Say $R$ is a ring, not necessarily a domain, and $M$ is an $R$-module. All rings are commutative with 1.  An element $m\in M$ is called <i>torsion</i> if $r.m=0$ for some regular element (non-zerodivisor) $r\in R$.  

(I learned this definition over non-domains from a lecture of Irena Swanson, and it's noted in the last paragraph of the <a href="http://en.wikipedia.org/wiki/Torsion_(algebra)#Definition">definition on Wikipedia</a>.)

It's easy to see that torsion elements localize to torsion elements (note $0$ is always torsion), because regular elements never disappear.  How about the converse?  If an element is locally torsion, is it torsion?  That is,

> If $f_1,\ldots,f_n$ generate the unit ideal in $R$, and $m\in M_{f_i}$ is torsion over $R_{f_i}$ for each $i$, then is $m$ torsion over $R$?

I started trying to make a high-dimensional variety as a counterexample, but instead wound up proving that <b>for $R$ Noetherian, torsionality is *stalk-local*</b> <sup><b>1</b></sup>, so in that case the answer is *yes*.  

So how about a proof or a counterexample for rings in general?  This will affect when and whether I think about sections of $O_X$-modules on a non-integral scheme $X$ as "torsion" or not... 

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<sup>1</sup> <b>Proof</b> of stalk-locality in Noetherian case: Say $m\in M$ is non-torsion, meaning $ann(m)$ is contained in the set of zero-divisors of $R$, which equals the [union of the associated primes of $R$](http://books.google.com/books?id=Fm_yPgZBucMC&pg=PA89&lpg=PA89&dq=union+of+the+associated+primes&source=bl&ots=7Ap1EuN2Oj&sig=ASntyjgExEkT4-4XV1g5s68aFb0&hl=en&ei=9bJjTImEAouosQPg3-jiBg&sa=X&oi=book_result&ct=result&resnum=10&ved=0CD8Q6AEwCQ#v=onepage&q=union%20of%20the%20associated%20primes&f=false).  By [prime avoidance](http://commalg.wiki-site.com/index.php/Prime_avoidance_lemma), $ann(m)$ is contained in some associated prime of $R$, say $p=ann(x)$ for $x\in R$.  I claim $m$ localizes to a non-torsion element of $M_p$.  If $r.m=0$ in $M_p$ for $r\in R$ (WLOG), it means $rs.m=0$ in $M$ for some $s\notin p$.  Now, $rs\in ann(m)\subseteq p$, so $r\in p$ and $rx=0$ in $R$.  But $x\neq 0$ in $R_p$ since $ann(x)$ is (contained in) $p$, so $r$ is not regular in $R_p$, as required.  Hence an element of $M$ which is torsion in every $M_p$ must be torsion in $M$.