Prove $R$ is a Field: Finitely Generated Modules are Free

  • Context: MHB 
  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Field
Click For Summary

Discussion Overview

The discussion revolves around the question of whether a commutative ring \( R \) with unit, where every finitely generated \( R \)-module is free, can be proven to be a field. Participants explore various implications and definitions related to finitely generated modules and their properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that showing \( R \) is a simple \( R \)-module suffices to conclude that \( R \) is a field.
  • It is suggested that if \( J \) is a proper ideal of \( R \), then \( R/J \) being a finitely generated \( R \)-module implies it is free by the hypothesis.
  • One participant questions the sufficiency of the condition that \( R/J \) is finitely generated and seeks clarification on why this follows.
  • Another participant raises a concern about the treatment of the zero module, arguing that if \( 0 \) is considered a finitely generated \( R \)-module, it leads to a contradiction regarding the assumption that \( R \neq 0 \).
  • There is a discussion about the definition of finitely generated modules and whether the zero module should be included in this definition.
  • Some participants assert that a finitely generated \( R \)-module need not have a basis, while those that do are defined as free.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the definitions and properties of finitely generated modules. There is no consensus on whether the inclusion of the zero module affects the validity of the original problem statement.

Contextual Notes

There are unresolved questions regarding the definitions of finitely generated modules and the implications of considering the zero module as finitely generated. The discussion also highlights the dependence on specific assumptions about ideals and modules.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

Let $R$ be a commutative ring with unit.

I want to show that if $R\neq 0$ such that each finitely generated $R$-module is free then $R$ is field. Could you give me some hints how we could show that? (Wondering)
 
Physics news on Phys.org
Suppose that if $I$ is an ideal of the commutative ring $R$ and it is a free $R$-module, then it has a basis, i.e., a generating set consisting of linearly independent elements.

Does a finitely generated $R$-module is free when the basis is finite? (Wondering)
 
mathmari said:
Hey! :o

Let $R$ be a commutative ring with unit.

I want to show that if $R\neq 0$ such that each finitely generated $R$-module is free then $R$ is field. Could you give me some hints how we could show that? (Wondering)

It suffices to show that $R$ is a simple $R$-module. Let $J$ be a proper ideal of $R$. Then $R/J$ is a finitely generated $R$-module, hence free by hypothesis. Free modules have trivial annihilators, so $\operatorname{Ann}_R(R/J) = 0$. On the other hand, $\operatorname{Ann}_R(R/J) = J$. Hence, $J = 0$.
mathmari said:
Suppose that if $I$ is an ideal of the commutative ring $R$ and it is a free $R$-module, then it has a basis, i.e., a generating set consisting of linearly independent elements.

Does a finitely generated $R$-module is free when the basis is finite? (Wondering)

That does not make sense. A finitely generated $R$-module need not have a basis, but the ones that do are free by definition.
 
Euge said:
It suffices to show that $R$ is a simple $R$-module.

Why does this suffice? (Wondering)
Euge said:
Let $J$ be a proper ideal of $R$. Then $R/J$ is a finitely generated $R$-module, hence free by hypothesis.

Why does it follow that $R/J$ is a finitely generated $R$-module? (Wondering)
 
It suffices to show that $R$ is simple, since then $aR = R$ for every nonzero $a\in R$, and consequently, all nonzero elements are invertible.

Since $R/J$ is a quotient of a finitely generated $R$-module (namely, $R$ itself), then $R/J$ is finitely generated as an $R$-module.
 
Hi mathmari,

I don't think this is debatable, but is $0$ considered not to be a finitely generated $R$-module in your class? If that's not the case, then there is an issue with the problem statement

mathmari said:
Let $R$ be a commutative ring with unit.

I want to show that if $R\neq 0$ such that each finitely generated $R$-module is free then $R$ is field.

If we say an $R$-module $M$ is finitely generated if there is a surjection $R^n \to M$ for some positive integer $n$, then $0$ is finitely generated. In the hypothesis of your problem, $0$ would also be a free $R$-module, which would imply $0 \cong R^n$ for some positive integer $n$; this forces $R = 0$, contrary to assumption. The statement would be correct if "each finitely generated $R$-module is free" is replaced with "each finitely generated nontrivial $R$-module is free".
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
20
Views
4K
  • · Replies 24 ·
Replies
24
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K