Question of ultrafilters/prime ideals in finite Boolean algebras

Question

Let $U$ be an ultrafilter in a finite Boolean algebra $A$. Does $U$ necessarily contains an atom of $A$?

Let $I$ be a prime ideal in the same finite Boolean algebra $A$. Does $I$ necessarily contains the complement of an atom of $A$?

If these two statements are correct. Does this mean that there are as many ultrafilters (prime ideals) in a finite Boolean algebra as there are atoms (complements of atoms)?

Answer 1

Every ultrafilter on a finite Boolean algebra is fixed (principal) and therefore contains an atom; there is exactly one ultrafilter for each atom. The ultrafilters on a Boolean algebra are precisely the prime filters, so the prime ideals are the complements of the ultrafilters. Thus, both statements are correct.