## Main Question or Discussion Point

Notations:
R denotes a commutative ring with identity

Terms:
R-module: a module whose base ring is R

Question:
Prove that if a nonzero commutative ring R with identity has the property that every finitely generated R-module is free then R is a field.

Idon't know how to complete the proof. The only thing I know is that if every finitely generated R-module is free then R-module R is also free, since it's generated by {1}.

Thanks for any help!

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Hint: If I is an ideal of R, then R/I is a cyclic (in particular, finitely generated) R-module. If R/I is free, what does this tell you about I?

What exactly does the term cyclic mean here? I am not familar with some terms.

There may be some grammatical mistakes above, sorry for that, I've done my best to express what I want to say.

By cyclic, I mean that the module $$R/I$$ can be generated by just one element (i.e., the coset $$1 + I$$) over $$R$$. Thus, since $$R/I$$ is finitely generated, you can apply your assumption to the $$R$$-module $$R/I$$ to conclude that $$R/I$$ must be free.

One way of defining a field is as a ring with no nontrivial ideals. Thus, all you need to show is that $$R$$ can't have any nonzero ideals (besides itself, of course). Can you see why it would be contradictory for $$R/I$$ to be free if $$I$$ were not either $$0$$ or $$R$$?

VKint, I had some trouble in my work recently, so I didn't reply in time, I'm sorry.

I don't kown whether the following is right, may you can help me again, thanks a lot!

If v∈M\I and i∈I, then v and any i are relatively prime, if not, then there's a scalar s which divides both v and i, then s is contained in I, otherwise the scalar i divided by s doesn't belong to I.

So, there exist scalars s1 and s2 for which s1v + s2i = 1,
then s1v + s2i + I = 1 + I → s1v + I = 1 + I
since s2i∈I, so s1v=1, that is to say s1 IS the multiplicative inverse of v.

The left could be solved by induction.

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