Let $F_n=\mathbb{Q}(2^{1/2^n})$ for $n\in \mathbb{N}$. Let $I_n$ be the kernel of $\overline{\mathbb{Q}} \otimes_\mathbb{Q} \overline{\mathbb{Q}} \to \overline{\mathbb{Q}} \otimes_{F_n} \overline{\mathbb{Q}}$, by $u \otimes_{\mathbb{Q}} v \mapsto u \otimes_{F_n} v$. Show that $I_n \subset I_{n+1}$
We have $F_0 \subset F_1 \subset ...$. Let $u,v \in \overline{\mathbb{Q}}$ such that $u\otimes_{F_n} v =0$, how can we show that $u \otimes_{F_{n+1}} v =0$?
Does the following diagram commute? $$\begin{array}{c} & A \otimes_F B & \\ & \swarrow ~~~~~~~ \searrow \\ A \otimes_{F_n} B & \rightarrow & A \otimes_{F_m} B. \end{array}$$
By the universal property, $u \otimes_{F_n} v = 0$ if and only if for every $F_n$-bilinear map $\phi:\bar {\Bbb Q} \times \bar{\Bbb Q} \to \Bbb F_n$, we have $\phi(u,v) = 0$. Note, however, that an $F_{n+1}$-bilinear map must also be $F_{n}$-bilinear.
We could also apply this reasoning to answer your original question directly, since an element $\sum_i u_i \otimes_{F_n} v_i$ is zero if and only if for every $F_n$-bilinear map $\phi$ to $F_n$, we have $\sum_i F(u_i,v_i) = 0$.