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Consider an integral domain $R$, its field of fractions $K$ and an $R-$module $M$ that has at least one linearly independent element, i.e.:

$$ D:=\{ m \in M \mid \forall \ r \in R \backslash\{0\}, \ rm \neq 0\} \neq \emptyset.$$

Then it should be true that $M \otimes_R K \neq 0$, but I don't know how to show it. It seems logical that $d \otimes 1 \neq 0$ for all $d \in D$, but how to prove it? We could find an $R-$bilinear map $ \alpha\colon M \times K \to N$ for some $R-$module $N$ so that $\alpha(d,1) \neq 0$, but I cannot find such an $\alpha$.

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    $\begingroup$ The obvious sequence $0\to M_{tor}\to M\to K\otimes_RM$ is exact. $\endgroup$ Commented Apr 11, 2018 at 16:24
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    $\begingroup$ Isn't this just a reformulation of what I want to prove? My question could be rephrased as asking why $\ker(f) \subseteq M_{tor}$, where $f \colon M \to K \otimes_R M$ $\endgroup$
    – 57Jimmy
    Commented Apr 12, 2018 at 9:28
  • $\begingroup$ I wasn't sure I understood your question. There is a generalization: $m/1=0$ in $S^{-1}M$ iff $sm=0$ for some $s$ in $S$. ($S$ is a multiplicative system. (And $R$ is not necessarily a domain.)) This is in Bourbaki's Commutative Algebra. If you want I can try to understand the proof and describe it in an answer. $\endgroup$ Commented Apr 12, 2018 at 11:20
  • $\begingroup$ @Pierre-YvesGaillard Thanks, now I've got it. The key fact I was missing is that we can work directly with the localization. Or we can work with the tensor product and use the bilinear map to the localization. Thanks anyway :) $\endgroup$
    – 57Jimmy
    Commented Apr 12, 2018 at 13:09
  • $\begingroup$ The fact that $R$ is a domain makes sure that $S=R \backslash \{0\}$ is multiplicatively closed. $\endgroup$
    – 57Jimmy
    Commented Apr 12, 2018 at 13:21

2 Answers 2

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Here is a general outline:

The first thing to note is that $$K \otimes_R M \cong K \otimes M/M_{tor},$$

where $M_{tor}$ is the torsion submodule of $M$ (this is not hard to show). In your case we have that $M_{tor} \neq M$, so we can reduce to the case where $M$ is torsion-free.

Then the idea is, as you have said yourself, to construct a bilinear map from $K \times M$ to a nonzero $R$-module. To do this, we want to take a look at formal products of the type $x \cdot m$ with $x \in K$ and $m \in M$. Formally, define on $K \times M$ an equivalence relation by

$$(\frac{a}{b}, m) \sim (\frac{c}{d},n) \iff adm = bcn \in M.$$

To show that this is indeed well-defined and an equivalence relation, you need that $M$ is torsion-free. Once you have this, consider $S := K \times M / \sim$ and we have a scalar multiplication $x \cdot m := (x, m)$ as above. Define addition on it in the canonical way and we obtain that $S$ is a $K$-vector space, in particular $R$-module. Check easily that the canonical map $M \to S$ is injective, hence $S \neq 0$ as $M \neq 0$. Then get a bilinear map $K \times M \to S, (x, m) \mapsto x \cdot m$, which can easily be shown to be surjective, and you are done.

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  • $\begingroup$ Thanks a lot! Is there a general name for the construction of $S$ or is it just an ad hoc construction? $\endgroup$
    – 57Jimmy
    Commented Apr 12, 2018 at 9:47
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    $\begingroup$ Wait a second: isn't $S$ just the localization of $M$ at $R \backslash \{0\}$? $\endgroup$
    – 57Jimmy
    Commented Apr 12, 2018 at 10:39
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    $\begingroup$ Yes, and if you are aware of the relation $S^{-1}M \cong M \otimes_R S^{-1}R$ for localizations, then a one-line argument would be: $K \otimes_R M \cong M_{(0)}$, which is not zero as you have elements that aren't torsion. $\endgroup$
    – johnnycrab
    Commented Apr 12, 2018 at 14:04
  • $\begingroup$ Yes, got that. Thanks! $\endgroup$
    – 57Jimmy
    Commented Apr 12, 2018 at 14:37
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Let $m$ denote your non-torsion element. Because $K$ is a flat $R$-module (as it is a localization) the following sequence is exact: $$0\rightarrow \overbrace{mR\otimes_RK}^{K} \rightarrow M\otimes_R K$$

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