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

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In summary, the result of the thread says that if an ideal of a free $R$-module is a free $R$-module, then it is a principal ideal that is generated by an element that is not a zero divisor in $R$. This means that $R$ is a PID.
  • #1
mathmari
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Hey! :eek:

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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  • #2
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.
 
  • #3
Euge said:
a nonzero ideal of $R$ is an $R$-submodule of the free $R$-module $R$;

Why does this stand? (Wondering)
 
  • #4
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 $R$.
 
  • #5
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)
 
  • #6
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.)
 
  • #7
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)
 

1. What is a P.I.D.?

A P.I.D. (principal ideal domain) is a type of commutative ring in which every ideal can be generated by a single element. In other words, every ideal in a P.I.D. is "principal". Examples of P.I.D.s include the integers and polynomial rings over a field.

2. What is the significance of a ring being a P.I.D.?

P.I.D.s are important in abstract algebra and number theory because they have many useful properties. For example, in a P.I.D., every irreducible element is prime, and every ideal can be uniquely factored into prime ideals. This makes P.I.D.s useful for solving problems involving divisibility and factorization.

3. How can I tell if a given ring is a P.I.D.?

There are a few different methods for determining if a ring is a P.I.D. One approach is to check if every nonzero prime ideal is maximal. Another method is to use the structure theorem for finitely generated modules over P.I.D.s, which states that any finitely generated module over a P.I.D. is isomorphic to a direct sum of cyclic modules.

4. What are free modules and how do they relate to P.I.D.s?

A free module is a type of module that has a basis, meaning it can be generated by a linearly independent set of elements. In a P.I.D., every submodule of a free module is also free, and every finitely generated module is isomorphic to a quotient of a free module. This is one of the key properties that makes P.I.D.s useful for studying modules.

5. Can a non-commutative ring be a P.I.D.?

No, a non-commutative ring cannot be a P.I.D. because the definition of a P.I.D. requires the ring to be commutative. However, there are similar types of rings in non-commutative algebra, such as division rings and skew fields, that have properties similar to P.I.D.s.

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