abstract algebra - Linearly disjoint fields

We say that two field $E,F$, extending the same base field $K$, are linearly disjoint if every finite subset of $E$ that is $K$-linearly independent is also $F$-linearly independent.

Suppose $K = \mathbb{Q}$. Is this definition equivalent to say that $E \cap F = \mathbb{Q}$? And if so, why?

My attempt: Assuming that the extensions $E/\mathbb{Q}$, $F/\mathbb{Q}$ are finite, I tried using the primitive element theorem, so that $E=\mathbb{Q}(\alpha)$ and $F=\mathbb{Q}(\beta)$, for some $\alpha,\beta$ algebraic. Then the elements of these fields are just polynomials in these numbers, but from here i was not able to conclude.

Is is even true if the extensions are not finite?

Thanks in advance!

Answer

$\newcommand{\Q}{\mathbb{Q}}$No, it is not equivalent.

As a possibly typical example, take $K = \Q$, $E = \Q(\omega \alpha)$, $F = \Q(\alpha)$, where $\alpha = \sqrt[3]{2}$ and $\omega$ is a primitive third root of unity.

We have $E \cap F = K$, but while $1, \omega \alpha, \omega^{2} \alpha^{2} \in E$ are independent over $K$, you have
$$1 + \frac{\alpha^{2}}{2}( \omega \alpha) + \frac{\alpha}{2} (\omega^{2} \alpha^{2}) = 1 + \omega + \omega^{2} = 0,$$
so they are not independent over $F$.