A question on minimal idempotent ultrafilter on N^2 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:05:00Z http://mathoverflow.net/feeds/question/109060 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109060/a-question-on-minimal-idempotent-ultrafilter-on-n2 A question on minimal idempotent ultrafilter on N^2 js 2012-10-07T13:47:23Z 2012-10-07T13:57:04Z <p>Is there some minimal idempotent ultrafilter $q \in \beta( \mathbb{N}^2)$ (with respect to the law $"+"$) such that any $A \in q$ is a subset of $\mathbb{N} \times { 0 }$ ? (See for example <a href="http://www.math.osu.edu/~bergelson.1/vbkatsiveli20march03.pdf" rel="nofollow">http://www.math.osu.edu/~bergelson.1/vbkatsiveli20march03.pdf</a> for definitions).</p> <p>Motivation : Van der Waerden theorem can be quickly deduced from a negative answer to this question.</p> http://mathoverflow.net/questions/109060/a-question-on-minimal-idempotent-ultrafilter-on-n2/109061#109061 Answer by Andreas Blass for A question on minimal idempotent ultrafilter on N^2 Andreas Blass 2012-10-07T13:57:04Z 2012-10-07T13:57:04Z <p>Since <code>$(\mathbb N-\{0\})^2$</code> is a 2-sided ideal in <code>$\mathbb N^2$</code>, it follows that <code>$\beta((\mathbb N-\{0\})^2)$</code> is a 2-sided ideal in <code>$\beta(\mathbb N^2)$</code> and therefore contains all of the latter semigroup's minimal idempotents.</p>