I started to study set theory; my question is about definition of filters in Cori-Lascar's book Mathematical Logic.
Definition (Filter): A filter in a Boolean Algebra $\mathcal{A}=(A,+,\times,0,1)$ is a subset $F$ of $A$ such that the set $I=\{x\in A:x^C\in F\}$ is an ideal in $A$, where $x^C$ is the complement of $x$.
There is also a canonical association, that's an involution too, $f:\mathcal{A}\rightarrow \mathcal{A}:x\mapsto x^C$ such that $f(F)=I$ and $f(I)=F$; in other words, $I$ is the set of complements of elements of $F$ and $F$ is the set of complements of $I$.
If I understand correctly, when $I$ is a maximal ideal of $A$, then there is a canonically associated ultrafilter (i.e. a maximal filter) $U$ such that $A=I\cup U$; so the assiociation $f$ give me \begin{align*} f(I)=A-I \end{align*} where $A-I=U$ so in this particular case, the set $U$ of the complements of the elements of maximal ideal $I$ is the same of the complementary set of $I$ in $A$
Question: if $I$ is not maximal, could you exhibit an example where this is not true? That is where a filter $F$ in a Boolean Algebra $\mathcal{A}$ is not $A-I$; where $I$ is the ideal associated to $F$? Many Thanks.
Ideal $\{0\}$ and filter $\{1\}$ are associated but are only complements of eachother in the trivial case where $A=\{0,1\}$.
I hope this answers your question.
In fact it can be shown that a filter is an ultrafilter if and only if its complement is an ideal, and in that situation this ideal is automatically a maximal ideal.
Dually it can be shown that an ideal is maximal if and only if its complement is a filter, and in that situation this filter is automatically an ultrafilter.
So you could say that an ultrafilter goes always hand in hand with a maximal ideal (as complements of eachother).
This situation shows up if we have a homomorphism $\phi:\mathcal A\to\{0,1\}$. Then $\phi^{-1}(\{0\})$ is a maximal ideal and $\phi^{-1}(\{1\})$ is an ultrafilter.