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!

  • $\begingroup$ If you're actually assuming that $F\subset E$ then this doesn't make much sense. But if you're not assuming that then what does it mean to say that a subset of $E$ is $F$-independent? $\endgroup$ – David C. Ullrich May 18 '18 at 15:01
  • $\begingroup$ It means that if you take elements $\alpha_1,...,\alpha_n$ in $E$ that are $K$-linearly independent, then if $\beta_1,...,\beta_n$ are in $F$ such that $\alpha_1\beta_1+...+\alpha_n\beta_n=0$, then necessarily $\beta_1=...=\beta_n=0$. The linear combination is well defined in $EF$. $\endgroup$ – El.Gon.Zalo May 19 '18 at 14:33
  • $\begingroup$ What is $EF$? (The problem is what is the meaning of $\alpha\beta$ for $\alpha\in E$ and $\beta\in F$. Saying "in $EF$" doesn't help answer that - if I don't know what $\alpha\beta$ is I certainly don't know what $EF$ is...) $\endgroup$ – David C. Ullrich May 19 '18 at 15:09
  • $\begingroup$ Ok, i see what you mean. Consider both $E,F \subset \overline{K}$ in the algebraic closure. $\endgroup$ – El.Gon.Zalo May 19 '18 at 15:28

$\newcommand{\Q}{\mathbb{Q}}$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 $F$.

  • $\begingroup$ Thank you so much for the answer! What about the other implication? If they are linearly disjoint, is the intersection trivial? $\endgroup$ – El.Gon.Zalo May 19 '18 at 15:51
  • $\begingroup$ I think this implication is true: Let $\{\alpha_1,...,\alpha_n\}$ be a basis of the extension $E \cap F / \mathbb{Q}$. If one supposes by contradiction that $E \cap F \neq \mathbb{Q}$, then $n>1$. The $\alpha_i$ are $\mathbb{Q}$-linearly independent, and so, by assumption, there are $F$-linearly independent. This is impossible. $\endgroup$ – El.Gon.Zalo May 19 '18 at 15:57
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    $\begingroup$ @El.Gon.Zalo, $\newcommand{\Q}{\mathbb{Q}}$my argument would be the following. Let $a \in E \cap F$. Then $1, a \in E$ are linearly dependent over $F$, as $a = a \cdot 1$, with $a \in F$. But then by assumption they are linearly dependent over $\Q$, so that $a = c \cdot 1 = c$ for some $c \in \Q$. $\endgroup$ – Andreas Caranti May 19 '18 at 16:54

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