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Let $R$ be a commutative ring with identity and $M$ an $R$-module.

If $N_1\longrightarrow N_2$ is injective (resp. surjective), is the induced map $M\otimes_R N_1\longrightarrow M\otimes N_2$ necessarily injective (resp. surjective)?

I really do not know how to prove and cannot give counterexamples.

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2 Answers 2

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Injectivity is definitely not preserved. Take $N_1 = N_2 = \mathbb{Z}$ and and take the map from $N_1$ to $N_2$ given by multiplication by $2$. This map is injective, but after tensoring with $\mathbb{Z}/2\mathbb{Z}$, it is not.

Surjectivity is preserved.

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It's worth mentioning that this question is related to the half-exactness of the tensor product functor.

As this Proving that the tensor product is right exact shows that tensor product is right exact (and therefore the functor preserve surjective map), the question that the tensor product does not preserve injectiveness of the map is the same as saying that in general that the tensor product functor is not exact functor.

This is really the property of the module $M$, if $M$ makes the functor exact, we call it flat. As we see $\Bbb{Z}/2\Bbb{Z}$ is not flat.

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