# Alternative construction of the tensor product (or: pass this secret)

The paper Tensor products and bimorphisms by B. Banachewski and E. Nelson studies tensor products (defined by classifying bimorphisms) in concrete categories. It is quite interesting that their main existence theorem gives an alternative, quite explicit construction of the tensor product of two modules (or any other algebraic structures).

If $M,N$ are $R$-modules with underlying sets $|M|,|N|$, consider $$P=\bigoplus_{m \in |M|} N \oplus \bigoplus_{n \in |N|} M$$ with the natural inclusions $i_m : N \to P$ for $m \in |M|$ and $j_n : M \to P$ for $n \in |N|$. Let $U=\langle i_m(n)-j_n(m) : (m,n) \in |M| \times |N| \rangle$. Then $P/U$ is a model for $M \otimes_R N$.

Question 1. Is there any other paper or book at all which mentions this construction? Or is it well-known?

Question 2. Is there a textbook introducing tensor products and gives this construction as a proof that it exists?

Question 3 (subjective): Isn't this construction more explicit than the usual one (which starts with the free module on $|M| \times |N|$ and mods out bilinear relations)? It only uses direct sums, generated submodules, and quotients, no free modules are needed. What do you think, do you favor it? Is it suited for the use in textbooks and classes? If you are a teacher or professor, would you consider using this construction in your class? What are your reasons?

I have found a smiliar "free module"-free construction of the module of differentials $\Omega^1_{A/R}$ for an $R$-algebra $A$: The $R$-linear map $A \otimes_R A \to A \otimes_R A$, $a \otimes b \mapsto ab \otimes 1 - b \otimes a - a \otimes b$ extends to an $A$-linear map $(A \otimes_R A) \otimes_R A \to A \otimes_R A$, when $A$ acts on the right. Let $\Omega^1_{A/R}$ be its cokernel, and $d(a)$ the image of $a \otimes 1$. The universal property is immediate. I would like to ask the same questions as above.

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This construction involves a massive direct sum, so it seems a bit misleading to say no free module is needed since that direct sum has the same flavor as a free module (another type of direct sum). –  KCd Feb 1 '13 at 5:18
Maybe, but it is conceptually different, and it remains true that the constructions avoid free objects (in any category where it is applied to). –  Martin Brandenburg Feb 1 '13 at 9:57

Answer 1. The closest thing to this construction I have seen is the Eilenberg-Watts theorem, which says that for any right exact functor $F\colon R$-mod$\to Ab$ that commutes with arbitrary direct sums, we have a natural isomorphism $F(-)\cong F(R)\otimes_R-$, where $F(R)$ is given its natural structure as a right $R$-module.

The key observation to Eilenberg's original proof is that given an $R$-module $M$, the canonical module homomorphism $\bigoplus_{m\in |M|}R\twoheadrightarrow M$ is in fact an $R$-bilinear function when considered as a two-variable function, and that consequently so is the image $\bigoplus_{m\in|M|}F(R)\twoheadrightarrow F(M)$ of the map under $F$. Then a little bit of diagram chasing shows that the induced map $F(R)\otimes_R M\to F(M)$ is in fact an isomorphism.

Hence, you can obtain constructions of the tensor product $M\otimes_R N$ from any right exact, direct-sum preserving functor $F_M$ for which $F_M(R)=M$. Thus, one should not be surprised at there being a ton of different constructions of the tensor product.

It is not the result of the theorem that's relevant here, however, but rather the idea behind the proof. Adapting, it seems to boil down to the observation that $\bigoplus_{m\in|M|} N$ has $|M|_{Ab}\otimes_\mathbb Z N$ as the natural quotient by the additive relations of $|M|_{Ab}$ that $|M|_{Set}$ has forgotten, and that $\bigoplus_{n\in|N|} M$ has $M\otimes_\mathbb Z |N|_{Ab}$ as the natural quotient by the additive realtions of $|N|_{Ab}$ forgotten by $|N|_{Set}$ (the proofs of these facts should be the same as in Eilenberg's proof). Then all your construction does is realize $M\otimes_R N$ as the pushout of the two.

Answer 2. I do not know of a textbook that does this stuff.

Answer 3. What is wrong with free modules? Your distaste for them mystifies me since I perceive algebraic objects are by any reasonable definition algebraic by virtue of being given as quotients of free objects (that's what an equation is). If I were teaching, what I would do is show how the (classical) explicit construction of the tensor product is nothing more than expanding the definitional hom-tensor adjunction, the internal hom to hom-set relationship, and the cartesian product to set-hom adjunction. Since I do not think the explicit construction is ever helpful for computational purposes, I would focus on the categorical properties from which one can both deduce the construction, and actually use for computation (e.g. preservation of direct sums and right exactness).

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