a counterexample in Boolean algebra First let me recall that a Boolean subalgebra $\Bbb{B}$ of a Boolean algebra $\Bbb{A}$ is called regular if for every subset $E\subseteq \Bbb{B}$ which has a supremum in $\Bbb{B}$, then $\vee^{\Bbb{B}}E=\vee^{\Bbb{A}}E$. 
Question: Can someone help me to find a Boolean algebra $\Bbb{A}$ with a subalgebra $\Bbb{B}$ such that there exists a subset $E\subseteq \Bbb{B}$ which has a supremum in $\Bbb{B}$ but has not a supremum in $\Bbb{A}$?
 A: I think I found an answer !
For completeness let me prove what I said in the comments :

Example of boolean algebras where there's a supremum in $\mathbb{A}$ but not in $\mathbb{B}$ : clearly some must exist, otherwise the concept of completion wouldn't exist. I'm going to give an example (not the easiest one, nor the most natural one, but I like it) : $\mathcal{B}(\mathbb{R})$ (borelian sets) as a subalgebra of $\mathcal{P}(\mathbb{R})$. If $X$ is any non borel subset of $\mathbb{R}$, then $\{\{x\}, x\in X\}$ has a sup in the latter (it's complete !), but not in the former. Indeed, if $Y$ were a supremum, then $Y\neq X$ and so for some $y\in Y$, $X\subset Y\setminus\{y\} \in \mathcal{B}(\mathbb{R})$.
Example of a boolean algebra/subalgebra pair where the suprema don't match : again, there must be many examples (otherwise the notion of regular subalgebras would be useless); but the one I'm giving here is to prepare for the example that answers the question. For a set $X$, denote $F_B(X)$ the free boolean algebra on $X$. My example has $\mathbb{B}= F_B(\mathbb{N})$ (for simplicity of notations, we'll agree that $0,1\notin \mathbb{N}$ throughout the answer) as a subalgebra of $\mathbb{A}= F_B(\mathbb{N}\cup\{a\})$, $a\notin \mathbb{N}$. But actually this won't work, so we actually consider a quotient of $\mathbb{A}$ by the ideal generated by $\neg a \land n $, $n\in \mathbb{N}$. Essentially we're saying $a\geq n$ for all $n$. Then in $\mathbb{B}$ (which can clearly be seen as a subalgebra of $\mathbb{A}$), the supremum of $\mathbb{N}$ is $1$ (one can prove this by the universal property of the free boolean algebra), whereas in $\mathbb{A}$, it's $a$ : here we have suprema that don't match !

Now comes the actual content of my answer. We consider a set $\tilde{\mathbb{N}}$ consisting of elements $\tilde{n}$ for $n\in \mathbb{N}$ such that $n\mapsto\tilde{n}$ is a bijection and $\mathbb{N}\cap\tilde{\mathbb{N}}=\emptyset$. Then our $\mathbb{B}$ is the same (up to isomorphism) and our $\mathbb{A}$ is $F_B(\mathbb{N}\cup\tilde{\mathbb{N}})$ quotiented by a suitable ideal so that $ \tilde{n} > m$ for all $n,m$. Then we consider again $\mathbb{N}$, which has a supremum in $\mathbb{B}$, but not in $\mathbb{A}$, which is what we wanted (assume the polynomial $p(v_1...,v_n, \tilde{w_1},...,\tilde{w_m})$ is an upper bound. Then consider $\tilde{k}\notin \{\tilde{w_1}, ..., \tilde{w_m}\}$ : $\tilde{k}\cap p(v_1...,v_n, \tilde{w_1},...,\tilde{w_m})$ is still an upper bound, and it's strictly smaller so there's no supremum)
