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If $R$ is a nonzero commutative ring with identity and every submodule of every free $R$-module is free, then $R$ is a principal ideal domain.

What I don't know is how to show that every ideal is free. Once an ideal is free, for nonzero $u,v\in I$, $uv-vu=0$ shows that the ideal has only one basis. that is, the ideal is principal. Any help?

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  • $\begingroup$ Need to show that no two elements $u$, $v$ of $I$ are linearly independent. If one of them is zero, clear. If none, you just wrote some equality above. Btw, $R$ PID $\iff$ submodules of free modules are also free. $\endgroup$ – Orest Bucicovschi Oct 6 '14 at 6:13
  • $\begingroup$ @orangeskid but how to show that under the condition "submodules of free modules are also free", ideal is free? $\endgroup$ – 김김김 Oct 6 '14 at 6:44
  • $\begingroup$ Oh, an ideal is a submodule of $R$, which is free, with basis the element $1$. $\endgroup$ – Orest Bucicovschi Oct 6 '14 at 6:47
  • $\begingroup$ @orangeskid thank you so much. $\endgroup$ – 김김김 Oct 6 '14 at 6:54
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Consider $R$ as $R$-module, then this is free, with basis $\lbrace 1 \rbrace $.

Suppose $I \subseteq R $ is an ideal, $I \neq 0 $; then $I$ is a submodule, whence it is free. If $u, v \in I $ they are linearly dependent over $R$, because $vu -uv = 0 $.

This implies that if $B= \lbrace u_1, u_2, \ldots \rbrace $ is a basis for $I$ as $R$-module, then $|B| = 1 $, so $I = < w > $ and I is a principal ideal.

For the domain part, if $a \in R $ , $a \neq 0 $, suppose $ba = 0 $ with $b \neq 0 $. Then consider $I = <a> $. By the previous part $I$ has a basis $\lbrace w \rbrace $. We have $w = k a $ for $ k \in R $. Then $$ bw = bka = k (ba) = 0$$ But $w $ is linearly independent and $b \neq 0 $. This is a contradiction and so $ba \neq 0 $.

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  • $\begingroup$ @김김김: I've added the domain part $\endgroup$ – WLOG Oct 6 '14 at 7:01

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