Let $M'$ be a submodule of $\mathbb{Z}$-module $M$, and let $i:M'\rightarrow M$ be a natural monomorphism.

How to prove the following theorem ? :

$i\otimes 1_N:M'\otimes_{\mathbb{Z}} N \rightarrow M\otimes_{\mathbb{Z}} N$ is a monomorphism for all $\mathbb{Z}$-modules $N$ iff for all $q\in \mathbb{Z}$ we have $M'\cap qM=qM'$.

If $N$ is finitely generated then we can use fact that $N$ is direct sum of cyclic modules and then we know that $\ker\left(M'\otimes \left(\mathbb{Z}/q\mathbb{Z}\right)\rightarrow M\otimes \left(\mathbb{Z}/q\mathbb{Z}\right)\right)\cong \left(M'\cap qM\right)/qM'$, so in this case the proof is easy. How to do it in general case ? Is it still true ?


It is still true because:

  1. Any $A$-module is the direct limit of its finitely generated submodules (for any ring $A$).
  2. Tensor products commute with direct limits.
  3. Direct limit is an exact functor.

Btw, a submodule of an $A$-module with this property (the morphism $M'\hookrightarrow M$ is universally exact) is called a pure submodule of $M$, and the morphism is said to be pure.

Standard examples of pure submodules:

  • A direct summand is a pure submodule
  • If $M/M'$ is flat, $M'$ is a pure submodule of $M$.

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