# Tensor product equivalent definitions

I'm studying tensor products right now and I've came across multiple definitions. The one I'm confused with is when we have vector spaces $V$ and $W$ and we define the tensor product as the quotient of the free module $F$ of $V \times W$ and $E$, where $E$ can either be:

Definition 1: $E$ is the subspace of $F$ spanned by all elements: $$(v_1+v_2,w)-(v_1,w)-(v_2,w)$$ $$(v,w_1+w_2)-(v,w_1)-(v,w_2)$$ $$a(v,w)-(av,w)$$ $$a(v,w)-(v,aw)$$

or definition 2: $E$ is the subspace of $F$ spanned by all elements: $$(v_1+v_2,w)-(v_1,w)-(v_2,w)$$ $$(v,w_1+w_2)-(v,w_1)-(v,w_2)$$ $$(av,w)-(v,aw)$$

Are these definitions equivalent? (namely, what I'm asking is, does: $(al_1 \otimes l_2) = (l_1 \otimes al_2), = a(l_1 \otimes l_2)$ hold in definition 2)?

Thanks very much

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None of these is a definition. The correct definition of the tensor product is the universal property: the tensor product classifies bilinear maps. Your "definitions" are actually constructions (but there are also other constructions of the tensor product; of course isomorphic). Now to answer your question: The first construction is the correct one, the second is not correct. If $\beta : V \times W \to U$ is a map which is additive in each variable and balanced in the sense that $\beta(av,w)=\beta(v,aw)$ for $a \in K$, $v \in V$ and $w \in W$, then there is no reason why we can conclude $a \beta(v,w) = \beta(a v,w)$. Try to find an example for $V=W=U=K=\mathbb{C}$.
Thanks for your help with this. The issue I am having is that some notions of the two constructions seem too different. For example the proof (at least the one I've seen) that if $v_1,v_2...$ is a basis for V and $w_1,w_2,...$ is a basis for $W$ then the tensors $v_iw_j$ are a basis for the tensor product seem to not work (namely we can't express any tensor in terms of linear combinations of the $(v_i \otimes w_j)$ because nothing "goes outside" the tensors as in definition 1)? If not, can you direct me to a proof? – user112532 Dec 1 '13 at 23:34