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Let $B, C, D, E$ be $A$-modules. Is there a way to show that $(B \otimes C) \otimes (D \otimes E)$ is isomorphic to $B \otimes C \otimes D \otimes E$ using the result that $(M \otimes N) \otimes P$ is isomorphic to $M \otimes N \otimes P$ for any $A$-modules $M, N$ and $P$ ?

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Do you also know that $M\otimes(N\otimes P)\cong M\otimes N\otimes P$? Or perhaps that $M\otimes N\cong N\otimes M$? –  Arturo Magidin May 3 '12 at 16:15
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What is even the definition of $M\otimes N\otimes P$ if you don't already know the tensor product is associative? –  Tara B May 3 '12 at 16:15
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@TaraB (one can construct ternary products on one big swoop just as one constructs the usual binary products, and then show that $(AB)C$ is isomorphic to it...) –  Mariano Suárez-Alvarez May 3 '12 at 16:19
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@AndréCaldas: Not quite. $M\otimes N\otimes P$ does not mean "the result of doing binary products in either order", it means "the universal object relative to multilinear functions from $M\times N\times P$." Similarly, $B\otimes C\otimes D\otimes E$ does not mean "the result of doing the binary tensors in any order" (the way it would if you were talking about multiplication of real numbers, say), but rather it means "the universal object associated to the multilinear functions from $B\times C\times D\times E$." –  Arturo Magidin May 3 '12 at 18:55
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@user25470: Several users have observed in flags that you could probably be much more explicit about what you mean. As it stands, it does not seem to answer the question! –  Mariano Suárez-Alvarez May 4 '12 at 21:06
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To answer your direct question, yes, there is certainly a way to prove this, becuase it's certainly true. How to do it? Anytime tensor products show up, your best bet is appealing to the universal property, because it's the only thing you really know about the tensor product (especially if you're doing this over an arbitrary commutative ring $A$).

I don't think using the result you mention will help, because it doesn't say anything about the quaternary product. But if you have the proof of the ternary result, the proof you want (for the quaternary case) will be pretty much identical: use universal properties to construct unique homomorphisms in each direction, and then show (trivially) that they are inverses of each other.

Edit: Obviously it has been too long since I did any tensor algebra myself. The real way to prove this is just to show that $(B \otimes C) \otimes (D \otimes E)$ satisfies the quatrilinear universal property (by appealing to the various bilinear universal properties involved), and must therefore be isomorphic to the quaternary tensor product, because the tensor product is unique up to isomorphism. The end! (No mucking about with trying to show inverses or anything like that.)

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I'm still puzzling over which ring the tensor products on both sides are supposed to live. I guess that $B \otimes_{A} C \otimes_{A} D \otimes_{A} E$ would at least make sense for every ring $A$, but I still don't see how $(B \otimes_{A} C) \otimes_{A} (D \otimes_{A} E)$ makes sense when $A$ is not assumed to be commutative, because in the latter case $(B \otimes_{A} C)$ is not necessarily an $A$-module in a natural way. –  Nils Matthes May 3 '12 at 19:34
    
@NilsMatthes I'm betting $A$ is assumed to be commutative, otherwise you get into trouble (as you say) with left- or right-modules. In general tensor algebra over non-commutative rings is sort of a fringe science because of problems like you mention. –  Paul Z May 3 '12 at 19:40
    
@Paul: Thanks, i already did it that way. I was just wondering whether it is possible to obtain the result without having to define multilinear maps the way you mention. –  Manos May 3 '12 at 20:56
    
@Manos Oh, no, probably there is not any other way (or rather, any other way is essentially equivalent). When you're dealing with objects defined by a universal property (like tensor products), pretty much the only tool you have is the universal property itself, so you might as well use it. –  Paul Z May 3 '12 at 21:06
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I will assume that you already know that $$ (M \otimes N) \otimes P \cong M \otimes (N \otimes P). $$ Then, I will show that "whatever order" the parenthesis appear, the result of calculating $X_1 \otimes X_2 \otimes \dotsb \otimes X_n$ (parenthesis were ommited since I did not have enough creativity to create a notation for it) is always the same as associating from left to right. That is, it is always isomorphic to $((((X_1 \otimes X_2) \otimes X_3) \dotsb) \otimes X_n)$.

We use induction on $n$, and we know that the statement is true for $n = 3$. Take an "innermost term". That is, some $(X_k \otimes X_{k+1})$, and treat it as a single term. Then, the inductive hypothesis implies that the product is isomorphic to $$ ((((X_1 \otimes X_2) \otimes X_3) \dotsb \otimes (X_k \otimes X_{k+1})) \otimes X_n). $$

Case 1 ($k = 1$): Nothing to do.

Case 2 ($k > 2$): From the equation above, we can take $k = 1$ and reduce this to Case 1.

Case 3 ($k = 2$): The equation becomes $$ ((((X_1 \otimes (X_2 \otimes X_3)) \dotsb ) \otimes X_n). $$ Now, just use the case for $n = 3$, that is, $X_1 \otimes (X_2 \otimes X_3) \cong (X_1 \otimes X_2) \otimes X_3$, to get to the desired conclusion.

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This proves associativity of the binary tensors; but the multi-tensor object is not an abbreviation of sequence of binary tensors (the way it is with, say, multiplication) but rather an object defined by a universal multilinear property. You've given a perfectly good proof of something else, but I think not of the given question. –  Arturo Magidin May 3 '12 at 18:55
    
You can define it using its universal property, you can define it using generators and relations, you can define it as a quotient of the multilinear maps or you can define $X_1 \otimes \dotsb \otimes X_n$ as anything that is isomorphic to $(((X_1 \otimes X_2) \otimes \dotsb) \otimes X_n$. I don't know if what you are saying is a consensus, but I don't see any problem with using this as the definition. But I do AGREE with you that the universal property should be the prefered definition. –  André Caldas May 4 '12 at 19:20
    
What I'm saying is that this question is asking for a proof of equivalence of two particular definitions. But that your answer does not establish the equivalence of those two particular definitions. Instead, your answer establishes a property of one of the two definitions (iterated binary product), not an equivalence between the two distinct definitions (iterated binary products vs. universal property). I am making no statement about which one should be prefered, I'm saying that your post is a correct answer to a different question. –  Arturo Magidin May 4 '12 at 19:22
    
@ArturoMagidin: I am saying "prefered" because the OP does not say what the definition is. We are all conjecturing about the OP's intention. I am used to the definition I have used. :-) I do agree with you that my answer is not good because it does not show the equivalence of the two definitions... and the equivalence is probably the intention of the OP. –  André Caldas May 4 '12 at 19:52
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