Examples proving why the tensor product does not distribute over direct products? In fact the canonical map is not surjective; can you give me a simple example?
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2People would be happier around here if you showed that you tried to solve the problem and is not just crowd sourcing the solution to your school problems. Also, paying more attention to your question would make us pay more attention to you. – Leo Azevedo Feb 07 '14 at 07:57
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9I'm surprised at the overwhelmingly negative response this particular question has received so far. This user is brand new and has asked a perfectly legitimate question, which to me seems to have much more mathematical content than most other "school problems" on this site. – zcn Feb 07 '14 at 08:08
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1I think you should write exactly in what category you want to find an example or add ``in some category'' to the question. It is not quite clear which question are you asking. – user68061 Feb 07 '14 at 18:48
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The canonical map can fail to be injective: if $A$, $B_n$ are $R$-modules, then tensoring the canonical projections $\prod B_n \to B_n$ with $A$ give maps $A \otimes \prod B_n \to A \otimes B_n$, which induce the canonical map $A \otimes \prod B_n \to \prod (A \otimes B_n)$. It can happen that $A \otimes B_n = 0$ for all $n$, but $A \otimes \prod B_n \ne 0$.
Take $R = \mathbb{Z}$, $A = \mathbb{Q}$, and $B_n = \mathbb{Z}/n\mathbb{Z}$: then $A \otimes B_n = 0$ for all $n$, but $A \otimes \prod B_n = S^{-1} \prod B_n$, where $S = \mathbb{Z} \setminus \{0\}$, and the element $(1, 1, \ldots) \in \prod B_n$ is not annihilated by any element of $S$, so $A \otimes \prod B_n \ne 0$ (since for any $\mathbb{Z}$-module $M$, $S^{-1}M = 0$ iff every element of $M$ is annihilated by some element of $S$).
Notice that this is an example where localizations do not commute with infinite direct products. In particular, even tensoring with a flat module does not commute with direct products.
zcn
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