I want to prove that the following statements are equivalent:
(1) Every non-trivial Boolean algebra has a prime ideal.
(2) In every non-trivial Boolean algebra, every ideal is contained in a prime ideal.
(3) For every set $X$, is a filter on $X$ contained in an ultrafilter on $X$.
I only know about the definitions of all these concepts, and nothing more. Is there an elementary proof I can understand? A reference is fine. I checked Herrlich's book "The axiom of choice" but the proofs there used Lattice theory, which I don't know.
(1) to (2) is simple. If $B$ is a Boolean algebra, and $I$ is an ideal, then $B/I$ is a Boolean algebra, so it contains a prime ideal $J$, but we can take its pullback to $B$, and it will be a prime ideal in $B$ containing $I$.
(2) to (3) is also simple. If $X$ is a set, then $\mathcal P(X)$ is a Boolean algebra, and a filter has a dual ideal, which is then contained by a prime ideal, whose dual is an ultrafilter.
Finally, (3) to (1) is the tricky part, and for the life of me I do not recall a direct proof. This part is proved in Herrlich's book without referring to lattice theory, by topological means. You can find a proof through the completeness theorem for first-order logic in Jech's "The Axiom of Choice" in Chapter 2. You can find another proof through analysis in Eric S. Schechter's "Handbook of Analysis and its Foundations" in Chapter 13.