Tensor product of modules: Difference between revisions

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==Additional structure==
{{confusing|The whole paragraph at the end is confusing. Also it seems to repeat what is already mentioned earlier.}}
{{see also|Free product of associative algebras}}
If ''S'' and ''T'' are commutative ''R''-algebras, then, as noted in [[#For equivalent modules]], {{math|''S'' ⊗<sub>''R''</sub> ''T''}} will be a commutative ''R''-algebra as well, with the multiplication map defined by {{math|1=(''m''<sub>1</sub> ⊗ ''m''<sub>2</sub>) (''n''<sub>1</sub> ⊗ ''n''<sub>2</sub>) = (''m''<sub>1</sub>''n''<sub>1</sub> ⊗ ''m''<sub>2</sub>''n''<sub>2</sub>)}} and extended by linearity. In this setting, the tensor product become a [[fibered coproduct]] in the category of ''R''-algebras.<!--Note that any ring is a '''Z'''-algebra, so we may always take {{math|''M'' ⊗<sub>'''Z'''</sub> ''N''}}.-->
 
If ''M'' and ''N'' are both ''R''-modules over a commutative ring, then their tensor product is again an ''R''-module. If ''R'' is a ring, ''<sub>R</sub>M'' is a left ''R''-module, and the [[commutator]]