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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 for definitions).

Motivation : Van der Waerden theorem can be quickly deduced from a negative answer to this question.

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How does van der Waerden theorem follow from this? – Joel Moreira Oct 12 at 15:48

1 Answer 1

Since $(\mathbb N-\{0\})^2$ is a 2-sided ideal in $\mathbb N^2$, it follows that $\beta((\mathbb N-\{0\})^2)$ is a 2-sided ideal in $\beta(\mathbb N^2)$ and therefore contains all of the latter semigroup's minimal idempotents.

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What do you mean by $\beta( ( \mathbb{N} - \{0 \})^2 )$ ? This cannot be the set of ultrafilters on $( \mathbb{N} - \{0 \})^2$ since this is not a subset of $\beta( \mathbb{N}^2)$ (for example : an ultrafilter on $( \mathbb{N} - \{0 \})^2$ does not contain $\mathbb{N}^2$). – js21 Oct 8 '12 at 4:58
Whenever a set $X$ is a subset of a set $Y$, every ultrafilter on $X$ can be identified with an ultrafilter on $Y$, namely its closure under supersets in $Y$. In other words, the inclusion map $X\to Y$ induces an injection $\beta(X)\to\beta(Y)$, which one uses to identify $\beta(X)$ with a subset of $\beta(Y)$. – Andreas Blass Oct 8 '12 at 11:52
Ok, thanks. Amusingly, I found by googling your name this article, which essentially contains the proof I had in mind (Theorem 8), with the same argument you give here. – js21 Oct 8 '12 at 13:53

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