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Given a field $K$ (for example $K = \mathbb Q$) and a polynomial $p \in K[X]$ with cyclic Galois group $C_n$ then one only needs to adjoin an $n$th root of some element $k \in K(\zeta_n)$ to $K(\zeta_n)$ to get the splitting field $L=K(\zeta_n,\sqrt[n]{k})$ of $p$. It is possible to actually find the number $k$.

It is said that the roots of $p$ lie in this field $L$, I would like to know how you can find these roots?

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Let $G\cong C_n$ be the Galois group of $L/K(\zeta_n)$ with generator $\sigma$, where $\zeta_n$ is a fixed primitive $n$-th root of unity. Let me also assume without loss of generality that $p$ is irreducible over $K(\zeta_n)$. If you think about it for a little bit, you will realise that the elements $\alpha=\sqrt[n]{k}$ are characterised by the fact that $\sigma^i(\alpha)=\zeta_n^i\alpha$ for all $i$. This is essentially Kummer theory. That gives you a hint how to find $k$:

Let $\beta$ be a root of $p$. Then, you will easily convince yourself that

$$

\alpha=\sum_{i=0}^{n-1}\zeta_n^{-i}\sigma^i(\beta)

$$

does satisfy the required transformation property. That property implies in particular that $\sigma$ fixes $\alpha^n$, and so we will have $k=\alpha^n\in K(\zeta_n)$. So how do we explicitly compute $k$ as a $K$-linear combination of $\zeta_i$? Well, as $i$ runs from 0 to $n-1$, $\sigma^i(\alpha)$ runs through the roots of $p$ (that’s where the irreducibility assumption comes in). So you easily see that in $\alpha^n$, the coefficient of any $\zeta_n^i$ is a ~~symmetric~~ polynomial in the roots of $p$ that is invariant under cyclic permutations ~~and can therefore be expressed explicitly in terms of elementary symmetric polynomials in the roots~~.

**Edit** (thank you, chandok):

To express these coefficients explicitly as elements of $K(\zeta_n)$, you need to essentially repeat the trick of resolvent cubics that will be familiar to you from solving quartics by radicals.

To show you how this works, let me take the example of cubic extensions. Let the roots of $p$ be denoted by $\beta_1$, $\beta_2$, $\beta_3$, and denote a primitive cube root of unity by $\omega$. Then $k=\alpha^3$ is

$$

\left(\sum_i\omega^{i-1}\beta_i\right)^3 = (\sum_i\beta_i^3

+ \beta_1\beta_2\beta_3)1 + 3(\beta_1^2\beta_2 + \beta_2^2\beta_3 + \beta_3^2\beta_1)\omega + 3(\beta_2^2\beta_1 + \beta_3^2\beta_2 + \beta_1^2\beta_3)\omega^2.

$$

The coefficient of 1 is completely symmetric in the roots, so you can express it in terms of elementary symmetric polynomials, i.e. in terms of coefficients of $p$. The coefficients $\gamma_1$ and $\gamma_2$ of $\omega$ and of $\omega^2$ respectively are not symmetric, but rather are swapped by any 2-cycle. But that means that any expression that is symmetric in the $\gamma_i$ will also be symmetric in the $\beta_i$. So you consider the polynomial

$$

f(x)=(x-\gamma_1)(x-\gamma_2).$$

Its coefficients are symmetric in the $\beta_i$, so you can express them in terms of the coefficients of $p$. Also, since the Galois group of $p$ is cyclic, $f$ splits completely over $K(\zeta_3)$. So you factor it and find the coefficients of $\omega$ and of $\omega^2$ in $k$.

A similar trick works for any $n$. Given an expression in the roots of $p$ that is symmetric under cyclic permutations, but not quite symmetric, you consider its orbit under the entire action of $S_n$ and concoct a polynomial, whose roots will consist of that orbit. You can then compute that polynomial from $p$, and since the Galois group of $p$ is cyclic, that polynomial will completely factor and thus give you the coefficients for $k$.

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