$f(X):= X^n- x\in K[X]$ is irreducible and $Gal(K(y)/K)\cong \mathbb{Z}/n\mathbb{Z}$ if $y$ is a root of $f$ Hello I need help with solving the following task
Let $K$ be a field with algebraic closure $\overline{K}$ and $K^{\times}:=K\backslash \{0\}$. Let $2\le n \in \mathbb{Z}$ such that $K^{\times}$ contains a primitive nth root of unity. If $charK=p>0$ assume that $p\nmid n$ holds and $K^{\times n}:=\{y^{n} \mid y\in K^{\times}\}$.
i) Let $x\in K^{\times}$ be an element with the property that if $0<i\in \mathbb{Z}$ and $y\in K^{\times}$ exists such that $x^{i} =y^n$ then $n\mid i $ holds. Show that the polynomial $f(X):= X^n- x\in K[X]$ is irreducible.
ii) Now let $y\in \overline{K}$ be a root of $f$. Show that $K\subset K(y)$ is a galois extension with $Gal(K(y)/K)\cong \mathbb{Z}/n\mathbb{Z}$.
Thanks in advance for any help. 
 A: Your problem constitutes the preliminaries of the so called Kummer theory, for which you can consult any textbook on Galois theory. The particular case considered here is simple enough to admit a direct approach. To avoid overlapping with classical proofs, let us follow more or less the hint given by @Reuns.
1) I keep your notations and hypotheses for the base field $K$, except that I prefer to write  $a\in K^*$ instead of $x$. Let $\alpha$ be a fixed $n$-th root of $a$ in $\bar K$. All the $n$-th roots of $a$ are of the form $\zeta \alpha$, where $\zeta$ runs through the group $\mu_n$ of $n$-th roots of $1$, which is contained in $K^*$, hence $K(\alpha)$ is the splitting field of $X^n-a$ over $K$. These roots are distinct because of the hypotheses on char($K$), so finally the extension $L/K$ is galois. To compute its degree, let $f(X)$ be the minimal polynomial of $\alpha$ over $K$, of degree $d$ say. As $f(X)$ divides $X^n-a$, it decomposes in $\bar K[X]$ into a product of $d$ distinct linear factors of the form $(X-\zeta \alpha)$. The computation of the constant term $f(0)$ shows immediately that ${\alpha}^d \in K^*$. From $n=dq+r$, where $r<d$ or $r=0$ (euclidian division), it follows that $\alpha^r \in K^*$, hence $a^r\in {{K}^*}^n$, and the third hypothesis (not used yet) implies that $r=0, n=dq$. But then ${\alpha}^d =b\in K^*$ implies $a^d={\alpha}^{nd}=({\alpha}^d)^{dq}=b^n$, and the third hypothesis again implies $n\mid d$, hence $n=d$. NB: This is purely an academic exercise. In Kummer theory, it's much easier to show first the property 2), from which 1) follows. 
2) We have seen in 1) that $K(\alpha)/K$ is galois, and its degree is $n$ if $a$ mod ${K^*}^n$ has multiplicative order $n$ (this is the meaning of the third hypothesis). Let $G$ denote the Galois group. Any $s\in G$ permutes the roots of $X^n-a$, hence $s(\alpha)=\zeta_s\alpha$, with $\zeta_s\in \mu_n$, and the map $s \to \zeta_s$ is obviously an injective homomorphism of groups $G\to\mu_n$. Since $G$ has order $n$, this is actually an isomorphism. NB: The third hypothesis is necessary. In general, one has only that the splitting field of $X^n-a$ is cyclic of degree $d$ over $K$, where $d$ is the multiplicative order of $a$ mod ${K^*}^n$ .
