I have come across a problem that shows E ∩ K = F if [EK : F] = [E : F][K : F]. I wonder if the converse holds. I know this reduce to showing the the F-basis of E is a K-basis if E ∩ K = F (if this is true), but is stuck here. Thanks.
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You may want to study the concept ot linearly disjoint extensions. Pete L. Clark's lecture notes give everything you want to know. My "first aid" notes for a past study group may or may not help, depending. – Jyrki Lahtonen May 01 '17 at 06:41
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No. Consider the degree $6$ extension $L=\Bbb Q(\sqrt[3]2,e^{2\pi i/3})$ of $\Bbb Q$, the splitting field of $x^3-2$ over $\Bbb Q$. Then $L=EF$ where $E=\Bbb Q(\sqrt[3]2)$ and $F=\Bbb Q(e^{2\pi i/3}\sqrt[3]2)$. Both of these have degree $3$ and $E\cap F=\Bbb Q$.
Angina Seng
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Thanks. But will it be true if we further require that [E : F], [K : F] coprime? – assking May 01 '17 at 07:06
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