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→Definition: uniform notation for the scalar products |
→Definition: explicitly stating the precondition used in the following |
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Proof: For the first statement, let ''L'' be the subgroup of <math>M \otimes_R N</math> generated by elements of the form in question, <math>Q = (M \otimes_R N) / L</math> and ''q'' the quotient map to ''Q''. We have: <math>0 = q \circ \otimes</math> as well as <math>0 = 0 \circ \otimes</math>. Hence, by the uniqueness part of the universal property, ''q'' = 0. The second statement is because to define a [[module homomorphism]], it is enough to define it on the generating set of the module. <math>\square</math>
The proposition says that one can work with explicit elements of the tensor products instead of invoking the universal property directly each time. This is very convenient in practice. For example, if ''R'' is commutative and the left and right actions by ''R'' on modules are considered to be equivalent, then <math>M \otimes_R N</math> can naturally be furnished with the ''R''-scalar multiplication by extending
:<math>r \cdot (x \otimes y) := (r \cdot x) \otimes y = x \otimes (r \cdot y)</math>,
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