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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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    Ok, thanks for the reference, although I was hoping to find a simpler argument!2012-06-20

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