So I started reading the chapter "Saturated Ideals I" from Kanamori's book. And I realized that I needed to refresh my knowledge on Boolean algebras and their properties in order to understand the following chapters better. And below is a statement which I think is necessary in order to accept some facts that Kanamori mentions and I need to see if my proof is correct. The reason I'm asking this, is that wherever I have encountered the below statement it is always used for complete algebras and it is totally likely that I'm making a mistake. So I would be really grateful if you could point any flaws in the argument.
The statement in the book(Jech's book):
$(I). \text{ }$Suppose $B$ is a complete Boolean algebra, then $B$ has the $\kappa\text{-c.c.}$ if and only if it has no descending $\kappa\text{-chains}$.
This is the one which I (think I) need:
$(II). \text{ }$Suppose $B$ is an arbitrary Boolean algebra, then $B$ has the $\kappa\text{-c.c.}$ if and only if it has no descending $\kappa\text{-chains}$.
My proof:
We first prove $(I)$:
Let $B$ be as above and assume it has the $\kappa\text{-c.c.}$ and let $\langle a_\alpha : \alpha \lt \kappa \rangle$ be a descending chain. Now let $b_\alpha = a_\alpha \cdot -a_{\alpha+1}$. Then it can be seen that $\langle b_\alpha : \alpha \lt \kappa \rangle$ is an antichain of size $\kappa$ which is a contradiction.
For the converse, let $\langle a_\alpha : \alpha \lt \kappa \rangle$ be an antichain in $B$. And let $b_\alpha = \Pi_{\beta \le \alpha} (-a_\beta)$, which exists by the completeness of $B$. Let $\gamma \lt \alpha$, now obviously $b_\alpha \le b_\gamma$. But suppose $b_\alpha = b_\gamma$, then $$-a_\alpha \cdot b_\gamma = -a_\alpha \cdot b_\alpha = b_\alpha = b_\gamma,$$
where the first equality holds because of our assumption and the second because of the definition of $b_\alpha$. Now we get $b_\gamma \lt -a_\alpha$ which in turn gives us: $$a_\alpha \lt -b_\gamma = \Sigma_{\beta \le \gamma} a_\beta,$$
which is a contradiction since $\langle a_\alpha : \alpha \lt \kappa \rangle$ was assumed to be an antichain. So we have that $b_\alpha \lt b_\gamma$. And now
$\langle b_\alpha : \alpha \lt \kappa \rangle$ is a descending chain in $B$, which again is a contradiction.
Now to prove $(II)$:
Let $B$ be an arbitrary Boolean algebra and let $C$ be the Boolean completion of $B$. Suppose $B$ has the $\kappa\text{-c.c.}$, then since $B$ is dense in $C$ it can easily be seen that $C$ has the $\kappa\text{-c.c.}$ too. So there are no descending $\kappa\text{-chains}$ in $C$ and so there are none in $B$.
For the converse assume that $B$ has no $\kappa\text{-chains}$, then $C$ can have no $\kappa\text{-chains}$ either since any $\kappa\text{-chain}$ in $C$ can be translated to a $\kappa\text{-chain}$ of $B$ because of the density of $B$ in $C$. So $C$ has the $\kappa\text{-c.c.}$ and again because of density $B$ has the $\kappa\text{-c.c.}$ too.
EDIT I:
As Henno Brandsma pointed out in the comments my "proof" for the converse of $(II)$ has a gap. As in, it is not that trivial that we can translate chains of $C$ to chains of $B$. Now I would like to convert my question to: Is $(II)$ even correct?
If $B$ is the finite-cofinite algebra on $\Bbb R$ then any chain (well-ordered or otherwise) in $B$ is at most countable so $B$ has no $\aleph_1$-chains. But it does have a $\aleph_1$-antichain of singletons.
So it seems some completeness is needed to get the equivalence. Except for $\kappa=\aleph_0$ of course, where the equivalence does always hold.