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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?

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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?

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  • $\begingroup$ Thanks a lot.This really helps. $\endgroup$ – Sky Oct 6 '18 at 15:16
  • $\begingroup$ 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? $\endgroup$ – Sky Oct 8 '18 at 6:58
  • $\begingroup$ @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. $\endgroup$ – egreg Oct 8 '18 at 7:24
  • $\begingroup$ I find an interesting proof in GTM13(rings and categories of modules) proposition 17.14: Every non-zero projective module contains a maximal submodule. $\endgroup$ – Sky Oct 8 '18 at 9:19
  • $\begingroup$ @Sky Then use it, I was too lazy for looking into the book. ;-) $\endgroup$ – egreg Oct 8 '18 at 9:28

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