In Thomas Jech's Set Theory book, he states that
If $A$ is a complete subalgebra of a complete Boolean algebra $B$ then $A$ is a regular subalgebra of $B$.
Also, if $A$ is a dense subalgebra of $B$ then $A$ is a regular subalgebra.
From the definition, $A$ is a regular subalgebra of $B$ if and only if, if $W$ is a maximal antichain in $A$ (i.e., $(\forall u, v\in W)((u\cdot v = 0)\vee(u = v))$ and $\sum W = 1\in A$), it is also a maximal antichain in $B$.
Now since $W$ also satisfies the maximal antichain conditions on $B$ (i.e., $\sum W = 1\in A\subseteq B$), it seems obvious that $W$ is also a maximal antichain of $B$. But I saw this post: maximal antichain and I understood that I was wrong. But I still did not understand how does completeness prevent something like the counterexample in maximal antichain? Also I do not have any idea about the second one... How should I approach these problems?
The key is that we need to distinguish between $\sum^A$ and $\sum^B,$ which may be different (and in general, $\sum^A$ may not even exist). It's clear that a $A'\subseteq A$ is an antichain in $A$ iff it is an antichain in $B,$ but as the counterexample shows, we can have $\sum^A A'=1$ but $\sum^B A'\ne 1.$ In other words, there can be a $z\in B$ with $z\ne 1$ that is an upper bound for $A',$ while there is no such $z\in A.$
However, in a complete subalgebra, there is no difference between $\sum^A$ and $\sum^B.$ This is precisely what it means to be a complete subalgebra. So, we have $\sum^A A' = 1\implies \sum^B A' =1,$ and therefore a maximal antichain in $A$ is also maximal in $B.$
For the second one, again, the goal should be to show $\sum^A A' = 1\implies \sum^B A' =1,$ but this time it doesn't quite immediately follow from the definition.
But, if we have a $z\in B$ that is an upper bound for $A'$ such that $z\ne 1,$ then we can see by density that there must be one in $A$ as well. Let $u\in A^+$ such that $u\le -z.$ Then if we take $z' = -u,$ then $z'\ge z$, so $z'\in A$ is an upper bound for $A'$ with $z'\ne 1.$