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

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 $V\otimes W$.
