# Is there a different construction of “the” tensor product of two modules?

It may be a pseudo question. But I still decide to ask. Given two $k$-modules $M$ and $N$，it seems to me that in the literature the tensor product $M\bigotimes_kN$ is always defined as the quotient of the free module generated over the set $M\times N$ modulo the ideal generated by the bi-linearity relations. But I am curious to see a different construction of $M\bigotimes_kN$ which is of course isomorphic to the one mentioned above.

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You don't only want to construct an object, but also a projection map. The universal property says any such gadgets doing the right thing will all be isomorphic. So I am not sure what more you can really ask... – Steven Gubkin Jun 13 '12 at 20:00
The tensor product is usually constructed in the manner you mentioned. It is usually defined, on the other hand, via the universal property: briefly, any billinear map $M\times N\to P$ lifts to a unique map $M\otimes N\to P$; this is the universal property, and by this any construction of the tensor product is isomorphic to any other. – Jason Polak Jun 13 '12 at 20:38
For $\mathbb Z$-modules, or abelian groups, it's the dual to the space of bilinear maps $M \times N \to S^1$. – Will Sawin Jun 13 '12 at 20:41
Well, you could consider the set of all pairs $(P,B:M\times N\to P)$ where $P$ is a $k$-module and $B$ is bilinear, and then define the tensor product as a certain submodule of the product, over all such pairs, of all of the modules $P$ (or rather you could use a modified version of this construction that does not violate the rules of set theory). – Tom Goodwillie Jun 13 '12 at 21:10
Mariano: only if the vector spaces are finite-dimensional. – darij grinberg Jun 14 '12 at 7:16

In my first encounter with the tensor product of modules (in a course on representation theory by prof. Lenstra), it was done in the following spirit:

First, the tensor product $M\otimes _RE$ is defined using the universal property. Next, we prove the following (and other) elementary properties (here I am concentrating on the object and leaving out the morphism):

• $M\otimes _RR$ exists, and equals $M$.
• '$\otimes$ commutes with $\oplus$': if $(M\otimes E_i)_i$ exist, then $M\otimes \oplus_iE_i$ exists and equals their direct sum.
• '$\otimes$ commutes with $coker$ (right-exactness of $M\otimes_R-$)': Let $f:E\to F$ be $R$-linear, and assume $M\otimes_RE$ and $M\otimes_RF$ exist. Then $M\otimes_R coker f$ exists and equals $coker(M\otimes_RE\xrightarrow{1\otimes f}to M\otimes_RF)$

Theorem: The tensor product $M\otimes_RE$ exists.

Proof: Take a generating set S of E, i.e. the natural map $f:R^{(S)}\to E$ is surjective. Next pick a generating set T of ker(f), so the natural map $h:R^{(T)}\to R^{(S)}$ has image ker(f). Now $coker(h)=R^{(S)}/\ker f$ is (isomorphic via f to) $E$.

By property 1 and 2, $M\otimes_RR^{(T)}$ and $M\otimes_RR^{(S)}$ exist. By property 3, we conclude that $M\otimes_RE$ exists.

Remark: I guess a didactical merit of this approach (compared to the standard construction as the free abelian group on the product modulo bilinear relations) is that it forces you to think and reason in terms of the universal property and exact sequences. I am not sure if this is also the reason my teachers had in mind.

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Dear Yongheng Zhang,

This doesn't address your question directly but rather a special case of it: If you would like to see the construction of the tensor product of finite dimensional vector spaces as the dual space of bilinear maps, you may be interested to look at this post here:

http://math.stackexchange.com/questions/138715/showing-that-the-dual-space-of-bilinear-maps-v-times-w-to-mathbbr-satisfi

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Hi Benjamin, thank you for pointing me to the above discussion. – Yongheng Zhang Jun 14 '12 at 19:10

A rather general definition of tensor products, far beyond the algebraic setup can be found in the arxiv papers 0808.0095, 0807.1436. A recent application to quanta is in the arxiv paper 1203.0412.

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I will look more into the three papers especially the last one. Thank you. – Yongheng Zhang Jun 14 '12 at 19:12