Tensor product of modules: Difference between revisions

Revision as of 09:39, 18 July 2022

Revision as of 09:42, 18 July 2022

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{{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, assimilar noted into [[#For equivalent modules]], {{math|''S'' ⊗_''R'' ''T''}} will be a commutative ''R''-algebra as well, with the multiplication map defined by {{math|1=(''m''₁ ⊗ ''m''₂) (''n''₁ ⊗ ''n''₂) = (''m''₁''n''₁ ⊗ ''m''₂''n''₂)}} and extended by linearity. In this setting, the tensor product become a [[fibered coproduct]] in the category of commutative ''R''-algebras. (But it is not a coproduct in the category of ''R''-algebras.) <!--Note that any ring is a '''Z'''-algebra, so we may always take {{math|''M'' ⊗_'''Z''' ''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, ''_RM'' is a left ''R''-module, and the [[commutator]]