After trying to understand tensor product of modules, there are a few questions:

  1. Generally how the construction of the tensor product of modules in (3) works. The Wikipedia description doesn't make much sense to me and introduces much that isn't elaborated on or derived.
  2. Where the "quotient of a free abelian group with a basis" in (3) came from. I didn't see that in the definition.
  3. If those 3 subgroup generators have names. Wondering where they came from, and if they have names like "associative" or anything.

My attempt to write out a definition based on (2) and Wikipedia.

(1) Suppose we have a ring and a right and left R-module. Then the tensor product over the ring is an abelian group along with a map mapping combinations of the two modules to elements of the abelian group. The map is called a balanced product. The group and map are universal in the sense that, for every second group and map, there is a unique group homomorphism, such that the group homomorphism composed with the first map generates the second map.

(2) Formally, suppose we have a ring $R$, and a right and left $R$-module $M$ and $N$. The tensor product over $R$, denoted $M \otimes_R N$, is an abelian group along with a map (balanced product). The map is denoted $\otimes : M \times N \to M \otimes_R N$. The set of all maps over $R$ from $M \times N$ to $G$ is denoted $L_R(M, N, G)$. The set of all groups is denoted $G^*$ and all maps is $\otimes^*$. The group and map are universal in the sense that $\forall G \in G^*,\forall f : M\times N \to G, \exists! \tilde{f} : M \otimes_R N \to G\ \ s.t.\ \ \tilde{f} \circ \otimes = f$.

Wikipedia then says:

(3) The construction of $M \otimes N$ takes a quotient of a free abelian group (an abelian group with a basis) with a basis as the symbols $m \ast n$, used here to denote the ordered pair $(m, n)$, for $m \in M$ and $n \in N$ by the subgroup generated by all elements of the form

  1. $−m \ast (n + n') + m \ast n + m \ast n'$
  2. $−(m + m') \ast n + m \ast n + m' \ast n$
  3. $(m · r) \ast n − m \ast (r · n)$

where $m, m' \in M, n, n' \in N$, and $r \in R$. The quotient map which takes $m \ast n = (m, n)$ to the coset containing $m \ast n$; that is,

$$\otimes :M\times N\to M\otimes _{R}N,\,(m,n)\mapsto [m*n]$$

is balanced, and the subgroup has been chosen minimally so that this map is balanced. The universal property of $\otimes$ follows from the universal properties of a free abelian group and a quotient.

Not sure where all of this came from. How they knew to use quotients of abelian groups, and these subgroup generators. Wondering if this is somehow related to free objects.

  • $\begingroup$ The simplest introduction I know is in Atiyah-Macdonald, Introduction to Commutative Algebra, ch. 2, pp.24-25. $\endgroup$
    – Bernard
    May 8, 2018 at 0:49

1 Answer 1


The construction of the most general tensor product is a quotient of a free product, as stated. A tensor product, at minimum, has left and right additivity, and the action works on the left or the right. The construction quotients out exactly what is necessary to meet this definition.

In practice, tensors can have more properties, such as being antisymmetric, which require further quotienting. That's why this construction is most general.

The subgroup generators are arbitrary, this quotient is for all elements, otherwise we wouldn't have a module structure.


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