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Let $(F_{i})_{i\in I}$ be a family of left R-modules and E is a right R-module.

(a) Define a homomorphism $\pi :E\otimes_{R}(\prod_{i\in I} F_{i})\to \prod_{i\in I}(E\otimes_{R} F_{i})$

question: Show that if E is a free right R-module, then $\pi$ is injective.

My thoughts: set $\pi(e\otimes(\prod_{i\in I}f_{i}))=\prod_{i\in I}(e\otimes_{R}f_{i})$, I think this definition is reasonable,but I still have no idea how to prove it is injective. Can someone help me solve this problem?

python3
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  • Are you going to prove injective or surjective? – AG learner Mar 06 '16 at 05:05
  • prove that $\pi$ is injective – python3 Mar 06 '16 at 05:10
  • For injectivity, simply take $E$ for $\mathbb Z$ since direct sum commutes with tensor product. For surjectivity, it is not true in general, see this post:http://math.stackexchange.com/questions/118790/examples-proving-why-the-tensor-product-does-not-distribute-over-direct-products – AG learner Mar 06 '16 at 05:10

1 Answers1

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As summarized in the comment, to prove injectivity, one only need to show that it is true for $R$.

Now, as map as defined, for $r\in R$, one has $ r\otimes(\prod_{i\in I}f_{i})=1\otimes r(\prod_{i\in I}f_{i})=1\otimes(\prod_{i\in I}rf_{i})=0$ if and only if $\prod_{i\in I}rf_{i}=0$, if and only if $rf_i=0$ for all $i$, if and only if $1\otimes rf_i=0$ for all $i$, if and only if $\prod_ir\otimes f_i=0$

AG learner
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