MHB Wondering if $R$ is a P.I.D.: Free Modules and Ideals

[mathmari]

Hey!

Let $R$ be a commutative ring with unit.
I want to show that if each $R$-submodule of a free $R$-module is free then $R$ is P.I.D..

From the thread http://mathhelpboards.com/linear-abstract-algebra-14/how-can-we-conclude-i-principal-ideal-18593.html we have that if an ideal of $R$ is a free $R$-module then it is a principal ideal that is generated by an element $a$ that is not a zero-divisor in $R$.

Do we conclude from that that $R$ is P.I.D. ? (Wondering)

 

[Euge]

Hi mathmari,

Yes, you can use that fact to conclude $R$ is a PID. After all, a nonzero ideal of $R$ is an $R$-submodule of the free $R$-module $R$; it is free by hypothesis. So by the result you quoted, the ideal is principal generated by an element that is not a zero divisor.

 

[mathmari]

a nonzero ideal of $R$ is an $R$-submodule of the free $R$-module $R$;

Why does this stand? (Wondering)

 

[Euge]

A nonzero ideal $N$ of $R$ must closed under addition and satisfy the condition that $rn\in N$ for all $r\in R$ and $n\in N$. Thus $N$ is closed under addition and scalar multiplication, making $N$ an $R$-submodule of $R$.

 

[mathmari]

A nonzero ideal $N$ of $R$ must closed under addition and satisfy the condition that $rn\in N$ for all $r\in R$ and $n\in N$. Thus $N$ is closed under addition and scalar multiplication, making $N$ an $R$-submodule of $R$.

So, we have that each nonzero ideal of $R$ is an $R$-submodule of $R$.
Is also each $R$-submodule of $R$ a nonzero ideal of $R$ ? (Wondering)

 

[Euge]

Not quite. Every ideal (yes, including $0$) is an $R$-submodule. So, to put it plainly, the $R$-submodules of $R$ are the ideals of $R$. (I just focused on nonzero ideals for the sake of this particular problem.)

 

[mathmari]

Not quite. Every ideal (yes, including $0$) is an $R$-submodule. So, to put it plainly, the $R$-submodules of $R$ are the ideals of $R$. (I just focused on nonzero ideals for the sake of this particular problem.)

I understand! Thank you very much! (Yes)