Let $R$ be a commutative ring and $M$ an $R$-module such that for every ideal $I \subset R$ the natural map $I \otimes_R M \rightarrow IM$ is an isomorphism.

Why is $M$ flat ?

This result is taken from Wikipedia.


This result should appear in many commutative algebra books, e.g. Matsumura, Commutative Ring Theory, Theorem 7.7, p.50. The point is that the given condition implies that $\mathrm{Tor}^1 (R/I, M)$ vanishes for any ideal $I$.

To show that $M$ is flat, we want to show that tensoring $M$ preserves the exact sequence of $R$-modules $0 \to N \to N'$. $N'$ is a direct limit of $N + F$, where $F$ is a finitely generated $R$-submodule of $N'$. By exactness of taking direct limit, we may assume that $N' = N+F$ for one such $F$. By inducting on number of generators of $F$, we may assume that $F$ is generated by one element, say $a$. Then $N' = N + Ra$ and
$$0 \to N \to N+Ra \to (N+Ra)/N \cong R/I \to 0$$ where $I = \{r \in R: ra \in N\}$. Vanishing $\mathrm{Tor}^1(R/I,M)$ then implies that this sequence is exact after tensoring $M$.

  • $\begingroup$ Thank you. Indeed the proof was in a book that I have at hand, but I wasn't looking at the right chapter... $\endgroup$ – user10676 Jan 22 '13 at 9:13

For an $R$-module $B$ we denote by $B^*$ the $R$-module $Hom_{Ab}(B,\mathbb{Q}/\mathbb{Z})$, which is called the Pontrjagin dual of $B$. You can prove the following:

1- A map $f:B\rightarrow C$ is injective if and only if the dual map $f^*:C^*\rightarrow B^*$ is surjective.

2- Since the functor $\otimes_R B$ is left adjoint to $Hom_{Ab}(B,-)$ then for any morphism of $R$-modules $A'\rightarrow A$ we have a commutative diagram: \begin{array}[lll] \ Hom(A,B^*)&\rightarrow&Hom(A',B^*)\\ \ \ \ \downarrow\simeq&&\ \ \ \downarrow\simeq\\ (A\otimes B)^*&\rightarrow&(A'\otimes B)^* \end{array}

Using this diagram and 1 you can prove that an $R$-module $B$ is flat if and only if $B^*$ is injective. Using this diagram again with $A'\rightarrow A$ replaced by the inclusion $I\rightarrow R$ then 1 and the Baer's criterion for injectivity show that $B^*$ is injective if and only if $I\otimes B\rightarrow IB$ is an isomorphism.


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