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Let $R$ be a commutative ring and $M$ a finitely generated torsion $R$-module. Then the ideal $\mathrm{Ann}(M)$ is nonzero?

If $R$ is integral domain, this trivially holds since any two ideals have nontrivial intersections.

Add: I say $m\in M$ is torsion, if there is a nonzero element $r\in R$ such that $rm=0$.

user26857
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J.Li
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  • I gambled that you would not use this definition, since it is relatively rare. As it happens, my dissertation was based on the notion of torsion you used, so it's a bit tragicomic that you wanted that version after all. For noncommutative rings, you can even have cyclic modules which are faithful, and yet every element has nonzero annihilator. – rschwieb Apr 15 '19 at 13:51

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With your definition of torsion, this is false. For instance, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$. Let $M$ be the $R$-module with generators $a,b$ and relations $ya=0$, $xb=0$, and $xa=yb$. Note that $M$ is 3-dimensional over $k$, with basis $\{a,b,c\}$ where $c=xa=yb$.

Every element of $M$ is annihilated by some nonzero element of $R$. Indeed, given an element $m=ra+sb+tc$ with $r,s,t\in k$, then $sx-ry$ annihilates $m$ and is nonzero as long as $r$ and $s$ are both not $0$. If $r=s=0$, then $m$ is annihilated by $x$.

However, there is no nonzero element of $R$ that annihilates all of $R$. Indeed, the annihilator of $a$ is $(y)$, the annihilator of $b$ is $(x)$, and the intersection $(x)\cap (y)$ is trivial.

Eric Wofsey
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  • This appears to be the example mentioned in Hutchins Examples of commutative rings example 173, attributed to Kaplansky. It didn't seem to be fairly well known (at least, not during the years I was asking mathematicians for an example of such a thing) but it fell into my lap at last. May I ask how you found out about it? – rschwieb Apr 15 '19 at 13:50
  • I just found it by trial and error. I started with elements $a$ and $b$ which were annihilated by $y$ and $x$ respectively but not annihilated by any common element since $xy=0$. Then I fiddled with making sure that every linear combination of $a$ and $b$ is also torsion, and came upon the relation $xa=yb$ which would ensure that. – Eric Wofsey Apr 15 '19 at 14:41
  • Interesting. It's just a little surprising, since nearly all of the notation matches! I think there is a transposition though, rather $ann(a)=(y)$ and $ann(b)=(x)$. – rschwieb Apr 15 '19 at 16:27