The standard proof of the Hahn-Banach theorem makes use of Zorn's lemma. I hear that, however, Hahn-Banach is strictly weaker than Choice. A quick search leads to many sources stating that Hahn-Banach can be proven using the ultrafilter theorem, but I cannot seem to find an actual proof. So...


1. What is the ultrafilter theorem; and
2. How does the said theorem imply the Hahn-Banach theorem?


Any easily-accessible reference would be quite enough; thanks in advance!



Hahn-Banach without Choice
The ultrafilter theorem is the statement that any filter on a set can be extended to an ultrafilter. It is perhaps more common to see it sated as the (Boolean) Prime ideal theorem: Every Boolean algebra admits a prime ideal. 


The Hahn-Banach theorem is actually equivalent to the statement that every Boolean algebra admits a real-valued measure, but this is not entirely straightforward (see Luxemburg, "Reduced powers of the real number system and equivalents of the Hahn-Banach extension theorem", Intern. Symp. on the applications of model theory, (1969) 123-127).


For a discussion of Hahn-Banach vs. Choice and some additional remarks and references, see Jech "The axiom of choice", North-Holland, 1973.


You may also be interested in the references I include in [this answer](https://mathoverflow.net/questions/42215/does-constructing-non-measurable-sets-require-the-axiom-of-choice/42226#42226). 


