Does a section that vanishes at every point vanish? Let $R$ be the coordinate ring of an affine complex variety (i.e. finitely generated, commutative, reduced $\mathbb{C}$ algebra) and $M$ be an $R$ module.
Let $s\in M$ be an element, such that $s\in \mathfrak{m}M$ for every maximal ideal $\mathfrak{m}$. Does this imply $s=0$?

Not in general, no.  For example, if $R = \mathbb C[T]$ and $M$ is the field of fractions
of $R$, namely $\mathbb C(T)$, then (a) every maximal ideal of $R$ is principal; (b) every
element of $M$ is divisible by every non-zero element of $R$.  Putting (a) and (b) together
we find that $M = \mathfrak m M$ for every maximal ideal $\mathfrak m$ of $R$, but certainly
$M \neq 0.$
Here is a finitely generated example: again take $R = \mathbb C[T]$, and take $M = \mathbb C[T]/(T^2).$  Then $s = T \bmod T^2 \in \mathfrak m M$ for every $\mathfrak m$, because
$\mathfrak m M = M$ if $\mathfrak m$ is a maximal ideal other than $(T)$, and this
is clear from the choice of $s$ if $\mathfrak m = (T)$.
The answer is yes if $M$ is finitely generated and torsion free.  For let $S$ be the total quotient ring of 
$R$ (i.e. the product of functions fields $K(X)$ for each irreducible component $X$
of the variety attached to $R$).
Then $M$ embeds into $S\otimes_R M$ (this is the torsion free condition), which in turn embeds into
a finite product of copies of $S$ (since it is finite type over $S$, which is just a product
of fields).
Clearing denominators, we find that in fact $M$ then embeds into $R^n$ for some $n$.
Thus it suffices to prove the result for $M = R^n$, and hence for $R$, in which case
it follows from the Nullstellensatz, together with the fact that $R$ is reduced.
Finally, note that for any finitely generated $R$-module, if $M = \mathfrak m M$ for all $\mathfrak m$
then $M = 0$ (since Nakayama then implies that $M_{\mathfrak m} = 0$ for all
$\mathfrak m$).  Thus if $M$ is non-zero it can't be that every section lies
in $\mathfrak m M$ for all $\mathfrak m$.
