# Restricting a module to the irreducible components of an affine curve is injective

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 $$\dim R = 1$$.

$$\newcommand{\ass}{\operatorname{Ass}} \newcommand{\p}{\mathfrak{p}} \newcommand{tm}{\subseteq}$$

Proposition: Let $$R$$ be a reduced ring and let $$M$$ be a torsion-free $$R$$-module. Then the map $$M \to \prod_{P \in \ass(R)} M/PM$$ is injective.

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 $$\prod_{P \in \ass(M)} M/PM$$.

$$M\to S^{-1}M$$ is injective, where $$S$$ is the set of regular elements. But $$S^{-1}R$$ is just a product of fields and the rest should be clear.

• I am not sure what I miss, but what you're saying is that the morphism $M \to \prod_i M_{P_i}$ is injective. So far I can follow, but where is the connection to the morphism $M \to \prod_i M/P_iM$? – windsheaf Sep 27 '18 at 15:12
• $M_{P_i}=(M/P_iM)_{P_i}$. – Mohan Sep 27 '18 at 18:12
• Okay, I think I got it. From stacks.math.columbia.edu/tag/0AVL we know that we only need to check the injectivity at the associated primes of $M$ and those are a subset of the minimal primes of $R$ by the torsion-free assumption on $M$. Is this correct? – windsheaf Sep 28 '18 at 12:09