Torsion Module M with ann(M)={0} and Integral Domain R | R-Module Notations

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The discussion focuses on identifying a torsion module M for which the annihilator ann(M) equals {0}. It establishes that if M is finitely generated and has this property, the base ring R cannot be an integral domain. The example provided involves the non-Noetherian ring R = k[x, √x, √[3]{x}, ..., √[n]{x}, ...], where the R-module R/⟨x⟩ demonstrates the desired characteristics. The conversation highlights the relationship between torsion modules and the structure of the underlying ring.

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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.
 
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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).
 
Thanks! VKint
 

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