I have posted several questions about the tensor product of modules before and this post would be the final one.

I have read wikipedia,mathSE,Dummit&Foote and Bourbaki for the definition of the tensor product and I found that each defines the tensor product in slightly different ways, and here is a definition I'm planning to take as my definiton.

Let $R_1,...,R_n$ be rings

Let $M_1,...,M_{n+1}$ be abelian groups where $M_1$ is a right $R_1$-module and $M_{n+1}$ is a left $R_n$-module and $M_i$ be an $(R_{i-1},R_i)$-bimodule for $2\leq i\leq n$.

Let $F(\prod_{i=1}^{n+1} M_i)$ be the free abelian group on $\prod_{i=1}^{n+1} M_i$.

Define $S_1=\bigcup_{j=1}^{n+1} \{(a_1,...,a_j+b,...,a_{n+1})-(a_1,...,a_j,...,a_{n+1})-(a_1,...,b,...,a_{n+1}): a\in \prod_{i=1}^{n+1} M_i , b\in M_j\}$

Define $S_2=\bigcup_{j=1}^n \{(a_1,...,a_j r,...,a_{n+1})-(a_1,...,r a_{j+1},...,a_{n+1}) : a\in \prod_{i=1}^{n+1} M_i, r\in R_j\}$

Let $G$ be the subgroup of $F(\prod_{i=1}^{n+1})$ generated by $S_1\cup S_2$

Define $M_1\otimes_{R_1}...\otimes_{R_n} M_{n+1}$ as the quotient abelian group $F(\prod_{i=1}^{n+1} M_i)$ by $G$ and call it the tensor product.

Can this be the definition of the tensor product?

Moreover, why do we allow $ar\otimes b\otimes c$ to be only equal to $a\otimes r b\otimes c$ but not equal to $a\otimes b \otimes rc$? What's an motivation for restricting this action to adjacent modules?


1 Answer 1


This a correct construction of the tensor product, but it's not really a definition. The definition of the tensor product of modules as above is as the abelian group $A$, unique up to isomorphism, with a multilinear map $\alpha:M_1\times...\times M_{n+1}\to A$ such that there is a natural bijection of abelian group homomorphisms $A\to B$ with multilinear maps $M_1\times...\times M_{n+1}\to B$ given by composing with $\alpha$. To check that your construction is correct, simply prove that it satisfies this definition.

We don't require $ar\otimes b\otimes c=a\otimes b\otimes rc$ because $R_1$ might not be the same as $R_2$, in which case such an equality is meaningless.

  • $\begingroup$ Even when $R_1=R_2$, it's generally not the case the equation holds. Why? $\endgroup$
    – Rubertos
    Mar 9, 2015 at 9:48
  • $\begingroup$ And by Multilinear map, do you mean multiadditive and $\alpha(a_1,...,a_j r,...,a_{n+1}) = (a_1,...,ra_{j+1},...,a_{n+1})$? Is that term more standard than the term "R-balanced map"? IMO, the term multilinear may also mean $\alpha(a_1,...,a_j r,...,a_{n+1})=r(a_1,...,a_j,...,a_{n+1})$ when the tensor product is viewed as a multimodule since the similar term bilinear means this.. $\endgroup$
    – Rubertos
    Mar 9, 2015 at 9:55
  • 1
    $\begingroup$ The two modules have different definitions. The tensor product of $M_1,M_2,M_3$ modules on appropriate sides for one ring $R$ is universal for $\mathbb{Z}$-multilinear maps $f$ out of $M_1\times M_2\times M_3$ such that $f(m_1r,m_2,m_3)=f(m_1,rm_2,m_3)$ and $f(m_1,m_2r,m_3)=f(m_1,m_2,rm_3)$. Your modification is the module universal for $f$ with the same properties, as well as $f(m_1r,m_2,m_3)=f(m_1,m_2,rm_3)$, which implies $f(m_1,rm_2,m_3)=f(m_1,m_2r,m_3)$. These aren't the natural maps to consider if the left and right $R$-module structure on $M_2$ differ. $\endgroup$ Mar 9, 2015 at 9:55
  • $\begingroup$ By multilinear I mean multiadditive and $\alpha(a_1...a_jr...a_{n+1})=\alpha(a_1...ra_{j+1}...a_{n+1})$. $\endgroup$ Mar 9, 2015 at 9:56
  • $\begingroup$ Thank you so much you saved me. $\endgroup$
    – Rubertos
    Mar 9, 2015 at 9:59

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