Let $R$ be a commutative ring with unit.

If $R\neq 0$ such that each finitely generated $R$-module is free then $R$ is a field.

In my notes there is the following proof:

We need to show that $\forall a\neq 0, Ra=R$.

Let $Ra\neq R$.

The $R$-module $R/Ra$ is $\neq 0$ since $Ra<R$ and it is finitely generated since it is generated over $R$ by $1+Ra$.

So, it is free $R$-module.

From an other proposition we have that $Ra=0$, so $a=0$. Contradiction.

So, $R=Ra$.


I haven't really understood that proof...

Why does it stand that $R/Ra$ is $\neq 0$ when $Ra<R$ and why is $R/Ra$ generated over $R$ by $1+Ra$ ?

  • $\begingroup$ If $Ra=R$ then $a$ has an inverse. $\endgroup$ – David C. Ullrich Jun 23 '16 at 19:01
  • 1
    $\begingroup$ The "we assume" at the start should be "we need to show that", it seems to me. $\endgroup$ – David C. Ullrich Jun 23 '16 at 19:01
  • $\begingroup$ That first "We assume" is odd. $\endgroup$ – DonAntonio Jun 23 '16 at 19:02
  • $\begingroup$ Ok, I will edit it... @DavidC.Ullrich $\endgroup$ – Mary Star Jun 23 '16 at 19:02
  • $\begingroup$ for understanding what they mean, can someone give an example of a finitely generated $\mathbb{Z}$ module which is not free ? $\endgroup$ – reuns Jun 23 '16 at 19:03

How about another proof? If $M$ is a free module over a commutative ring with identity $A$, then any two bases of $M$ have the same cardinality. This is a general fact about commutative rings ("invariant basis number"). That cardinality is called the rank of $M$.

If $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence of free $A$-modules, then you can show that $$\textrm{Rank } M' + \textrm{Rank } M'' = \textrm{Rank } M$$ One way to show this is to use the fact that free modules are projective, and an exact sequence of projective modules must split.

Now assume that all finitely generated modules over a commutative ring with identity $R$ are free. Let $0 \neq a \in R$...

What I'm writing in this paragraph isn't actually necessary for the proof; it's just related to the questions you asked: Suppose the $R$-module $Ra$ is properly contained in the $R$-module $R$. Then the $R$-module $R/Ra$ is not zero, because if $x$ is an element of $R$ which is not in $Ra$, then $x+Ra$ is not the zero element of $R/Ra$ (if it was, then $x+Ra = 0 + Ra$ implies $x \in Ra$).

...The modules $Ra, R/Ra$ are finitely generated: the $R$-module generated by $1+Ra$ is $R/Ra$, and the $R$-module generated by $a$ is $Ra$. The second claim is clear, the first one I believe you asked why in your question: if $r + Ra$ is any element of $R/Ra$ (for $r \in R$), then $r + Ra = r \cdot (1+Ra)$. This is what it means for $1+Ra$ to generate $R/Ra$.

Now the sequence $$0 \rightarrow Ra \rightarrow R \rightarrow R/Ra \rightarrow 0$$ is exact. So the rank of $R$, which is $1$, is equal to the rank of $Ra$ plus the rank of $R/Ra$. A free module over a commutative ring with identity is the zero module if its rank is zero; otherwise it has positive rank. This forces one of the modules $Ra, R/Ra$ to be zero. But $Ra$ is not the zero module.

So $R/Ra$ is the zero module. This means that $R = Ra$.

  • $\begingroup$ Why the downvote? I'd like to improve this answer if possible. $\endgroup$ – D_S Jul 2 '16 at 21:00

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