How to show that complement of prime filter is ideal?

Question

How to show that in any lattice L, F is a prime filter if an only if its complement L\F is an ideal?

Answer 1

Suppose $F$ is a prime filter. Since $F$ is a filter, $a,b\in F\implies a \wedge b\in F$ and $b\in F, a\geq b\implies a\in F$. Since $F$ is prime, $a\vee b \in F \implies a\in F$ or $b\in F$. Let $I=L-F$.

We want to show $I$ is an ideal. Let $a,b\in I$. If $a\vee b\in F$, then, since $F$ is prime, either $a\in F$ or $b\in F$. But $a,b\not\in F$ since they are in $I$. Thus we must have $a\vee b\in I$.

Let $a\in I,b\leq a$. If $b\in F$, then since $F$ is a filter and $a\geq b$, we would have $a\in F$. But $a\in I$ so we must have $b\in I$ as well. Thus $I$ is an ideal.

I believe the other direction is proved similarly.