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 torsion 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 definition on Wikipedia.)

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 for $R$ Noetherian, torsionality is stalk-local, so in that case the answer is yes.¹

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...

¹ Proof 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 $R$. By prime avoidance, $ann(m)$ 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 $M$.