2
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I know the following simple fact is true, but I can't find a good proof:

Over the naturals, the only ultrafilter $\mathcal U$ such that $\mathcal U \oplus \mathcal U = \mathcal U \odot \mathcal U$ is the principal ultrafilter $2$.

Thanks!

(sum and product are defined as:

$A \in \mathcal U \oplus \mathcal V \Leftrightarrow \{n \,|\, A - n \in \mathcal V \} \in \mathcal U$

$A \in \mathcal U \odot \mathcal V \Leftrightarrow \{n \,|\, A/n \in \mathcal V \} \in \mathcal U$

where $A-n=\{m \, |\, m+n \in A\}$ and $A/n=\{m \, | \, m \cdot n \in A \}$ )

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    The definition of $\oplus$ and $\otimes$ is the same as [here](http://math.stackexchange.com/questions/31147/sum-and-product-of-ultrafilters)? I think you should give some explanation for your notation, since if $\mathcal U$ is an ultrafilter on $S$, then you are comparing two ultrafilters, one of them on $S$ and the second one of them on $S\times S$.2012-06-20
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    Sorry that is not the definition, now I wrote the correct definition (and changed notation to avoid confusions)2012-06-20
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    This is given as Corollary 17.17, [p.346](http://books.google.com/books?id=KYXgdiegKDsC&pg=PA346), in Hindman-Strauss: Algebra in the Stone-Cech compactification.2012-06-20
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    Ok, thanks for the reference, although I was hoping to find a simpler argument!2012-06-20

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