$\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is a Boolean algebra. Prove that:
- For any prime filter $F$, there is a homomorphism $h:B\to\{0,1\}$ such that $F = \{x\in B:h(x)=1\}$
- For a homomorphism $h:B\to\{0,1\}$, the set $F = \{x\in B:h(x)=1\}$ is a prime filter in $\mathcal{B}$.
A filter $F$ is a prime filter if it is proper and if $a \lor b \in F$ implies either $a \in F$ or $b \in F$.
My work: Isn't (i) obvious? I'm unsure how to prove (i) formally. It seems clear that we should be able to define a homomorphism $h$ such that if $f \in F$ then $h(f) = 1$, and $h(f) = 0$ if $f \notin F$. What am I missing?
If $h$ is a homomorphism, we know that $h(0) = 0$ so $0 \notin F$, i.e. $F$ is proper. If $a \lor b \in F$, then $h(a\lor b) = h(a)\lor h(b) = 1$. So either $h(a) = 1$ or $h(b) = 1$, hence $a \in F$ or $b \in F$. $F$ is a prime filter! Is this proof complete?
P.S. For reference, $F \subseteq B$ is a filter if:
- $F \neq \phi$
- If $x,y \in F$ then $x\land y\in F$
- If $x\in F$ and $x\leq y$ then $y\in F$
For (i), you still have to prove that $h$ thus defined is a homomorphism.
Now, if $h(a \vee b) = 0$, then $a \vee b \notin F$, whence $a,b \notin F$, yielding $h(a),h(b) = 0$ and $h(a) \vee h(b) = 0$.
If $h(a \vee b) = 1$, then $a \vee b \in F$, whence $a \in F$ or $b \in F$, yielding $h(a) = 1$ or $h(b) = 1$, and so $h(a) \vee h(b) = 1$.
So $h(a \vee b) = h(a) \vee h(b)$.
Likewise you can prove that $h(a \wedge b) = h(a) \wedge h(b)$.
It is quite clear that $h(0)=0$ (since $F$ is proper) and $h(1)=1$ (since $F$ is a filter).
Now, a lattice homomorphism between Boolean lattices which sends $0$ to $0$ and $1$ to $1$ is a BA homomorphism (i.e., it also preserves complements); this is very easy to verify.