Equal tensor product implies equal basis element Let $\beta$ and $\gamma$ be bases for finite dimensional vector spaces U and V respectively.  Prove that if $u_1, u_2 \in \beta$ and $v_1, v_2 \in \gamma$ and $u_1 \otimes v_1 = u_2 \otimes v_2$, then $u_1 = v_1$ and $v_1 = v_2$.  Is it true if $u_1, u_2 \in U$ and $v_1, v_2 \in V$
Proof
Because $u_1$ and $v_1$ are bases element of U and V it follows that $u_1 \otimes v_1$ is a basis element of $U \otimes V$.  Similarly $u_2 \otimes v_2$ is a basis element of $U \otimes V$.  Since $u_1 \otimes v_1 = u_2 \otimes v_2$,
they are the same basis element, therefore $u_1 = v_1$ and $v_1 = v_2$
If $u_1, u_2 \in U$ and $v_1, v_2 \in V$, then this need not be true.  Given any nonzero scalar k, $(ku_1) \otimes v_1 = u_1 \otimes (kv_2)$ but obviously $(ku_1) \neq u_1$ and $v_1 \neq kv_1$.
Is my argument correct?  It looks so simple that I feel like I'm overlooking something.

A good way to show that a tensor $t$ is nonzero in $U \otimes V$ is to show that there is a bilinear map $U \times V \to W$ that is nonzero on $t$. In this case, you'd like to show that $u_1 \otimes v_1 - u_2 \otimes v_2 \neq 0$. Try building a bilinear map (I suggest a map $U \times V \to F$, the underlying field) that is nonzero on this, assuming for example that $u_1 \neq u_2$. Use that it is friendly to build a bilinear form on a finite dimensional vector space when you're given a basis.
