Problem statement: Given $F \subseteq K,L \subseteq \bar{F}$. $K,L$ finite field extensions and $\bar{F}$ is the algebraic closure of $F$. Show that $K \otimes L$ is a field if and only if any $F$-linearly independent elements $k_1,...,k_n \in K$ is still linearly independent over $L$.
Suppose $K \otimes L$ is a field. So $KL$ is well defined as a subfield of $\bar{F}$ and we have a natural injective homomorphism $f: K \otimes L \to KL$. Suppose we have $\sum_i k_il_i = 0$, we can pull it back through $f$ and $\sum_i k_i \otimes l_i = 0$, but then what algebraic manipulation should I do...?
Assume $k_i$ are linearly independent over $F$, then we can extend them by further elements, if necessary, to obtain a basis (letting $l_i=0$ for the further elements).
Also, let $(b_j)_j$ be a basis for $\,{}_FL$.
Then $(k_i\otimes b_j)_{i,j}$ is a basis for $K\otimes L$, and expressing each $l_i$ by the basis elements, $l_i=\sum_i \lambda_{i,j}b_j$, we obtain $$0\ =\ \sum_i k_i\otimes l_i\ =\ \sum_{i,j}\lambda_{i,j}\,k_i\otimes b_j$$ but $(k_i\otimes b_j)$ are linearly independent, so each $\lambda_{i,j}=0$, implying $l_i=0$.
Conversely, because of finite dimensions (or surjectivity of $f$), it's enough to show that $f:K\otimes L\to KL\subseteq\bar F,\ \,(k\otimes l)\mapsto k\cdot l\ $ is injective.
Let again $k_i$ and $b_j$ be bases for $K$ and $L$, respectively.
The kernel of $f$ consists of the elements $\sum_{i,j}\lambda_{i,j}\,k_i\otimes b_j$ such that $\,\sum_{i,j}\lambda_{i,j}(k_i\cdot b_j)=0\ $ in $\bar F$.
Using the hypothesis, since $k_i$ was a basis, they are also linearly independent over $L\,$ (within $\bar F$), so for each $i$, we get $\sum_{i,j}\lambda_{i,j}b_j=0$ thus all $\lambda_{i,j}=0$ for the kernel elements.