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Let $B = (B, \nabla, \eta, \Delta, \epsilon )$ be a bialgebra over a commutative ring $k$. Let $M$ and $N$ be two left $B$-modules. Then the tensor product $M \otimes_k N$ becomes a left $B$-module with multiplication rule given by

$$ B \otimes_k M \otimes_k N \xrightarrow{\Delta \otimes 1} B \otimes_k B \otimes_k M \otimes_k N \xrightarrow{1 \otimes \tau \otimes 1} B \otimes_k M \otimes_k B \otimes_k N \xrightarrow{\mu \otimes \mu} M \otimes_k N$$

where the $\mu$'s are the $B$-module multipication on $M$ and $N$. I am trying to show that $B$-$\mathsf{Mod}$ forms a monoidal category with tensor product $- \otimes_k -$. We have a $k$-linear associator $$ \alpha_{MNP} \colon (M \otimes_k N) \otimes_k P \xrightarrow{\sim} M \otimes_k (N \otimes_k P)$$ and I think this is supposed to be the associator in $B$-$\mathsf{Mod}$, so I need to show it is $B$-linear, i.e. that $$b [m \otimes (n \otimes p)] = \alpha_{MNP} (b[(m \otimes n) \otimes p])$$ but I am not really getting anywhere with it. Does anyone have any advice or at least be able to point me to a souce where this is covered? I am currently using Pareigis's Advanced Algebra, but proving monoidal structure is left as an exercise.

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    What you want should follow straight from the coassociativity of $B$, which you have not used yet. In Sweedler notation, both terms are equal to $\sum b_{(1)}m \otimes b_{(2)}n \otimes b_{(3)}p$, although depending on your style, you may want a proof that is more categorical.2012-09-14
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    @Aaron I can see by coassociativity we have $$ \sum_{(b)}\sum_{(b_{(2)})} b_{(1)} \otimes \left( (b_{(2)})_{(1)} \otimes (b_{(2)})_{(2)} \right) = \sum_{(b)}\sum_{(b_{(1)})} (b_{(1)})_{(1)} \otimes \left( (b_{(1)})_{(2)} \otimes (b_{(2)}) \right) $$ but why is it true if I slot in $m, n$ and $p$?2012-09-14
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    Are we just using the fact that $M \otimes_k N \otimes_k P$ is a $B \otimes_k B \otimes_k B$ module, and the action of the two algebra elements above on $m \otimes n \otimes p$ gives what we want?2012-09-14
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    Yes, exactly. Essentially, we're using that $\mu \otimes (\mu \otimes \mu) = \alpha \circ (\mu \otimes \mu) \otimes \mu$2012-09-14
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    @Aaron Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-11

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