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

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Discussion Overview

The discussion revolves around the properties of a commutative ring \( R \) and its ideals, specifically exploring the relationship between free modules and whether \( R \) is a principal ideal domain (P.I.D.). The scope includes theoretical aspects of module theory and ideal properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant suggests that if each \( R \)-submodule of a free \( R \)-module is free, then \( R \) is a P.I.D.
  • Another participant agrees, stating that a nonzero ideal of \( R \) is an \( R \)-submodule of the free \( R \)-module \( R \), which is free by hypothesis.
  • A participant questions the reasoning behind why a nonzero ideal is considered an \( R \)-submodule of \( R \).
  • Responses clarify that a nonzero ideal \( N \) is closed under addition and scalar multiplication, thus qualifying as an \( R \)-submodule.
  • There is a query about whether every \( R \)-submodule of \( R \) is a nonzero ideal, leading to a clarification that while every ideal is an \( R \)-submodule, not every \( R \)-submodule is a nonzero ideal.

Areas of Agreement / Disagreement

Participants generally agree on the properties of ideals as \( R \)-submodules, but there is some uncertainty regarding the implications of these properties for concluding that \( R \) is a P.I.D. The discussion remains unresolved regarding the broader implications of the initial hypothesis.

Contextual Notes

There is an implicit assumption that the properties of ideals and submodules are well understood, but the discussion does not resolve the implications of these properties for the classification of \( R \) as a P.I.D.

mathmari
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Hey! :o

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)
 
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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.
 
Euge said:
a nonzero ideal of $R$ is an $R$-submodule of the free $R$-module $R$;

Why does this stand? (Wondering)
 
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 $R$.
 
Euge said:
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)
 
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.)
 
Euge said:
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)
 

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