As we all know, every ring with unit has left maximal ideals; then we can define the radical of a ring as the intersection of all left maximal ideal.

But for modules, not every module has maximal submodules, for instance the $$\mathbb{Z}$$-module $$\mathbb{Q}$$ doesn't have maximal submodules. But in almost every book, the definition of radical of a module is as the intersection of all maximal submodules. This has confused me for a long time.

1. Does this really make sense for all modules? How to explain this?

2. For the $$\mathbb{Z}$$-module $$\mathbb{Q}$$, what is $$\operatorname{Rad}\mathbb{Q}$$?

There are many questions about radical of modules in Mathmatics like Question about radical of a module.. In the proof of this question, it is also used the existence of maximal submodules. If we use the equivalent result in this question, we have $$\operatorname{Rad}\mathbb{Q}=\mathbb{Q}$$.

3. Every nonzero projective module has maximal submodules. How to prove this?

By convention, the intersection of the empty family of submodules of $$M$$ is $$M$$.

Therefore, the radical of $$\mathbb{Q}$$ is $$\mathbb{Q}$$ because the module has no maximal submodules.

This is consistent with the characterization of $$\operatorname{Rad}M$$ as the sum of all inessential submodules of $$M$$. A submodule $$L$$ is inessential (also called superfluous) if, for every submodule $$X$$ of $$M$$, $$L+X=M$$ implies $$X=M$$.

The family of all inessential submodules is never empty, because clearly $$\{0\}$$ is inessential. Suppose $$L$$ is inessential and $$N$$ is a maximal submodule. Then either $$L\subseteq N$$ or $$L+N=M$$; the latter case leads to a contradiction, so $$L\subseteq N$$. Therefore $$N$$ contains every inessential submodules, hence also the sum thereof.

If we call $$S$$ the sum of all inessential submodules, we have proved that $$S\subseteq\operatorname{Rad}(M)$$.

Conversely, if $$x\in\operatorname{Rad}(M)$$, then $$Rx$$ is inessential. Suppose $$Rx+X=M$$, but $$X\ne M$$. If $$x\in X$$, then $$X=M$$. Suppose $$x\notin X$$. Then $$M/X=(Rx+X)/X\cong Rx/(X\cap Rx)$$ is a nonzero finitely generated module, which has a maximal submodule; its inverse image $$N$$ in $$M/X$$ is then a maximal submodule and $$x\notin N$$. Contradiction. This proves that $$Rx\subseteq S$$. As $$x$$ is an arbitrary element of the radical, we have proved that $$\operatorname{Rad}M\subseteq S$$.

In the case $$M$$ has no maximal submodule, the proof above shows that, for every $$x\in M$$, $$Rx$$ is inessential, hence the sum of all inessential submodules is $$M$$.

Note that in no part of the proof we relied on the assumption that $$M$$ actually has maximal submodules. We just used that a finitely generated module has them. Similarly does the proof you're referring to: it doesn't state that a maximal submodule exists; it says that *if $$N$$ is a maximal submodule, then…”.

Why do projective modules have maximal submodules? If $$P$$ is projective, then $$F=P\oplus Q$$ is free, for a suitable module $$Q$$. Free modules do have maximal submodules (easy proof). Let $$N$$ be maximal in $$F$$; then $$P/(P\cap N)\cong(P+N)/N$$ If $$P+N=F$$ for some $$N$$, we are done, because then $$P/(P\cap N)$$ is simple. Otherwise $$P$$ is contained in the radical of $$F$$. Is this possible?

• Thanks a lot.This really helps. – Sky Oct 6 '18 at 15:16
• Sorry.what I thought before is not right.I findI can't prove why $P$ can't contain in $Rad F$ now.If $P\subset Rad F$,then we can show $P=Rad P$.Hence $P=rP$,where $r$ is the radical of ring.So how to get a contradiction?I still have two question :1. Is there an example of module $M$ such that $M=rM$ where $M$ has maximal submodules? 2.If $M$ is a module which has maximal submodules,is $Rad M$ superflous? – Sky Oct 8 '18 at 6:58
• @Sky Use the fact that $F$ is free, so it is a direct sum of copies of $R$; take a direct sum of $R$, except at a coordinate where you take a maximal left ideal. This is a maximal submodule. – egreg Oct 8 '18 at 7:24
• I find an interesting proof in GTM13(rings and categories of modules) proposition 17.14: Every non-zero projective module contains a maximal submodule. – Sky Oct 8 '18 at 9:19
• @Sky Then use it, I was too lazy for looking into the book. ;-) – egreg Oct 8 '18 at 9:28