1. May 11, 2009

### sanctifier

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.

2. May 11, 2009

### VKint

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 $$\langle x,y \rangle$$ in the $$k[x,y]$$-module $$k[x,y]/\langle xy \rangle$$, 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
$$R = k [ x,\sqrt{x}, \sqrt[3]{x}, \ldots, \sqrt[n]{x}, \ldots ]$$.
This is a non-Noetherian domain. The R-module $$R/\langle x \rangle$$ has the desired property (I think).

3. May 20, 2009

### sanctifier

Thanks! VKint