# The condition that a ring is a principal ideal domain

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?

• 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. – Orest Bucicovschi Oct 6 '14 at 6:13
• @orangeskid but how to show that under the condition "submodules of free modules are also free", ideal is free? – 김김김 Oct 6 '14 at 6:44
• Oh, an ideal is a submodule of $R$, which is free, with basis the element $1$. – Orest Bucicovschi Oct 6 '14 at 6:47
• @orangeskid thank you so much. – 김김김 Oct 6 '14 at 6:54

## 1 Answer

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 $ba \neq 0$.

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