in the section "Evaluation map and tensor contraction" above: This result implies which automatically gives the important fact that forms a basis for where are bases of and . Furthermore, given three vector spaces , , the tensor product is linked to the vector space of all linear maps, as follows: This is an example of adjoint functors: the tensor product is "left adjoint" to Hom. Tensor products of modules over a ring The tensor product of two modules and over a commutative ring is defined in exactly the same way as the tensor product of vector spaces over a field: where now is the free -module generated by the cartesian product and is the -module generated by the same relations as above. More generally, the tensor product can be defined even if the ring is non-commutative. In this case has to be a right--module and is a left--module, and instead of the last two relations above, the relation is imposed. If is non-commutative, this is no longer an -module, but just an abelian group. The universal property also carries over, slightly modified: the map defined by is a middle linear map (referred to as "the canonical middle linear map".); that is, it satisfies: The first two properties make a bilinear map of the abelian group For any middle linear map of a unique group homomorphism of satisfies and this property determines within group isomorphism. See the main article for details. 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 where is a free abelian group over and G is the subgroup of generated by relations The universal property can be stated as follows. Let G be an abelian group with a map that is bilinear, in the sense that Then there is a unique map such that for all and Furthermore, we can give a module structure under some extra conditions: If A is a (S,R)-bimodule, then is a left S-module where If B is a (R,S)-bimodule, then is a right S-module where If A is a (S,R)-bimodule and B is a (R,T)-bimodule, then 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