Recall that a right $R$-module $M_R$ is called square-free if whenever $A,B\leq M$ are submodules with $A\cap B=0$ and $A\cong B=0$ then $A=B=0$.

A submodule $N$ of a module $M$ is called a summand (written $N \leq^{\oplus}M$) if $M$ is the internal direct sum $M=N\oplus K$ for some submodule $K\leq M$.

$M_R$ is called summand-square-free if whenever $A,B$ are summands with $A\cap B=0$ and $A\cong B$ then $A=B=0$.

If $M$ is a module, by a factor module of $M$ we mean a quotient $M/N$ for some $N\leq M$.

Indeed, any square-free module is summand-square-free but the converse is not necessarily true.

We have the following characterization:

$M_R$ is square-free $\iff$ Every factor module of $M$ is summand-square-free.

How can I prove this statement?!.

I appreciate any help. Thanks in advance.

  • $\begingroup$ $M\cong M/0$. @MarianoSuárez-Álvarez $\endgroup$ Oct 1, 2022 at 19:04
  • $\begingroup$ Of course not. @MarianoSuárez-Álvarez $\endgroup$ Oct 1, 2022 at 22:21

1 Answer 1


Only if part

Assume that $M_R$ is summand-square-free.

Let $A,B\leq M$ be submodules such that $A\cap B=0$ and $A\cong B$.

Then $\frac{A+B}{B}$ and $\frac{A+B}{A}$ are factor modules and they are isomorphic since $\frac{A+B}{B}\cong A$ and $\frac{A+B}{A}\cong B$.

It also follows that $\frac{A+B}{B}\cap\frac{A+B}{A}=0$ since $A\cap B=0$.

Then, by assumption, $\frac{A+B}{B}=\frac{A+B}{A}=0$. This implies $A=B=0$.

If part

Assume that $M_R$ is square-free.

Let $A=\frac{N_1}{M_1}$ and $B=\frac{N_2}{M_2}$ be two factor modules with $A\cap B=0$ and and $A\cong B$.

Then, if $A'=\frac{N_1+N_2}{M_1+N_2}$ and $B'=\frac{N_1+N_2}{M_2+N_1}$ then $A'\cong B'$ and $A'\cap B'=0.$ So without loss of generality we may assume that $A=\frac{N}{M_1}$ and $B=\frac{N}{M_2}$.

Then, if $A''=\frac{\frac{N}{M_1\cap M_2}}{\frac{M_1}{M_1\cap M_2}}$ and $B''=\frac{\frac{N}{M_1\cap M_2}}{\frac{M_2}{M_1\cap M_2}}$ then $A''\cong B''$ and $A''\cap B''=0.$ So without loss of generality we may assume that $A=\frac{N}{M_1}$ and $B=\frac{N}{M_2}$ such that $M_1\cap M_2=0$.

Without loss of generality we assumed that $A=\frac{N}{M_1}$ and $B=\frac{N}{M_2}$ such that $M_1\cap M_2=0$. We have also the conditions $A\cong B$ and $A\cap B=0$. It then follows that $M_1\cong M_2$.

Now, $M_1\cong M_2$ and $M_1\cap M_2=0$ imply that $M_1=M_2=0$ since $M_R$ is square-free.

Then, $A\cap B=N\cap N=N=0$. Hence, $A=B=\frac{0}{0}=0$.

  • 1
    $\begingroup$ Thanks dear for you efforts. But I do not think that the proof is true for the following reasons: In the proof of the Only if part you assumed in the first line that $M_R$ is summand-square-free whereas $M$ is assumed in the statement of the problem to have summand-square-free factors. Also, what is the factor module of $M$ (a quotient of the form $M/N$) of which $\frac{A+B}{A}$ and $\frac{A+B}{B}$ are summands to apply the hypothesis?!. $\endgroup$ Oct 1, 2022 at 19:02
  • $\begingroup$ Is this rubbish proof?:) $\endgroup$
    – Bob Dobbs
    Oct 1, 2022 at 20:44
  • $\begingroup$ I didn't say that. I'm just skeptical of your proof. $\endgroup$ Oct 1, 2022 at 22:22

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