Let $A$ be a boolean algebra and let $B\leq A$ be a boolean sub-algebra which is dense (for all $0\neq a\in A$, there is a $0\neq b\in B$ such that $b\leq a$). We suppose also that $B$, as a partially ordered set, has Krull dimension. Is it true that $A$ has Krull dimension? Can we bound this dimension by the dimension of $B$?
In case in the above situation $A$ has not Krull dimension, can we say at least that it has Gabriel dimension?
The example I have in mind is the case when $A$ is the completion of $B$, an answer in this case would be enough, even if I do not see why this should be simpler.
EDIT:
Definition[Krull dimonesion] Let $(L,\leq)$ be a lattice. The Krull dimension $K.dim(L)$ of $L$ is defined as follows:
-- $K.dim(L)=-1$ if and only if $L=\{0\}$;
-- if $\alpha$ is an ordinal and we already defined what it means to have Krull dimension $\beta$ for any ordinal $\beta<\alpha$, $K.dim(L)=\alpha$ if and only if $K.dim(L)\neq \beta$ for all $\beta<\alpha$ and, for any descending chain $$x_1\geq x_2\geq x_3 \geq \ldots \geq x_n\geq \dots$$ in $L$, there exists $\bar n\in \mathbb N_+$ such that $K.dim([x_n,x_{n+1}])=\beta_n$ for all $n\geq \bar n$ and $\beta_n<\alpha$.
If $K.dim(L)\neq \alpha$ for any ordinal $\alpha$ we set $K.dim(L)=\infty$.
Definition[Gabriel dimonesion] Let $(L,\leq)$ be a frame. We define the Gabriel dimension $G.dim(L)$ of $L$ by transfinite induction:
-- $G.dim(L)=0$ if and only if $L$ is trivial. A frame $S$ is $0$-simple (or just simple) if it is an atom;
-- let $\alpha$ be an ordinal for which we already know what it means to have Gabriel dimension $\beta$, for all $\beta\leq\alpha$. A frame $S$ is $\alpha$-simple if, for all $0\neq a\in S$, $G.dim([0,a])\nleq \alpha$ and $G.dim([a,1])\leq\alpha$;
-- let $\sigma$ be an ordinal for which we already know what it means to have Gabriel dimension $\beta$, for all $\beta<\sigma$. Then, $G.dim(L)=\sigma$ if $G.dim(L)\not<\sigma$ and, for all $1\neq a\in L$, there exists $b>a$ such that $[a,b]$ is $\beta$-simple for some ordinal $\beta<\sigma$.
If $G.dim(L)\neq \alpha$ for any ordinal $\alpha$ we set $G.dim(L)=\infty$.
(Notice that any complete Boolean algebra is a frame, furthermore it can be proved that having Krull dimension is a sufficient (but not necessary) condition for having Gabriel dimension)