Tensor product of modules: Difference between revisions

Revision as of 09:45, 18 July 2022

Revision as of 10:02, 18 July 2022

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<math display="block">\begin{cases}\operatorname{Hom}_{\Z} (M \otimes_R N, G) \simeq \operatorname{L}_R(M, N; G) \\ g \mapsto g \circ \otimes \end{cases}</math>

This is a succinct way of stating the universal mapping property given above. (ifIf a priori one is given this is natural isomorphism, then <math>\otimes</math> can be recovered by taking <math>G = M \otimes_R N</math> and then mapping the identity map.)

Similarly, given the natural identification <math>\operatorname{L}_R(M, N; G) = \operatorname{Hom}_R(M, \operatorname{Hom}_{\Z}(N, G))</math>,<ref>First, if <math>R=\Z,</math> then the claimed identification is given by <math>f \mapsto f'</math> with <math>f'(x)(y) = f(x, y)</math>. In general, <math>\operatorname{Hom}_{\Z }(N, G)</math> has the structure of a right ''R''-module by <math>(g \cdot r)(y) = g(r y)</math>. Thus, for any <math>\Z</math>-bilinear map ''f'', ''f''′ is ''R''-linear <math>\Leftrightarrow f'(xr) = f'(x) \cdot r \Leftrightarrow f(xr, y) = f(x, ry).</math></ref> one can also define {{math|''M'' ⊗_''R'' ''N''}} by the formula