Editing Tensor product (section)

=== As a quotient space ===
A construction of the tensor product that is basis independent can be obtained in the following way.

Let {{mvar|V}} and {{mvar|W}} be two [[vector space]]s over a [[field (mathematics)|field {{mvar|F}}]]. 

One considers first a vector space {{mvar|L}} that has the [[Cartesian product]] <math>V\times W</math> as a [[basis (linear algebra)|basis]]. That is, the basis elements of {{mvar|L}} are the [[ordered pair|pairs]] <math>(v,w)</math> with <math>v\in V</math> and {{tmath|1= w\in W }}. To get such a vector space, one can define it as the vector space of the [[function (mathematics)|functions]] <math>V\times W \to F</math> that have a finite number of nonzero values and identifying <math>(v,w)</math> with the function that takes the value {{math|1}} on <math>(v,w)</math> and {{math|0}} otherwise.

Let {{mvar|R}} be the [[linear subspace]] of {{mvar|L}} that is spanned by the relations that the tensor product must satisfy. More precisely, {{mvar|R}} is [[Linear span|spanned by]] the elements of one of the forms:
: <math>\begin{align}
(v_1 + v_2, w)&-(v_1, w)-(v_2, w),\\
(v, w_1+w_2)&-(v, w_1)-(v, w_2),\\
(sv,w)&-s(v,w),\\
(v,sw)&-s(v,w),
\end{align}</math>
where {{tmath|1= v, v_1, v_2\in V }}, <math>w, w_1, w_2 \in W</math> and {{tmath|1= s\in F }}.

Then, the tensor product is defined as the [[quotient space (linear algebra)|quotient space]]:
: <math>V\otimes W=L/R,</math>
and the image of <math>(v,w)</math> in this quotient is denoted {{tmath|1= v\otimes w }}.

It is straightforward to prove that the result of this construction satisfies the [[universal property]] considered below. (A very similar construction can be used to define the [[tensor product of modules]].)