Theorem 1 (Sikorski's extension theorem)   Let $A$ be a Boolean subalgebra of a Boolean algebra $B$ , and $f:A\to C$ a Boolean algebra homomorphism from $A$ to a complete Boolean algebra $C$ . Then $f$ can be extended to a Boolean algebra homomorphism $g:B \to C$ .

Proof. We prove this using Zorn's lemma. Let $M$ be the set of all pairs $(h,D)$ such that $D$ is a subalgebra of $B$ containing $A$ , and $h:D\to C$ is an algebra homomorphism extending $f$ . Note that $M$ is not empty because $(f,A)\in M$ . Also, if we define $(h_1,D_1)\le (h_2,D_2)$ by requiring that $D_1\subseteq D_2$ and that $h_2$ extending $h_1$ , then $(M,\le)$ becomes a poset. Notice that for every chain $\mathcal{C}$ in $M$ , $$(\bigcup \lbrace h\mid (h,D)\in \mathcal{C}\rbrace, \bigcup \lbrace D\mid (h,D)\in \mathcal{C}\rbrace)$$ is an upper bound of $\mathcal{C}$ (in fact, the least upper bound). So $M$ has a maximal element, say $(g,E)$ , by Zorn's lemma. We want to show that $E=B$ .

If $E\ne B$ , pick $a\in B-E$ . Let $r$ be the join of all elements of the form $g(x)$ where $x\in E$ and $x\le a$ , and $t$ the meet of all elements of the form $g(y)$ where $y\in E$ and $a\le y$ . $r$ and $t$ exist because $C$ is complete. Since $g$ preserves order, it is evident that $r\le t$ . Pick an element $s\in C$ such that $r\le s\le t$ .

Let $F=\langle E,a\rangle$ . Every element in $F$ has the form $(e_1\wedge a)\vee (e_2\wedge a')$ , with $e_1,e_2\in E$ . Define $h:F\to C$ by setting $h(b)=(g(e_1)\wedge s)\vee (g(e_2)\wedge s')$ , where $b=(e_1\wedge a)\vee (e_2\wedge a')$ . We now want to show that $h$ is a Boolean algebra homomorphism extending $g$ . There are three steps to showing this:

1. $h$ is a function. Suppose $(e_1\wedge a)\vee (e_2\wedge a')=(e_3\wedge a)\vee (e_4\wedge a')$ . Then, by the last remark of this entry, $e_2\Delta e_4 \le a \le e_1\leftrightarrow e_3$ , so that $g(e_2)\Delta g(e_4) = g(e_2\Delta e_4)\le s \le g(e_1\leftrightarrow e_3) = g(e_1)\leftrightarrow g(e_3)$ , which in turn implies that $(g(e_1)\wedge s)\vee (g(e_2)\wedge s') = (g(e_3)\wedge s)\vee (g(e_4)\wedge s')$ . Hence $h$ is well-defined.
2. $h$ is a Boolean homomorphism. All we need to show is that $h$ respects $\vee$ and $'$ . Let $x=(e_1\wedge a)\vee (e_2\wedge a')$ and $y = (e_3\wedge a)\vee (e_4\wedge a')$ . Then $x\vee y = (e_5\wedge a)\vee (e_6\wedge a')$ , where $e_5=e_1\vee e_3$ and $e_6=e_2\vee e_4$ . So \begin{eqnarray*} h(x\vee y) &=& (g(e_5)\wedge s)\vee (g(e_6)\wedge s') \\ &=& ((g(e_1)\vee g(e_3))\wedge s)\vee ((g(e_2)\vee g(e_4))\wedge s') \\ &=& (g(e_1)\wedge s)\vee (g(e_2)\wedge s') \vee (g(e_3)\wedge s)\vee (g(e_4)\wedge s') \\ &=& h(x)\vee h(y), \end{eqnarray*}so $h$ respects $\vee$ . In addition, $h$ respects $'$ , as $x'= (e_2'\wedge a)\vee (e_1\wedge a')$ , so that \begin{eqnarray*} h(x') &=& h((e_2'\wedge a)\vee (e_1\wedge a')) = (g(e_2')\wedge s)\vee (g(e_1)\wedge s') \\ &=& (g(e_2)'\wedge s)\vee (g(e_1)\wedge s') = ((g(e_1)\wedge s)\vee (g(e_2)\wedge s'))' \\ &=& h(x)'. \end{eqnarray*}
3. $h$ extends $g$ . If $x\in E$ , write $x=(x\wedge a)\vee (x\wedge a')$ . Then $$h(x)= (g(x)\wedge s)\vee (g(x)\wedge s') = g(x).$$

This implies that $(g,E)< (h,F)$ , and with this, we have a contradiction that $(g,E)$ is maximal. This completes the proof.

Corollary 1   Every Boolean ideal of a Boolean algebra is contained in a maximal ideal.

Proof. Let $I$ be an ideal of a Boolean algebra $A$ . Let $B=\langle I\rangle$ , the Boolean subalgebra generated by $I$ . The function $f:B \to \lbrace 0,1\rbrace$ given by $f(a)=0$ iff $a\in I$ is a Boolean homomorphism. First, notice that $f(a)=0$ iff $a\in I$ iff $a'\notin I$ iff $f(a')=1$ . Next, if at least one of $a,b$ is in $I$ , $a\wedge b\in I$ , so that $f(a\wedge b)=0=f(a)\wedge f(b)$ . If neither are in $I$ , then $a',b'\in I$ , so $(a\wedge b)'= a'\vee b'\in I$ , or $a\wedge b\notin I$ . This means that $f(a\wedge b)=1=f(a)\wedge f(b)$ .

Now, by Sikorski's extension theorem, $f$ can be extended to a homomorphism $g: A\to \lbrace 0,1\rbrace$ . The kernel of $g$ clearly contains $I$ , and is in addition maximal (either $a$ or $a'$ is in the kernel of $g$ ).