Help with a proof that every ultrafilter on a distributive lattice is also prime Let $(L,\leq,\wedge,\vee)$ be a distributive lattice. Let $F$ be an ultrafilter on $L$, show that if $a,b\in L$ and $a\vee b\in F$ the either $a\in F$ or $b\in F$.
This is how I tackled it, but it seems very messy and long-winded for a supposedly easy result, that is relegated to easy exercises in many texts so any neater proofs will be much appreciated, and also if there any mistakes in my proof let me know how to correct it. 
Define $F_a:=\{z:z\geq a\wedge x$ for some $x\in F\}$. It is straightforward to see that $F_a$ is a filter containing $F$ and as $a\vee b\in F$ we have $a=a\wedge(a\vee b)\in F_a$. We can similarly define $F_b$.
Now suppose and $a,b\notin F$, so $F\subsetneq F_a,F_b$.
It is again plainly obvious that $(a\wedge b]:=\{z:z\leq a\wedge b\}$ is an ideal on $L$ and $F\cap(a\wedge b]=\emptyset$, otherwise $a,b\in F$
Claim: $F_a\cap(a\wedge b]=\emptyset$ or $F_b\cap(a\wedge b]=\emptyset$.
Otherwise if not there are $z_1\in F_a\cap(a\wedge b]$ and $z_2\in F_b\cap(a\wedge b]$ where for some $x_1,x_2\in F$ we have $z_1\geq a\wedge x_1$ and $z_2\geq b\wedge x_2$. Then $z_1\geq a\wedge y$ where $y=x_1\wedge x_2$ and as $(a\wedge b]$ is downward closed we have $a\wedge y\in(a\wedge b]$. Similarly $b\wedge y\in (a\wedge b]$.
Then as ideals are upward directed we have $(a\wedge y)\vee(b\wedge y)\in(a\wedge b]$. Then as $L$ is distributive $(a\wedge y)\vee(b\wedge y)=(a\vee b)\wedge y\in F$ as $y,a\vee b \in F$, contradicting that $F\cap(a\wedge b]=\emptyset$, so the claim is correct.
But then the claim would imply that either $F_a$ or $F_b\subsetneq L$ which then contradicts the maximality of the ultrafilter $F$, so  the supposition is wrong and so if $a\vee b\in F$ it must then be that $a\in F$ or $b\in F$ 
 A: Your proof is fine. Actually when you say $a \in F_a$, there is no need to use $a\vee b \in F$, because $a \geqslant a \wedge x$ for any $x \in F$ (so the reasoning here the same as for proving that $F_a$ contains $F$).
Here is a variation of your proof, but it is essentially the same (because it is quite a standard proof of this fact). We use maximality of $F$ right away and write $x \leqslant a \wedge b$ instead of $x \in (a \wedge b]$, so your claim and usage of $(a, b]$ notation "disappear". We use distributivity two times instead.
From $F \subsetneq F_a, F_b$ and maximality of $F$ we get that $F_a = F_b = L$. In particular, $a \wedge b\in F_a$ and $a \wedge b \in F_b$. That means $\exists x, y \in F$ such that 
$$a \wedge b \geqslant a \wedge x, \quad a \wedge b \geqslant b \wedge y.$$ 
Using distributivity we then have $$a \wedge b \geqslant (a\wedge x)\vee(b\wedge y) = ((a\wedge x)\vee b) \wedge((a\wedge x)\vee y) =\\= ((a\vee b)\wedge(x\vee b))\wedge((a\vee y)\wedge(x\vee y)).$$
But $a\vee b \in F$ by our assumption and $x\vee b, a\vee y, x \vee y \in F$ because $x, y \in F$. So $a \wedge b \in F$, contradicting $a, b \notin F$.
A: More concise:
Let $F$ be an ultrafilter with $a,b\notin F$. 
Then $F_a=L=F_b$ so that $a\wedge b\in F_{a}$ and $a\wedge b\in F_{b}$ where $F_a$ and $F_b$ are defined as in your question.
Then $a\wedge u\leq a\wedge b$ and $b\wedge v\leq a\wedge b$
for $u,v\in F$. 
Then $a\wedge w\leq a\wedge b$ and $b\wedge w\leq a\wedge b$
for $w=u\wedge v\in F$.
Then by distributivity: $\left(a\vee b\right)\wedge w=\left(a\wedge w\right)\vee\left(b\wedge w\right)\leq a\wedge b$. 
Now assumption $a\vee b\in F$ leads to $a\wedge b\in F$ and consequently $a,b\in F$.
This is not true so we are allowed to conclude that $a\vee b\notin F$.
Proved is now that $a,b\notin F\implies a\vee b\notin F$ or equivalently:$$a\vee b\in F\implies a\in F\text{ or }b\in F$$
