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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 $P$ ?

rschwieb
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Manos
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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 03 '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 03 '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-Álvarez May 03 '12 at 16:19
  • @Mariano: I see. I've forgotten pretty much everything I once knew about tensor products (which wasn't much). – Tara B May 03 '12 at 16:23
  • I believe, as suggested by @wxu in the comments to my previous (now deleted) answer, that only using the fact that $M\otimes (N\otimes P)\cong M\otimes N\otimes P$, you will always be stuck with some parentheses. The easiest way to show your result would be using the universal property of $B\otimes C\otimes D\otimes E$. – M Turgeon May 03 '12 at 17:13
  • Yes, if we have not the definition of $B\otimes C\otimes D\otimes E$, what are we doing here? So maybe we will get a result that $(B\otimes C)\otimes (D\otimes E)$ is isomorphic to a module $M$ given by adding some parentheses into $B\otimes C\otimes D\otimes E$, this module $M$ cannot be $B\otimes C\otimes D\otimes E$, since we donot know what does it mean when we write $B\otimes C\otimes D\otimes E$. Now we want to show $M$ is isomorphic to $B\otimes C\otimes D\otimes E$, we must use the definition of $B\otimes C\otimes D\otimes E$...... – wxu May 03 '12 at 17:29
  • @Manos: Do you define $\otimes = \otimes_{\mathbb{Z}}$ or $\otimes=\otimes_{A}$? In the second case, if $A$ is not commutative the middle tensor product $(B \otimes_{A} C) \otimes_{A} (D \otimes_{A} E)$ does not make much sense in my opinion, because $(B \otimes_{A} C)$ is not (necessarily) naturally an $A$-module anymore, only an abelian group. – Nils Matthes May 03 '12 at 17:34
  • You just have to interpret the problem as "show that 'independently of the parenthesis', the products are always isomorphic". Then, this becomes the definition of $M \otimes N \otimes P$. This is how you define $abc$ in a (associative) ring. – André Caldas May 03 '12 at 17:35
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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 03 '12 at 18:55
  • @Manos: I think you can prove it along the same lines; however, I at least don't really see how to deduce it from "iterated binary tensor is isomorphic to ternary" – Arturo Magidin May 03 '12 at 18:58
  • @Nils: It is $\otimes=\otimes_A$ and $A$ is commutative. – Manos May 03 '12 at 20:51
  • @ArturoMagidin: I guess you can adopt the definition you prefer, as long as they are equivalent. You can even fix an order and define it based on this order. Then you can have a proposition telling you that other orders are isomorphic. – André Caldas May 04 '12 at 18:52
  • @AndréCaldas: Indeed; my point was just that you were proving that the definition via iterated binary products yields an object in which the order of the binary products does not matter (i.e., you were proving general associativity), whereas the question is actually asking you to prove that the iterated binary product construction is equivalent to the $4$-term multilinear construction. – Arturo Magidin May 04 '12 at 19:06
  • @ArturoMagidin: I agree that what you are saying is probably the intention of the OP. :-) – André Caldas May 04 '12 at 19:46
  • If you wanna make life real easy it follows from associativity of composed adjoints – user25470 May 04 '12 at 10:49
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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-Álvarez May 04 '12 at 21:06

2 Answers2

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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.)

Paul Z
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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 03 '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 03 '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 03 '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 03 '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.

André Caldas
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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 03 '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 04 '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 04 '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 04 '12 at 19:52