The question is originally coming from a non-affine case, but I will only consider the affine one here.
Let $R$ be a ring which is reduced and has only a finite number of minimal prime ideals $P_1,\ldots,P_r$. Let $M$ be a finitely generated and torsion-free $R$-module (where the latter means that no regular element in $R$ can annihilate a non-zero element of $M$). There is a canonical map
$$R \to \prod_{i=1}^r R/P_i$$
which is injective since its kernel is given by $\bigcap_{i=1}^r P_i = \operatorname{Rad}(R) = 0$.
If we tensor this morphism with $M$, is it still injective? That is, is the morphism
$$\varphi:M \to \prod_{i=1}^r M/P_iM$$
injective?
It certainly is (by definition) if $M$ is flat.
- But does the torsion-freeness suffice?
- If not, can you give a counter example?
- What are far less restrictive assumption on $M$ such that $\varphi$ is injective?
Of course, the reduceness of $R$ would help here again if $\bigcap_{i=1}^r P_iM \stackrel{??}{=} (\bigcap_{i=1}^r P_i)M = 0$. But I don't see why this should be true.
Edit: I forgot to mention that I work with curves and thus we also have $\dim R = 1$.
Best Answer
$\newcommand{\ass}{\operatorname{Ass}} \newcommand{\p}{\mathfrak{p}} \newcommand{tm}{\subseteq}$
Proof: By Tag 0AVL it suffices to show that the induced local homomorphism for all associated primes of $M$ is injective. The torsion-free assumption on $M$ provides $\ass(M) \tm \ass(R)$ (see Torsionfree and associated primes) and thus it suffices to show the injectivity at minimal primes of $R$. Let $\p \in \ass(R)$ be a minimal prime of $R$. Then the localized homomorphism is $M_{\p} \to \prod_{P \in \ass(R)} (M/PM)_\p$. But for every minimal $\p \neq P$ we have $(M/PM)_\p = 0$ and thus the morphism is simply given by $M_\p \to (M/\p M)_\p \cong M_\p$ where the last isomorphism holds for minimal primes.
Remark: As the proof shows, we see that the above injection factors through $\prod_{P \in \ass(M)} M/PM$.