# How Zorn's lemma is used here?

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.

• Zorn's lemma is equivalent to the maximum modulus principle. Aug 17, 2022 at 13:53
• @Noobie Which maximum modulus principle? Please tell in detail. Aug 17, 2022 at 13:56
• @Noobie What????? Aug 17, 2022 at 14:11
• Sorry, I meant the Hausdorff maximal principle en.m.wikipedia.org/wiki/…. Aug 17, 2022 at 14:16

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.

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.

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. $$\ \small\rm QED$$