About R-module

  • Thread starter sanctifier
  • Start date
  • #1
58
0

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!
 

Answers and Replies

  • #2
139
12
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?
 
  • #3
58
0
Thanks for reply! VKint

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.
 
  • #4
139
12
By cyclic, I mean that the module [tex] R/I [/tex] can be generated by just one element (i.e., the coset [tex] 1 + I [/tex]) over [tex] R [/tex]. Thus, since [tex] R/I [/tex] is finitely generated, you can apply your assumption to the [tex] R [/tex]-module [tex] R/I [/tex] to conclude that [tex] R/I [/tex] 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 [tex] R [/tex] can't have any nonzero ideals (besides itself, of course). Can you see why it would be contradictory for [tex] R/I [/tex] to be free if [tex] I [/tex] were not either [tex] 0 [/tex] or [tex] R [/tex]?
 
  • #5
58
0
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.
 
Last edited:

Related Threads for: About R-module

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
818
  • Last Post
Replies
2
Views
5K
Replies
12
Views
565
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
Top