On tensor product $U\otimes V$, $0\otimes w = 0$ for any $w\in V$

Question

I've tried using the definition of tensor product: $U\otimes V = \mathcal{F}_{U\times V}/\langle S\rangle$, where $$S = \{(\lambda_1u_1+\lambda_2u_2,v)-\lambda_1(u_1,v)-\lambda_2(u_2,v); (u,\lambda_1v_1+\lambda_2v_2)-\lambda_1(u,v_1)-\lambda_2(u,v_2)\}$$

and so I got that $$0\otimes w = (0,w)+(1-\lambda_1-\lambda_2)(\beta u,\alpha v)$$ for some $\alpha, \beta\in \mathbb{F}$. But I don't know how this can give me the zero element of $U\otimes V$. Any help?

I know that the converse holds, that is, if $u\otimes v = 0$, then $u=0$ or $v=0$.

Answer 1

I never liked this point of view, but here it goes: what you have is that $$0\otimes w=(0,w)+\langle S\rangle.$$ So the question is simply whether $(0,w)\in \langle S\rangle $. And it is: $$ (0,w)=(0_\mathbb F\,0,w)-0_\mathbb F\,(0,w)\in S $$ where $0_\mathbb F$ denotes the scalar zero, and $0$ denotes the zero vector.

Edit: usually there is no need to do this. One simply proves in general that $U\otimes V$ is a vector space, and that $\lambda(u,v)=(\lambda u,v)=(u,\lambda v)$.

Answer 2

Note that $0\otimes w = (0+0)\otimes w = 0\otimes w + 0 \otimes w$ by bilinearity, so you get that $0\otimes w$ is the zero element of $V\otimes W$.