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Let $K \subset L$ be an algebraic field extension, and let $\alpha$ and $\beta$ be non-conjugates elements of $L$. Is it true that $K(\alpha) \cap K(\beta) = K$? This is probably easy but I just can't think of a general solution, even if in simple cases like $\mathbb{Q}(\sqrt{2}) \cap \mathbb{Q}(\sqrt{3})$ it's obviously true.

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Here's a counterexample: $L=K((2)^{1/4})$, $\alpha= (2)^{1/2}$, and $\beta=(2)^{1/4}$. In this case, $K(\alpha) \cap K(\beta)=K(\alpha)$, but $\alpha$ and $\beta$ are not conjugate, since the minimal polynomial (over $\mathbb Q$) of $\alpha$ is $x^2-2$ and that of $\beta$ is $x^4-2$.

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