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Is it true that $[K(\zeta_n):K]=\phi(n)$ where $\zeta_n$ be primitive root of unity and K be field of char zero?

I think it should be equal to the degree of cyclotomic polynomial which has degree $\phi(n)$.

Can someone please elaborate that what is the procedure for finite fields

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    $\begingroup$ See here. It divides $\phi(n)$, but need not be equal. $\endgroup$ – Dietrich Burde Jan 7 '16 at 10:20
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No. This is true iff $\Phi_n$, the $n$th cyclotomic polynomial, is irreducible over $K$. The degree of $K(\zeta_n)$ over $K$ is equal to the degree of one irreducible factor of $\Phi_n$ over $K$. They all have the same degree.

Here are some examples: If $K$ is algebraically closed, the degree is always $1$. Over $K=\mathbb{Q}$, $\Phi_n$ is always irreducible. Over $K=\mathbb{R}$ or $K=\mathbb{Q}(\sqrt{5})$, $\Phi_5 = t^4+t^3+t^2+t+1$ is not irreducible, in fact $\Phi_5 = (t^2 + \frac{1 + \sqrt{5}}{2} t+ 1)·(t^2 + \frac{1 - \sqrt{5}}{2} t + 1)$. Here, the degree is $2$.

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