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=== Tensor product of modules over a non-commutative ring === Let ''A'' be a right ''R''-module and ''B'' be a left ''R''-module. Then the tensor product of ''A'' and ''B'' is an abelian group defined by: <math display="block">A \otimes_R B := F (A \times B) / G</math> where <math>F (A \times B)</math> is a [[free abelian group]] over <math>A \times B</math> and G is the subgroup of <math>F (A \times B)</math> generated by relations: <math display="block">\begin{align} &\forall a, a_1, a_2 \in A, \forall b, b_1, b_2 \in B, \text{ for all } r \in R:\\ &(a_1,b) + (a_2,b) - (a_1 + a_2,b),\\ &(a,b_1) + (a,b_2) - (a,b_1+b_2),\\ &(ar,b) - (a,rb).\\ \end{align}</math> The universal property can be stated as follows. Let ''G'' be an abelian group with a map <math>q:A\times B \to G</math> that is bilinear, in the sense that: <math display="block">\begin{align} q(a_1 + a_2, b) &= q(a_1, b) + q(a_2, b),\\ q(a, b_1 + b_2) &= q(a, b_1) + q(a, b_2),\\ q(ar, b) &= q(a, rb). \end{align}</math> Then there is a unique map <math>\overline{q}:A\otimes B \to G</math> such that <math>\overline{q}(a\otimes b) = q(a,b)</math> for all <math>a \in A</math> and {{tmath|1= b \in B }}. Furthermore, we can give <math>A \otimes_R B</math> a module structure under some extra conditions: # If ''A'' is a (''S'',''R'')-bimodule, then <math>A \otimes_R B</math> is a left ''S''-module, where {{tmath|1= s(a\otimes b):=(sa)\otimes b }}. # If ''B'' is a (''R'',''S'')-bimodule, then <math>A \otimes_R B</math> is a right ''S''-module, where {{tmath|1= (a\otimes b)s:=a\otimes (bs) }}. # If ''A'' is a (''S'',''R'')-bimodule and ''B'' is a (''R'',''T'')-bimodule, then <math>A \otimes_R B</math> is a (''S'',''T'')-bimodule, where the left and right actions are defined in the same way as the previous two examples. # If ''R'' is a commutative ring, then ''A'' and ''B'' are (''R'',''R'')-bimodules where <math>ra:=ar</math> and {{tmath|1= br:=rb }}. By 3), we can conclude <math>A \otimes_R B</math> is a (''R'',''R'')-bimodule.
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