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How would you prove this: If E is an extension of K and a and b are elements of E\K, a^m belongs to K, b^n belongs to K, gcd (m, n) =1, then K(ab) =K(a,b) ?? I don't know which strategy to use to prove it.

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We can assume that $m$ and $n$ are minimal such that $a^m,b^n \in K$ (if not replace $m$ and $n$ by the minimal such. They will still be coprime). Since $m$ and $n$ are coprime, there exists a $k,l \in \mathbb{N}$ such that $mk = 1 + ln$. Then $$(ab)^{mk} = (b^n)^l(a^m)^kb \in K(ab)$$ so since $a^m,b^n \in K$, this shows that $b \in K(ab)$. Similarly we can show that $a \in K(ab)$ (just consider $(ab)^{nl}$) and hence $K(a,b) \subseteq K(ab)$. Since $K(ab) \subseteq K(a,b)$, they must be equal.

You could also use degrees to prove this : if you can show that both $[K(ab):K]=mn=[K(a,b):K]$, then since $K(ab) \subseteq K(a,b)$, they must be equal.

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  • $\begingroup$ Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder $\endgroup$ – Luis Gimeno Sotelo Dec 30 '18 at 12:33

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