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Let $M=\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}$ as a $\mathbb{Z}$-module. Is M free?

How can I go about this? For $\mathbb{Z}/2\mathbb{Z}$ I know $0$ can't be a basis element and $1$ gives $2.1=0$ so cant be a basis element too.

For my problem this won't work. I think it might free with basis $(a,1 or 3)$.

Also second part says Give an example of a ring $R$ for which every finitely generated $R$-module $M$ is free.

Can I take the polynomial ring?

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In general, for any (commutative) ring $R$, a free $R$-module is torsion free. To see this, take a free $R$-module $M$, and without loss of generality assume $M = R^n$. Take $(r_1,...,r_n) \in M$ with $r \in R$ a nonzero divisor such that $r(r_1,...,r_n)=(rr_1,rr_2,...,rr_n)=0$. Then $rr_i=0$ for all $i$ so $r_i=0$, for all $1 \leq i \leq n$.

Now take $M= \mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}$ and consider $(0, 2) \in M$. Then $2(0,2)=0$. Since $2$ is not a zero divisor in $\mathbb{Z}$, it must be that $M$ has torsion, and hence is not free.

For the second part of the question, take $R=k$ a field. Then modules over $R$ are vector spaces so all modules are free.

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    $\begingroup$ @rschwieb My definition of torsion is the following: An element $m \in M$ is torsion if there exists a non zero divisor $r \in R$ such that $rm=0$. In the case of $\mathbb{Z}/4\mathbb{Z}$ as a module over itself, no regular element (non zero divisor) annihilates a nonzero element of the ring, as then it would be, well, a zero divisor. In the case of modules over a domain, you don't have to specify the extra condition that $r \in R$ should be regular $\endgroup$ – leibnewtz May 18 '17 at 3:15
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    $\begingroup$ ok, fair enough: that one works for what you want to do. Thanks for clarifying $\endgroup$ – rschwieb May 18 '17 at 3:19
  • $\begingroup$ no problem, I think I've seen a couple definitions of torsion so it's a fair question $\endgroup$ – leibnewtz May 18 '17 at 3:23
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A summand of a free module is projective. Hence $\mathbb{Z}/4\mathbb{Z}$ should be projective.

However, the canonical epimorphism $\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}$ doesn't split, so $\mathbb{Z}/4\mathbb{Z}$ is not projective.


The condition that every finitely generated left $R$-module is free is equivalent to the ring $R$ being a division ring. Hint: every finitely generated module is projective, so the ring is semisimple and the Wedderburn-Artin theorem applies.

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