Krull dimension of dense extensions 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)
 A: I claim that every infinite complete Boolean algebra does not have Krull dimension.
Let's begin with some definitions.
A poset $P$ is said to have Krull dimension $0$ if for each descending sequence $(x_{n})_{n\in\mathbb{N}}$, we have $x_{n}=x_{n+1}$ for some $n$ (i.e. $P$ satisfies the descending chain condition). If $\alpha$ is an ordinal, then a poset $P$ is said to have Krull dimension at most $\alpha$ if for each infinite descending sequence $(x_{n})_{n}$ there is some $\beta<\alpha$ and $N$ where for each $n\geq N$, the interval $[x_{n+1},x_{n}]$ has Krull dimension at most $\beta$. A poset $P$ has Krull dimension if there is some ordinal $\alpha$ where $P$ has Krull dimension at most $\alpha$.
Suppose to the contrary there is an infinite Boolean algebra $B$ that does have Krull dimension. Then let $\alpha$ be the least ordinal such that there is an infinite complete Boolean algebra $B$ with Krull dimension $\alpha$. Then it is easy to show that there is a descending sequence $(x_{n})_{n}$ where each interval $[x_{n},x_{n+1}]$ is infinite. Therefore, since each interval $[x_{n},x_{n+1}]$ is an infinite complete Boolean lattice, the Krull dimension of each interval $[x_{n},x_{n+1}]$ is at least $\alpha$. This is a contradiction. 
On the other hand, there are many Boolean algebras with Krull dimension such as the Boolean algebra of finite and cofinite subsets of $\mathbb{N}$.
