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 $F$.
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