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I'm looking for some simple counterexamples to prove that the following statements are false:

Let R be a commutative ring with 1 and M an R-module

a) M free ⇒ M indecomposable

b) M cyclic ⇒ M indecomposable

c) M cyclic ⇒ M free

d) M indecomposable ⇒ M cyclic

For a): I thought of ℤ² since ℤ² = ℤxℤ is free but ℤ² can be decomposed into ℤ² ≅ ℤ ⊕ ℤ

For b): I thought of ℤ/6ℤ as it is cyclic (because it is generated by the unity element) but decomposable into ℤ/6ℤ ≅ ℤ/3ℤ ⊕ ℤ/2ℤ

For c): maybe ℤ/2ℤ?

I can't think of a counterexample for c) and d) via ℤ. I would be very grateful if someone would help me!

  • When you give examples of a module (like $\mathbb Z/2\mathbb Z$ in your post, for example), you should always make it clear what underlying ring you are considering it a module over because the properties you are trying to verify can and do depend on this choice. For instance, $\mathbb Z/2\mathbb Z$ is free as a $\mathbb Z/2\mathbb Z$-module but not as a $\mathbb Z$-module. – Arkady Sep 07 '22 at 08:09

1 Answers1

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c) ℤ/2ℤ is cyclic but not free (it has torsion).

d) $(\mathbb{Q},+)$ is indecomposable ( Why is the additive group of rational numbers indecomposable?) but not cyclic.

Anne Bauval
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  • Thanks a lot! Could you maybe explain to me what a torsion in modules is? I haven't had this topic yet. So how can I prove that Z/2Z is not free? – Tinkerbell_ye Sep 07 '22 at 07:39
  • See https://en.m.wikipedia.org/wiki/Free_module or https://en.m.wikipedia.org/wiki/Free_abelian_group, and https://en.m.wikipedia.org/wiki/Torsion-free_module or https://en.m.wikipedia.org/wiki/Torsion-free_abelian_group. – Anne Bauval Sep 07 '22 at 07:49
  • Or simply: in a free $\mathbb Z$-module $\oplus_{i\in I}{\mathbb Z} e_i$, for any $x=\sum x_ie_i$, if $nx=0$ for some $n\in{\mathbb Z}^*$ then $nx_i=0$ hence $x_i=0$ (for all $i$) hence $x=0$, whereas in Z/2Z, $2[1\bmod2]=0$. – Anne Bauval Sep 07 '22 at 08:22