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It is well known that the Stone–Čech compactification $\beta \mathbb N^+$ of the positive natural numbers has the structure of a compact left semitopological semigroup and hence, by Ellis's lemma, has idempotents. The usual proof of Ellis's lemma uses Zorn's lemma. Idempotent ultrafilters are clearly non-principal. It is known that the existence of non-principal ultrafilters is weaker than the axiom of choice.

My question is whether the existence of idempotent ultrafilters in $\beta \mathbb N^+$ is still weaker than choice?

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    $\begingroup$ The general rule of thumb about assertions like this is "If it pertains a concrete set then it is weaker than the axiom of choice" since we can always ensure that choice is well preserved long after the set is generated but eventually violated in severe ways. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 20, 2012 at 4:10

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Yes, it's still weaker. To build a model of ZF in which choice fails but $\beta\mathbb N^+$ has idempotents, start with a model of ZFC (which will, of course, have idempotent ultrafilters in $\beta\mathbb N^+$). Add a lot of Cohen-generic subsets of some regular cardinal $\kappa$ well above the cardinal of the continuum; forcing conditions are partial functions of size $<\kappa$. No new reals are added, and your idempotent ultrafilters from the ground model are still idempotent ultrafilters (and AC still holds). Now pass to the symmetric submodel given by the group of automorphisms of your forcing that permutes the names of the added Cohen subsets, with the filter determined by supports of size $<\kappa$. That model violates choice, because you can't well-order the power set of $\kappa$. But the ground model's reals and ultrafilters haven't been touched, so you still have the same idempotent ultrafilters that you had to start with.

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    $\begingroup$ But does the existence of an ultrafilter imply the existence of an idempotent? This might have been the motivation of the original question. $\endgroup$
    – Juris Steprans
    Commented Apr 20, 2012 at 10:57
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    $\begingroup$ Juris: That's a good question. I don't know, but my guess is negative. Start with Solovay's Lebesgue measure model (or $L(\mathbb R)$ in the presence of large cardinals), where there are no ultrafilters on $\omega$, and adjoin an ultrafilter by forcing with $P(\omega)/$fin. I'd be surprised to see an idempotent ultrafilter in the resulting model. I'd expect that the only ultrafilters would be the selective one you adjoined, ultrafilters obtained from it by iterated summation, and isomorphic copies thereof; and I'd expect none of these to be idempotent. $\endgroup$
    – Andreas Blass
    Commented Apr 20, 2012 at 13:14
  • $\begingroup$ Yes, the question of Juris is what I had in mind. I see I was naive to think idempotents in compact left topological semigroups implies choice. Notice that since $\beta N^+$ is a free left compact left semitopological semigroup on one generator the existence of an idempotent in it implies the general case. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 20, 2012 at 14:16
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    $\begingroup$ I just found out a very recent paper which partially answers Juris's question above: ams.org/journals/proc/2018-146-01/S0002-9939-2017-13719-8/… Apparently it suffices to assume that every filter on $\mathbb R$ can be extended to an ultrafilter, and this gives us idempotent ultrafilters on $\mathbb N$ $\endgroup$
    – David Fernandez-Breton
    Commented May 18, 2018 at 3:16

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