# Prove that if a nonzero commutative ring R with identity has the property that every finitely generated R-module is free then R is a field.

Prove that a ring R having the property that every ﬁnitely generated R-module is free is either a ﬁeld or the zero ring.

This question has been asked before at the link above, but I am trying to prove it in a somewhat different vein, largely because I spent a while on it before looking for alternative answers and can't get past a certain point. Maybe some of my intermediate steps are incorrect.

Let $v\in R$, $v\neq0$. Consider the cyclic $R$-module $\langle\langle v \rangle\rangle = \{rv\mid\ v\in R\}$. This is a finitely generated $R$-module and so has a basis by assumption, specifically $\{v\}$. Then $\{v\}$ is linearly independent, and so for any $r\in R$, $rv=0$ implies that $r=0$. Since $v$ is an arbitrary non-zero element in $R$, we can apply the same method to any other non-zero element $u$ of $R$ and conclude that for any nonzero $u \in R$, $ru=0$ implies $r = 0$. So $R$ has no zero divisors and is thus an Integral Domain.

First of all, are the assertions I have made thus far correct? If not, please let me know. If this is correct to this point, is there a way to extend this to demonstrate the existence of multiplicative inverses in $R$ and thus to show that $R$ is a field in a way that does not rely on the $R/I$ approach used in the link above?

• Nothing wrong so far, but you may want to use $r.(s.v)=v$ is also going to have a solution. – David Hill Feb 21 '18 at 3:18
• Thanks for the response. I can't see how to establish the above relationship though. My guess is that if I could establish that for arbitrary nonzero $s$ there exists $r$ such that $r.(s.v)=v$, then this implies $(r.s-1).v=0$, and so $r.s=1$ since there are no zero divisors. Then $s$ has inverse $r$. First, if you could elaborate a little bit more on your comment, I would really appreciate it. Second, are the statements I have just made a valid approach, provided I can establish $r.(s.v)=v$? – mrmingus Feb 21 '18 at 10:22
• Yes, this is how you proceed. – David Hill Feb 21 '18 at 15:12
• OK. I still am having problems with establishing that $r.(s.v)=v$. My initial thought was to use a result from a previous theorem that if $R$ is an integral domain and $M$ is an $R-module$ with a subset $S$ of linearly independent vectors, then if I multiply each element in $S$ by a non-zero element $s$ in $R$, the set $sS$ will be linearly independent over $R$ as well, but no dice. Are you saying that, given the fact that $R$ is an integral domain and the assumptions of the problem as stated, I should be able to demonstrate that $v$ is contained in the submodule $<<sv>>$ for nonzero $s$? – mrmingus Feb 21 '18 at 17:37
• Line 2: can you explain why $\{v\}$ is a basis for $Rv$? – Delong Feb 22 '18 at 1:59