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Let $R$ be a commutative ring and let $M_{1},...M_{n},L$ be $R$-modules. Let $\iota:M_{1},...M_{n}\rightarrow M_{1}\bigotimes...\bigotimes M_{n}$ be defined by $\iota(m_{1},...,m_{n})\mapsto m_{1}\otimes...\otimes m_{n}$ and let $\phi:M_{1}\times...\times M_{n}\rightarrow L$ be an $n$-multilinear map. Claim: there is a unique $R$-module homomorphism $\Phi:M_{1}\bigotimes...\bigotimes M_{n}\mapsto L$ such that $\phi =\Phi \circ \iota$.

Assume this holds for $n=2$ (in other words, assume the universal property for the tensor product of pairs of modules). I want to show that this case implies the remaining cases, for example, $n=3$. Apparently this is immediate, but I have absolutely no idea why it is true. I have tried every way of pairing the modules but nothing has worked.

Edit: I am using the construction from free groups, etc. definition of tensor products.

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  • $\begingroup$ Normally one defines the $n$-fold tensor product to be a vector space $M_1 \otimes \ldots \otimes M_n$ equipped with a linear map $\iota \colon M_1 \times \ldots \times M_n \to M_1 \otimes \ldots M_n$ which satisfies the universal property that you describe, and then the notation $m_1 \otimes \ldots \otimes m_n$ is by definition $\iota(m_1, \ldots, m_n)$. With this definition your claim is a tautology. But one must prove that the $n$-fold tensor product actually exists - perhaps you're trying to show that the $3$-fold tensor product exists if the $2$-fold tensor product exists? $\endgroup$ – Paul Siegel Jan 1 '18 at 4:45
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Existence

$\Phi: m_1 \otimes \cdots \otimes m_n \mapsto \phi(m_1, \cdots, m_n)$ satisfies the condition.

Uniqueness

Let $\Phi$ and $\Phi'$ be maps satisfying the condition.

Then, I will prove that $\Phi = \Phi'$.

Let $m_1 \otimes \cdots \otimes m_n \in M_1 \otimes \cdots \otimes M_n$. Then:

$$\begin{array}{rcl} \Phi(m_1 \otimes \cdots \otimes m_n) &=& \Phi(\iota(m_1,\cdots,m_n)) \\ &=& \phi(m_1,\cdots,m_n) \\ &=& \Phi'(\iota(m_1,\cdots,m_n)) \\ &=& \Phi'(m_1 \otimes \cdots \otimes m_n) \end{array}$$

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  • $\begingroup$ Why is that $\Phi$ what you claim it is? For example, why is it well-defined? $\endgroup$ – SihOASHoihd Jan 1 '18 at 5:08
  • $\begingroup$ I don't think you can get around having to use the universal property for tensor products of pairs of modules. $\endgroup$ – SihOASHoihd Jan 1 '18 at 5:09
  • $\begingroup$ @SihOASHoihd what is it with not providing a definition and then downvoting others for using different definitions? $\endgroup$ – Kenny Lau Jan 1 '18 at 5:35
  • $\begingroup$ Sorry I tried upvoting you afterwards but it went to 1 instead of 0. $\endgroup$ – SihOASHoihd Jan 1 '18 at 5:48
  • $\begingroup$ Your answer sort of trivializes the question doesn't it? $\endgroup$ – SihOASHoihd Jan 1 '18 at 5:52

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