How Zorn's lemma is used here?

Question

I am studying the following theorem in Advanced modern algebra/ Joseph J. Rotman. - Third edition,(Graduate studies in mathematics ; volume 165),

A left $R$ module $M$ over a ring $R$ is semisimple if and only if every submodule of $M$ is a direct summand.

I have not understood how Zorn's lemma is used in the converse part of the theorem :

By Zorn's lemma, there is a family $(S_j)_{j \in I}$ of simple sub-modules of $M$ maximal such that the sub-module $U$ they generate is their direct sum: $U = \oplus_{j \in I} S_j.$

I have taken (I hope this is a right direction) $$F = \{ (S_j)_{j \in I} : \text{The submodule generated by }(S_j)_{j \in I} \text{ is their direct sum where each $S_j$ are simple}\}$$

The previous part of the argument tells :

Every non-zero sub-module $B$ contains a simple summand.

This implies that $M$ contains a simple summand and consequently $F$ is nonempty. But somehow I am not able to show rigorously that every chain has an upper bound. Any help is appreciated.

Answer 1

For convenience of notation, let us consider sets of simple submodules whose sum is direct rather than indexed families. (It makes no significant difference, since the simple submodules in such a family must be distinct or else the submodule the generate would not be their direct sum.) Let $F$ be the set of all sets of simple modules of $M$ whose sum is direct, ordered by inclusion. Suppose $\mathcal{C}\subseteq F$ is a chain, and let $A$ be the union of all the elements of $\mathcal{C}$. Then $A$ is a set of simple submodules of $M$, and we want to show that the submodule generated by the elements of $A$ is their direct sum so that $A\in F$ and $A$ upper bound of the chain $\mathcal{C}$.

To show this, it is helpful to use the following fact. Given a set $A$ of submodules of a module, the submodule they generate together is their direct sum iff for any distinct $S_1,\dots,S_n\in A$ and any $s_1\in S_1,\dots,s_n\in S_n$, $\sum_i s_i=0$ implies $s_i=0$ for all $i$. Applying this to our $A$, note that since $\mathcal{C}$ is a chain, for any finite sequence of elements $S_1,\dots,S_n\in A$ there is a single element $B\in\mathcal{C}$ which contains all of them. But then since $B\in F$, we know that for any $s_1\in S_1,\dots,s_n\in S_n$, $\sum_i s_i=0$ implies $s_i=0$ for all $i$, as desired.

More briefly, given a family of simple submodules, the condition that the submodule they generate is their direct sum can be tested by considering only finitely many of the simple submodules at a time. This means that a union of a chain of such families is another such family, since any finite subset of the union is contained in some element of the chain.

Answer 2

Worth mention: there are various (folklore) forms of Zorn's Lemma that prove more convenient for these type of inductive arguments. Below is one such, excerpted from: $ $ Anderson; Dobbs; and Zafrullah: Some applications of Zorn's lemma in algebra.

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The use of Zorn’s Lemma has been a part of the mainstream in virtually every area of algebra for more than $75$ years. The aim of this note is to indicate some new applications of Zorn’s Lemma to a number of algebraic areas by using a slightly different perspective. In Theorem $1$ and its aftermath, we show that a property $P$ holds for all the subobjects of a given object if and only if $P$ supports both the chain condition from Zorn’s Lemma and some finitistic conditions on subobjects that have the flavor of mathematical induction.

Theorem $1$. $ $ Let $R$ be a ring and let $E$ be a (left) $R$-module. Let $P$ be a property of modules. Then each nonzero submodule of $E$ satisfies property $P$ if and only if the following $3$ conditions hold.
$(1)\ \ P$ is true for every nonzero cyclic submodule of $E$.
$(2)\ \ $ If $A$ is a nonzero submodule of $E$ such that $A$ satisfies property $P$ and $x\in E$, then $A + Rx$ again satisfies property $P$.
$(3)\ \ $ If $\{H_\alpha\}$ is a chain of nonzero submodules of $E$ such that each $H_\alpha$ satisfies property $P$, then $\cup H_\alpha$ satisfies property $P$.

Proof. $ $ Suppose that $(1)$, $(2)$ and $(3)$ hold. If the “if” assertion fails, $E$ has a nonzero submodule $A$ such that $A$ does not have property $P$. Let $S$ be the set of all nonzero submodules $I$ of $A$ such that $I$ has property $P$. Then $S$ is nonempty because, by $(1)$, every nonzero cyclic submodule has property $P$. Moreover, $S$ can be partially ordered by inclusion. By $(3)$, S meets the requirements of Zorn’s Lemma, and so must contain a maximal element $J$. Since $A$ does not have property $P,\ J \subsetneq A$. Pick $\,x \in A\backslash J$. Note that $J + Rx \subseteq A$ has property $P$ (by $(2)$), contradicting the maximality of $J$, thus completing the proof of the “if” assertion. The converse is evident. $\ \small\rm QED$

Remark $2$. $ $ In the statement of Theorem $1$, we can replace $(2)$ by

$(2')$: $ $ If $A$ and $B$ are two nonzero submodules of $E$ such that $A$ and $B$ satisfy property $P$, then $A+B$ satisfies property $P$. We will emphasize the case where $P$ holds for all nonzero submodules and leave the case where it holds for all nonzero proper submodules to Theorem 3 and the comment that precedes it.

Recall that a (left) $R$-module $M$ is semisimple if $M$ is a direct sum of simple submodules (see, e.g., [5, page 11]). A semisimple module $M$ is characterized by the property that every submodule of $M$ is a direct summand of $M$ [5, Proposition 4.1, page 11]. Of course this characterization does not change if we replace “every submodule” by “every nonzero submodule”. Also, as noted in the proof of [5, Proposition 4.1, page 11] (cf. also [4, Theorem 2.3.2]), every submodule of a semisimple module is a semisimple module. Setting $P$ = “is a direct summand of $M$” in Theorem $1$ gets us the following new characterization of semisimple modules.

Proposition $5$. $ $ An $R$-module $M$ is a semisimple $R$-module if and only if the following $3$ conditions hold.
$(1)\ \ $ Every nonzero cyclic submodule of $M$ is a direct summand of $M$.
$(2)\ \ $ If $A$ is a nonzero submodule that is a direct summand of $M$, then for every nonzero cyclic submodule $B,\ A + B$ is a direct summand of $M$.
$(3)\ \ $ If $\{H_\alpha\}$ is a chain of nonzero submodules of $E$ such that each $H_\alpha$ is a direct summand of $M$, then $\cup H_\alpha$ is a direct summand of $M$.

Proof. $ $ That $M$ semisimple implies $(1),\, (2)$ and $(3)$ is clear from the above comments. For the converse, take $P$ = “is a direct summand of $M$” in Theorem $1$. Combining this with the fact that the zero submodule is trivially a direct summand of $M$, we have the conclusion that $M$ is semisimple. $\ \small\rm QED$