Tensor product of two complex vector spaces Take two vector spaces $V$ and $W$ over $\mathbb{C}$. We want $V {\otimes}_{\mathbb{C}} W$. As far as I know, we form this tensor product via quotienting the larger space $V \times W$ by the space spanning the linear combinations $$(v_1 + v_2, w_1)-(v_1, w_1)-(v_2, w_1)\:\:\:\:\:(1)$$ $$(v_1, w_1 + w_2)-(v_1, w_1)-(v_1, w_2)\:\:\:\:\:(2)$$ $$\lambda (v_1, w_1)-(\lambda v_1, w_1)\:\:\:\:\:(3)$$ $$\lambda (v_1, w_1)-(v_1, \lambda w_1)\:\:\:\:\:(4)$$
The confusion is about combinations (3) and (4) and what they imply. My understanding is that the scalars come from $\mathbb{C}$ so I can do this:
For $z \in \mathbb{C}$, $$z(v \otimes w) = zv \otimes w = v \otimes zw \:\:\:\:\:(*)$$
But someone told me this cannot happen.
I need to take tensor products of vector spaces and I was trying to apply the tensor product of modules from Dummit and Foote's Abstract Algebra (3rd Ed). Since vector spaces are modules, I thought it would work the same way.
I couldn't follow their explanation at the time for why (*) isn't true for vector spaces (if that's what they were even getting at). Can anyone help clear this up?

Think $v\otimes w$ as the coset
$$v\otimes w=(v,w)+R$$
where $R$ is the mentioned subspace, then the relations (3), (4) mean
$$z(v_1,w_1)+R=(z v_1,w_1)+R=(v_1,z w_1)+R,$$
which is your relation $(*)$.
