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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? 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 $S\times S$. –  Martin Sleziak Jun 20 '12 at 14:54
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Sorry that is not the definition, now I wrote the correct definition (and changed notation to avoid confusions) –  John Jun 20 '12 at 15:11
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This is given as Corollary 17.17, p.346, in Hindman-Strauss: Algebra in the Stone-Cech compactification. –  Martin Sleziak Jun 20 '12 at 15:14
    
Ok, thanks for the reference, although I was hoping to find a simpler argument! –  John Jun 20 '12 at 15:36
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