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!
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.