Proof. If $I$ is prime, then we are done. Let $S:=\lbrace J\mid J\mbox{ is an ideal in }L\mbox{, and }a\notin J\rbrace$ . Then $I\in S$ . Order $S$ by inclusion. This turns $S$ into a poset. Let $C$ be a chain in $S$ . Let $K=\bigcup C$ . If $x,y\in K$ , then $x\in J_1$ and $y\in J_2$ for some ideals $J_1,J_2\in C$ . Since $C$ is a chain, we may assume that $J_1\subseteq J_2$ , so that $x\in J_2$ as well. This means $x\vee y\in J_2\subseteq K$ . Next, assume $x\in K$ and $y\le x$ . Then $x\in J$ for some ideal $J\in C$ , so that $y\in J\subseteq K$ also. This shows that $K$ is an ideal. If $a\in K$ , then $a\in J$ for some $J\in C\subseteq S$ , contradicting the definition of $S$ . So $a\notin K$ and $K\in S$ also. This shows that every chain in $S$ has an upper bound. We can now appeal to Zorn's lemma, and conclude that $S$ has a maximal element, say $P$ .

We now want to show that $P$ is the candidate that we are seeking: $P$ is a prime ideal in $L$ and $a\notin P$ . Since $P\in S$ , $P$ is an ideal such that $a\notin P$ . So the only thing left to prove is that $P$ is prime. This amounts to showing that if $x\wedge y\in P$ , then $x\in P$ or $y\in P$ . Suppose not: $x,y\notin P$ . Let $Q_1$ be the ideal generated by elements of $P$ and $x$ , and $Q_2$ the ideal generated by $P$ and $y$ . Since $Q_1$ and $Q_2$ properly contain $P$ , $a\in Q_1$ and $a\in Q_2$ . Write $a\le p_1\vee x$ and $a\le p_2\vee y$ , where $p_1,p_2\in P$ . Then $a\vee p_2\le (p_1\vee p_2)\vee x$ and $a\vee p_1\le (p_1\vee p_2)\vee y$ . Take the meet of these two expressions, and we obtain $(a\vee p_2) \wedge (a\vee p_1)\le ((p_1\vee p_2)\vee x)\wedge((p_1\vee p_2)\vee y)$ . Since $L$ is distributive, on the left hand side, we get $a\vee (p_1\wedge p_2)$ . On the right hand side, we have $(p_1\vee p_2)\vee (x\wedge y)\in P$ . As the left hand side is less than or equal to the right hand side, we get that $a\vee (p_1\wedge p_2)\in P$ . Since $a\le a\vee (p_1\wedge p_2)\in P$ , $a\in P$ , a contradiction. Therefore, $P$ is prime and the proof is complete.