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Let $F$ be a field and $K$ a field extension of $F$. Suppose $a,b\in K$ are algebraic over $F$ with degrees $m$ and $n$, where $m,n$ are relatively prime. Then $F(a) \cap F(b) = F$.

I see that the intersection on the LHS must contain $F$, but I don't see why $F$ contains the LHS.

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    $\begingroup$ Since $K=F(a)\cap F(b)$ is a subfield of $F(a)$ and of $F(b)$, the degree of $K$ over $F$ must divide the degrees $m,n$, so must divide $(m,n)=1$. $\endgroup$ – Pedro Tamaroff Mar 27 '16 at 4:38
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Hint: $[F(a):F]=[F(a):F(a)\cap F(b)][F(a)\cap F(b):F]$

$[F(b):F]=[F(a):F(a)\cap F(b)][F(a)\cap F(b):F]$

Thus $[F(a)\cap F(b):F]$ divides $[F(a):F]$ and $[F(b):F]$ so $[F(a)\cap F(b):F]=1$ since $[F(a):F]$ and $[F(b):F]$ are relatively prime.

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    $\begingroup$ That's not a hint, it is a full solution. $\endgroup$ – Pedro Tamaroff Mar 27 '16 at 4:38

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