# The module is square-free iff every factor is summand-square-free

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.

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

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, $$A=B=\frac{0}{0}=0$$.

• 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?!. Oct 1, 2022 at 19:02
• Is this rubbish proof?:) Oct 1, 2022 at 20:44
• I didn't say that. I'm just skeptical of your proof. Oct 1, 2022 at 22:22