Dauns’ Proposition 3-4.5 (i) on essential extensions I am reading a part of the book “Dauns - Modules and Rings (1994)”, the part that deals with essential extensions. I want to understand the proof of proposition 3-4.5 part (i) on page 41.
Firstly let me give some definitions used by Dauns:
Definition. $M$ is a left $R$-module.
A submodule $K \leq M$ is a complement submodule of $M$ if there exists a (in general non-unique) submodule $B \leq M$ such that
(i) $K \cap B = 0$, and, 
(ii) if $K < A \leq M, K \neq A$ then $A \cap B \neq 0$ (in words: such that $K$ is maximal with respect to $K \cap B = 0$)
Definition. $M$ is a left $R$-module.
A submodule $V \leq M$ is large in $M$ if $M$ is an essential extension of $V$, i.e., 
$0 \neq A < M \Rightarrow V \cap A \neq 0$ or equivalently if $M’ \leq M$ and $M’ \cap V = 0$, then $M’ = 0$. 
I use the notation $V \leq_e M$ or $V <_e M$ (if $V \neq M$).
Part (i) of proposition 3-4.5 is as follows:
Let $M$ be a left $R$-module and $K < M$ a submodule, $K \neq M$.
Suppose that for each large submodule $L$ in $M$ we have that $(K+L) / K \leq_e M / K$.
Prove that $K < M$ is a complement submodule of $M$.
Dauns starts the proof with:
Suppose $K$ is not a complement, then $K$ is not maximal with respect to $K \cap B = 0$. 
Thus there exists a $N < M$ with $K < N$, $K \neq N$, and $N \cap B = 0$.
Then Dauns says:
By the maximality of $B$ we have that $K \leq_e N$ for otherwise we could enlarge $B$ so that it still would intersect $K$ trivially. 
Here I am stuck: what is $B$ and why it is maximal, and with respect to what? And why is $K \leq_e N$ ?
After this, the proof continues:
$L = K \oplus B <_e N \oplus B \leq M$ again, I do not understand this: why is $K \oplus B <_e N \oplus B$ ?
Accepting the above, I can understand the rest of the proof:
$N \cap (K \oplus B) = K + (N \cap B) = K$, by the modular law
$K < N$ thus $N / K \neq \bar{0} $
By hypothesis $(K + L) / K = (K \oplus B) / K \leq_e M / K$, however,
$$\frac{K+L}{K} \cap \frac{N}{K} = \frac{(N \oplus B) \cap N}{K} = \frac{K}{K} = \bar{0}$$
This is a contradiction because if
$(K+L) / K \leq_e M / K$ and 
$\bar{0} \neq N / K \leq M / K$
we must have $((K+L) / K) \cap (N / K) \neq \bar{0}$, by definition. 
Therefore $K$ is a complement submodule of $M$.
This ends the proof.
Please, can anyone help me with this proof and fill in my blank spots ?
 A: Thinking about this problem, I myself found this answer, hoping it is ok:
Let $0 \neq K < M, K \neq M$.
Let $\mathscr{S} = \{ L \leq M \mbox{   } | \mbox{   } K \cap L =0 \}$ then $\mathscr{S}$ is not empty and an inductive poset (with respect to inclusion), 
by Zorn’s Lemma it has a maximal element, say $B$. $B$ is maximal w.r.t. $K \cap B =0$.
Suppose $K$ is not a complement submodule in $M$, then $K$ is not maximal with respect to $K \cap B = 0$.
Thus there exists $N < M$ with $K < N, K \neq N$ and $N \cap B = 0$, note that $N \neq 0$.
Prove that $K <_e N$:
Let $N’ < N$ with $K \cap N’ = 0$, then $N’ \in \mathscr{S}$, 
$B$ is maximal in $\mathscr{S}$, thus $N’ \subset B$, thus $N \cap N’ = 0$, 
therefore $N’ = 0$ hence $K <_e N$.
Now prove that $K \oplus B <_e N \oplus B$:
Let $0 \neq x = n + b \in N \oplus B$, for some $n \in N$ and $b \in B$.
If $n = 0$ then $1x \in K \oplus B$, so suppose $n \neq 0$.
$K \leq_e N$, so there is an $r \in R$ such that $rn \in K$, also $b \in B \Rightarrow rb \in B$.
Thus $rx \in K \oplus B$ and therefore $K \oplus B <_e N \oplus B$.
Put $L = K \oplus B$.
Now follow the rest of the proof above.
Please comment on this proof.
**** EDIT ****
I think my verification of $K <_e N$ is wrong. From the maximality of $B$ in $\mathscr{S}$ one cannot deduce that $N’ \subset B$, I mixed up the notions “maximal” and “upper bound”.
I think this is a better verification of $K <_e N$:
Let $0 \neq x \in N$, we have to prove that $Rx \cap K \neq 0$.
$0 \neq x \in N$ thus $Rx \subset N$ and $Rx \cap B = 0$ (because $N \cap B = 0$).
Suppose $Rx \cap K = 0$, we have $B \cap K = 0$, therefore $(Rx + B) \cap K = 0$.
Thus $Rx + B = B$ by the maximality of $B$ w.r.t. $B \cap K = 0$,
thus $x \in B$, thus $x = 0$, contradiction,
therefore $Rx \cap K \neq 0$ and $K <_e N$.
Please, give me comments on my proof.
