburg for A $0$-dimensional ring that is not noetherian Martin Brandenburg 2012-04-06T08:43:57Z 2012-04-06T09:31:39Z <p>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.</p> <p>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.</p> <p>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 $I$.</p> <p>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.</p>