About a torsion module

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

Main Question or Discussion Point

Notations:
M denotes an abelian group under addtion
R denotes a commutative ring with identity
ann(?): Let M be an R-module and r∈R v∈M, then ann(v)={r∈R | rv=0}

Terms:
R-module: a module whose base ring is R
torsion element: A nonzero element v∈M for which rv=0 for some nonzero r∈R
torsion module: all elements of the module are torsion elements
integral domain: a commutative ring R with identity with the property that for r,s∈R, rs≠0 if r≠0 and s≠0

Question:
Find a torsion module M for which ann(M)={0}.

I wonder whether the base ring R of such M can only be an integral domain.
 

Answers and Replies

  • #2
139
12
On the contrary, if a commutative ring R has such a module M (and if M is finitely generated), then R cannot be an integral domain. Why? (Hint: In a domain, the intersection of nonzero ideals is nontrivial.)

The original example I thought of was the ideal [tex] \langle x,y \rangle [/tex] in the [tex] k[x,y] [/tex]-module [tex]k[x,y]/\langle xy \rangle[/tex], but that doesn't quite work. I tried tweaking it a bit, but to no avail. Thus, the example I finally settled on uses a non-Noetherian ring R to circumvent the above argument that R cannot be an integral domain. Consider
[tex] R = k [ x,\sqrt{x}, \sqrt[3]{x}, \ldots, \sqrt[n]{x}, \ldots ] [/tex].
This is a non-Noetherian domain. The R-module [tex] R/\langle x \rangle [/tex] has the desired property (I think).
 
  • #3
58
0
Thanks! VKint
 

Related Threads for: About a torsion module

  • Last Post
Replies
1
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
Replies
12
Views
565
Replies
3
Views
818
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
4K
Replies
2
Views
2K
Top