I am reading “Lam - Lectures on Modules and Rings (1999)”. I was reading Proposition (19.1) on page 506, in which Lam states four equivalent conditions for a right R-module to be finitely cogenerated. Conditions (1) and (2) are ($M$ is a right $R$-module):
(1) For any family of submodules $\{ N_i | i \in I \}$ of $M$, if $\cap_{i \in I} N_i = 0$ then $\cap_{i \in J} N_i = 0$ for some finite $J \subseteq I$
(2) For any family of submodules $\{ N_i | i \in I \}$ of $M$ which form a chain, $N_i \neq 0$ for all $i \in I$ implies $\cap_{i \in I} N_i \neq 0$
Lam does not give a proof of $(2) \Longrightarrow (1)$, but he states that the direct proof of $(2) \Longrightarrow (1)$ is interesting. He gives a proof in “Lam – Exercises in Modules and Rings (2007)” on page 381. I can follow most of this proof, but at one point I am stuck. And I want to ask if anybody can help me to finish this proof.
Proof of $(2) \Longrightarrow (1)$. Assume (2) holds, but (1) does not hold. Then there exists a family of submodules $\bar{N} = \{ N_i | i \in I \}$ of $M$ such that $\cap_{i \in I} N_i = 0$, but $\cap_{i \in J} N_i \neq 0$ for every finite $J \subseteq I$
By Zorn’s Lemma, there exists such a family that is maximal (w.r.t. inclusion). We may thus assume that $\bar{N} = \{ N_i | i \in I \}$ is such a maximal family. Clearly, this family must be closed with respect to finite intersections. Now, by Hausdorff’s maximal principle, there is a maximal chain $\bar{A} = \{ A_i | i \in L \}$ contained in the family $\bar{N} = \{ N_i | i \in I \}$. By (2) we have $A = \cap_{i \in L} A_i \neq 0$. Now there are two possibilities: (a) $A$ is a member of $\bar{N}$; (b) $A$ is not a member of $\bar{N}$.
I understand possibility (b). It leads to a contradiction. I will not repeat this part of the proof here.
I have problems with possibility (a), it leads to a contradiction. But I cannot figure out why.
So $A$ is a member of the family $\bar{N} = \{ N_i | i \in I \}$, thus $A = N_a$ for some $a \in I$. Then for any $j \in I$, $A \cap N_j = N_a \cap N_j$ belongs to $\bar{N}$, because $\bar{N}$ is closed under finite intersections.
Now, Lam states, by the maximality of the chain $\bar{A} = \{ A_i | i \in L \}$, we must have $A \cap N_j = A$. This yields $A \subseteq N_j$ for all $j \in I$ and so $\cap \{ N_j | j \in I \} \supseteq A \neq 0$, which is a contradiction.
I cannot figure this out. How is the maximality of $\bar{A}$ used ?
If $N_j \in \bar{A}$, then $A \subseteq N_j$, i.e., $A = A \cap N_j$.
Thus if there is a $b \in I$ with $A \cap N_b \neq A$ then $N_b \notin \bar{A}$.
Can we, in this case, say something about $\bar{A}’ = \bar{A} \cup \{N_b\}$ ? Is it a chain?
And what is the use of $N_a$ ?
Can anyone help me with this proof ?
It took me a while, but I think this is the answer.
$A$ is a member of the family $\bar{N} = \{ N_i | i \in I \}$, thus $A = N_a$ for some $a \in I$.
Then for any $j \in I$: $A \cap N_j = N_a \cap N_j$ belongs to $\bar{N}$, because $\bar{N}$ is closed under finite intersections.
For each $l \in L$ we have $A \cap N_j \subseteq A_l \in \bar{A}$, by the definition of $A$.
This makes $\bar{A} \cup \{ A \cap N_j \}$ a chain, so $ A \cap N_j \in \bar{A}$, by the maximality of $\bar{A}$
Then, again by the definition of $A$, we have $A \subseteq A \cap N_j \subseteq A$, i.e., $A=A \cap N_j$, and therefore $A \subseteq N_j$. This goes for all $j \in I$.
Then it follows that $0=\bigcap_{i \in I} N_j \supseteq A \neq 0$ which is a contradiction.
Hence, we proved $(2) \Rightarrow (1)$