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In a discussion on MO, I found someone claiming the following:

Proposition: For a commutative unital ring $R$, the following are equivalent:

(i) every submodule of a free $R$-module is free;
(ii) every ideal $I\!\trianglelefteq\!R$ is free as a $R$-module;
(iii) $R$ is a PID.

Proof: (i)$\Rightarrow$(ii) is obvious.

(iii)$\Rightarrow$(i) is theorem 8.6.1 in Grillet's Abstract Algebra.

But how can I prove (ii)$\Rightarrow$(iii)? I have to prove that $xy=0$ implies $x=0$ or $y=0$; and also that every $I\!\trianglelefteq\!R$ is of the form $I=(a)$. We know every $I\!\trianglelefteq\!R$ has a basis $B=\{b_i;i\!\in\!I\}$ and therefore $I\cong R^{(B)}$.

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1 Answer 1

up vote 5 down vote accepted

Re-written, for clarity

Lemma 1: A non-zero free $R$-module is either isomorphic to $R$ or contains a sub-module isomorphic to $R\oplus R$.

Proof: Obvious

Lemma 2: If $I\!\trianglelefteq\!R$, then $I$ cannot contain a sub-module isomorphic to $R\oplus R$.

Proof: If $I$ contains a sub-module isomorphic to $R\oplus R$, then there are $x_1,x_2\in I$ such that $\forall r_1,r_2\in R: r_1x_1 + r_2x_2 = 0$ if and only if $r_1=r_2=0.$ But we can see that $x_2(x_1) + (-x_2)x_1 = 0$, so such $x_1,x_2$ cannot exist.

Lemma 3: Given $r\in R$, if $I=(r)$ is isomorphic to $R$, then $r$ is not a zero-divisor of $R$.

Proof: Obvious.

Lemma 4: If $I\!\trianglelefteq\!R$ is a free $R$-module, then $I$ is principal.

Proof: By the lemmas 1 and 2 above, $I$ must be $(0)$ or isomorphic to $R$ as an $R$-module. But that means $I$ is principal.

Theorem: If every ideal of $R$ is a free $R$-module, then $R$ is a PID.

Proof: By Lemma 4, every ideal must be principal. By Lemma 3, $R$ cannot have zero divisors. So $R$ is a PID.

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Aha, this proves the basis has only $1$ element, so every ideal is principal. What about the non-existence of zero divisors? –  Leon Oct 25 '11 at 15:00
    
If $(r)$ is isomorphic to $R$ as $R$-modules, then $r$ cannot be a zero-divisor. –  Thomas Andrews Oct 25 '11 at 15:09
    
Okay, made the "no zero divisors" part more explicit. –  Thomas Andrews Oct 25 '11 at 15:26
    
Great, thank you very much :). –  Leon Oct 25 '11 at 15:27

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