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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 $\dim R = 1$.

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

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

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$M\to S^{-1}M$ 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.

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  • $\begingroup$ 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 $M \to \prod_i M/P_iM$? $\endgroup$ – windsheaf Sep 27 '18 at 15:12
  • $\begingroup$ $M_{P_i}=(M/P_iM)_{P_i}$. $\endgroup$ – Mohan Sep 27 '18 at 18:12
  • $\begingroup$ 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? $\endgroup$ – windsheaf Sep 28 '18 at 12:09

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