Linearly disjoint extensions

Remarks

The definition is symmetric in $E$ and $F$. It holds if and only if the $K$-linear map $E \otimes_K F \to \Omega$ with $x \otimes y \mapsto xy$ is injective, where $\otimes_K$ is the tensor product. If $E/K$ is finite, it holds if and only if $[EF:F] = [E:K]$, where $EF$ is the compositum. If $E/K$ is finite Galois, it holds if and only if $E \cap F = K$.

Example: $E = \Q(\sqrt{2})$ and $F = \Q(\sqrt{3})$ in $\C$ are linearly disjoint over $\Q$. Check: $EF = \Q(\sqrt{2}, \sqrt{3})$ has degree $4$ over $\Q$, so $[EF:F] = 2 = [E:\Q]$.

Non-example: let $\omega = e^{2\pi i/3}$, $E = \Q(\sqrt[3]{2})$ and $F = \Q(\omega\sqrt[3]{2})$. Then $E \cap F = \Q$, but $EF = \Q(\sqrt[3]{2}, \omega)$ has degree $6$, so $[EF:F] = 2 < 3 = [E:\Q]$. Hence the $\Q$-basis $1, \sqrt[3]{2}, \sqrt[3]{4}$ of $E$ is linearly dependent over $F$.