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Problem Of The Week # 350 - Jun 18, 2019

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Euge

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Jun 20, 2014
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Euge

MHB Global Moderator
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Jun 20, 2014
1,925
No one answered this week's problem. You can read my solution below.


By induction, it suffices to consider the tensor product of two flat modules, $M$ and $N$, over commutative ring $R$. Given a short exact sequence $0 \to X' \to X \to X'' \to 0$ of $R$-modules, tensoring with $N$ yields a short exact sequence $0 \to N \otimes_R X' \to N \otimes_R X \to N \otimes_R X'' \to 0$. Tensoring the latter sequence with $M$ results in the short exact sequence $0 \to M \otimes_R (N \otimes_R X') \to M \otimes_R (N \otimes_R X) \to M \otimes_R (N \otimes_R X'') \to 0$; by naturality of the associativity isomorphism, the sequence $0 \to (M \otimes_R N) \otimes_R X' \to (M \otimes_R N) \otimes_R X \to (M \otimes_R N) \otimes_R X'' \to 0$ is exact, as desired.
 
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