Let $R$ be an integral domain and $M$ an $R-$module. Let $(m_i)_{i \in I}$ be a maximal linearly independent subset of $M$. Define the rank of $M$ as the cardinality of $I$. I want to show that $(m_i \otimes 1)_{i \in I}$ is a $K-$basis of $M \otimes K$, where $K:=\mathrm{Frac}(R)$. This will also imply that the rank of $M$ is well-defined.

Note: every element of $M \otimes K$ can be written in the form $m \otimes \frac{1}{r}$.

First part: $(m_i \otimes 1)_{i \in I}$ is $K-$linearly independent.

If this were not the case, then we would have an equation of linear dependence over $K$. Clearing denominators, we get an expression of the form

$$\left( \sum_{j=1}^n r_j m_{i_j} \right) \otimes 1 = 0.$$

Does this imply that the sum in brackets is zero? If yes, then the first part is over. I think that, in general, if $m \neq 0$, then $m \otimes 1 \neq 0$, but to show it I would need an $R-$bilinear map $M \times K \to N$ that does not vanish on $(m,1)$; and I cannot find such a map...

Second part: if I add an element to $(m_i \otimes 1)_{i \in I}$, then it becomes linearly dependent.

Without loss of generality, we can add an element of the form $m \otimes 1$. Since I must have an equation of linear dependence over $R$ (by maximality), I also get an equation (the same) of linear dependence over $K$. Hence we are done.

So all in all my question is just: does $m\neq 0$ imply $m \otimes 1 \neq 0$? Why so? If not, how can I conclude the first part of the proof?

  • $\begingroup$ $\mathbb Q\otimes_{\mathbb Z}\mathbb Z/(2)=0$ $\endgroup$ – Pierre-Yves Gaillard Apr 10 '18 at 19:39
  • $\begingroup$ @Pierre-YvesGaillard Thanks, I am aware of such examples (I should have mentioned it in the question), but that’s exactly my difficulty: what is it, when the rank is not $0$, that guarantees that the tensor product is not $0$? $\endgroup$ – 57Jimmy Apr 10 '18 at 20:22

As explained here, we can argue using the bilinear map $f \colon M \times K \to M_{(0)}$, where $M_{(0)} = (R \backslash\{0\})^{-1}M$ is the localization of $M$ away from $0$. Indeed, $\frac{m}{r} = f(m,r) \in M_{(0)}$ is $0$ if and only if $m$ is torsion. In particular, if $m$ is not torsion, then $m \otimes \frac{1}{r}$ cannot be $0$.


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