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https://wj32.org/wp/wp-content/uploads/2012/12/advanced-linear-algebra.pdf

https://www.physicsforums.com/threads/about-r-module.313248/

http://isites.harvard.edu/fs/docs/icb.topic256346.files/Set%207.pdf

http://121.192.180.130:901/media/5225/homework13.pdf

http://121.192.180.130:901/media/5252/2015-01-07abstract%20algebra32.pdf

These are proofs that I can find so far. But among all of them, I cannot understand a point that:

I think all of these proofs only shows that if $R/I$ is a free module over R, than R must be either a field or the zero ring.

But I cannot see how does it imply the fact that: If every finitely generated R-module is free, then R must be either a field or the zero ring.

Could someone tell me how to show that? Thank so much!

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  • $\begingroup$ Is $R$ any ring or any commutative ring? $\endgroup$ Oct 16, 2016 at 4:26
  • $\begingroup$ R is any commutative ring. $\endgroup$
    – Y.X.
    Oct 16, 2016 at 4:26
  • $\begingroup$ In fact, the property can be restated as follows: If every cyclic R-module is free, then R must be either a field or the zero ring. $\endgroup$
    – user26857
    Oct 16, 2016 at 7:27
  • $\begingroup$ The link to isites.harvard.edu is broken. I'm also unable to find any copy saved on the Wayback Machine. $\endgroup$ Jun 20, 2022 at 9:18

1 Answer 1

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Let $I \subseteq R$ be an ideal. Then $R/I$ is an $R$-module. Moreover, it is generated by the element $\overline{1} \in R/I$. So $R/I$ is a finitely-generated $R$-module.

Then by assumption $R/I$ is free. Assume for contradiction that $I$ is not equal to $0$ or $R$. First, since $R/I \neq 0$, $R/I$ is not spanned by the empty set. So it must have a basis $\{\overline{r_1}, \ldots \overline{r_k}\}$ which is nonempty. But since $I \neq 0$, we can choose some $i \in I\setminus \{0\}$, and then $i\cdot \overline{r_1} = 0$. Thus, the "basis" is actually linearly dependent.

This is a contradiction, and thus we conclude that $I$ must have been $0$ or $R$, i.e. $0$ and $R$ are the only ideals of $R$. This implies that $R$ is either the zero ring or a field.

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  • $\begingroup$ Sorry but I haven't even know about the concept of torsion-free... Could you just tell me why "if R/I is a free module over R, than R must be either a field or the zero ring" implies "If every finitely generated R-module is free, than R must be either a field or the zero ring"? $\endgroup$
    – Y.X.
    Oct 16, 2016 at 4:21
  • $\begingroup$ @Y.X. An $R$-module $M$ is torsion-free if for every $r \in R$ and every $m \in M$, then $rm = 0$ iff $r=0$ or $m=0$. $\endgroup$
    – Alex G.
    Oct 16, 2016 at 4:22
  • $\begingroup$ @Y.X. Suppose that every finitely generated $R$-module is free. Then in particular, the $R$-modules $R/I$ are all free, since they are finitely generated as I showed above. Thus, my proof goes through that $R$ must be $0$ or a field. $\endgroup$
    – Alex G.
    Oct 16, 2016 at 4:23
  • $\begingroup$ But are the R/I modules exhaust all the modules over R? Or all the R-modules are isomorphic to some R/I? $\endgroup$
    – Y.X.
    Oct 16, 2016 at 4:25
  • $\begingroup$ Over a ring with zero divisors, a free module need not be torsion free. $\endgroup$ Oct 16, 2016 at 4:25

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