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

[mathmari]

Hey!

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)

 

[mathmari]

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)

 

[Euge]

Hey!

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, $J = 0$.

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.

 

[mathmari]

It suffices to show that $R$ is a simple $R$-module.

Why does this suffice? (Wondering)

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)

 

[Euge]

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.

 

[Euge]

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

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".