Let $K$ be a field, Splitting field $L$ over $K$ of polynomial $f(x)=x^n-\theta$ Let $K$ be a field, let $n\geq2$ be an integer such that the characteristic $ch(K)$ of $K$ does not divide $n$, and let $\theta \in K$. Consider the splitting field $L$ over $K$ of the polynomial $f(x)=x^n-\theta$
(a) Show that $f(x)$ is a separable polynomial over $K$.
(b) Show that if $a$ and $b$ are two roots of $f(x)$ in $L$, then $b=\eta a$ for some $n$th root of unity $\eta \in L$.
(c) Show that $L$ contains a primitive $n$th root of unity $\omega$ (i.e. $\omega=1_L$ and $\omega^k \neq 1_L$ for $1\leq k\leq n-1)$ and if $a\in L$ is a root of $f(x)$ then all roots of $f(x)$ are $a,\omega a,...,\omega^{n-1}a$. Show that $L=K(\omega,a)$.
(d) Show that for each $\sigma \in G=Aut(L\backslash K(\omega))$ we have $\sigma(a)=\omega^{j(\sigma)}a$ for a unique integer $0\leq j(\sigma)\leq n-1$ and that the map $\sigma \mapsto j(\sigma)+n\Bbb{Z}$ is an injective group homomorphism $j:G\rightarrow \mathbb{Z}/n\mathbb{Z}$.
(e) Use (d) to show that $G$ is a cyclic group of order dividing $n$.
I have barely started learning the concepts of fields and Galois Theory. I know that $f(X)\in K[X]$ is called separable when it has distinct roots in a splitting field over $K$.But Given some $\theta \in K$, I don't know how to go about this. Rest of them, I'm really lost... Any help is appreciated.
 A: We assume of course that
$\theta \ne 0; \tag 0$
then:
(a.) With
$f(x) = x^n - \theta, \; \theta \in K, \tag 1$
we have
$f'(x) = nx^{n -1}; \tag 2$
since
$\text{ch}(F) \not \mid n, \tag 3$
it follows that
$n \ne 0 \tag 4$
in $F$ or any extension of $F$.  Therefore 
$f'(x) \ne 0, \tag 5$
and the only zero of (2) is $0$ itself; indeed, by (4),
$f'(z) = nz^{n - 1} = 0 \Longleftrightarrow z^{n - 1} = 0 \Longleftrightarrow z = 0; \tag 6$
on the other hand, 
$f(0) = 0^n - \theta = -\theta \ne 0; \tag 7$
we clearly see that $f(x)$ and $f'(x)$ have no root in common.  Therefore, by the standard result, the zeroes of $f(x)$ are distinct in any field containing them.
(b.)  If $a, b \in L$ are such that
$f(a) = f(b) = 0, \tag 8$
then
$a^n - \theta = 0 = b^n - \theta, \tag 9$
whence
$a^n = \theta = b^n; \tag{10}$
thus, with
$\eta = \dfrac{b}{a}, \tag{11}$
we find
$\eta^n = \left ( \dfrac{b}{a} \right )^n = \dfrac{b^n}{a^n} = \dfrac{\theta}{\theta} = 1; \tag{12}$
we see that $\eta$ is an $n$-th root of unity and
$b = \eta a; \tag{13}$
we note that 
$\eta \ne 1 \tag{14}$
since $a \ne b$ are two zeros of $f(x)$.
(c.)  We have seen in (b) that $L$ contains an $n$-th root of unity $\eta \ne 1$; now let
$\Upsilon = \{ \eta_1 = \eta, \eta_2, \ldots, \eta_m \} \tag{15}$
be the set of non-unit $n$-th roots of unity in $L$, and suppose none of them are primitive; we see that
$\eta = \eta_1 \in \Upsilon \ne \emptyset; \tag{16}$
then for each $i$, $1 \le i \le m$ there is an integer $k_i$, $1 \le k_i \le n - 1$, with
$\eta_i^{k_i} = 1; \tag{17}$
let $k_{min}$ be the least of these $k_i$; with the aid of Euclidean division, we may write
$n = qk_{min} + r, \; 0 \le r < k_{min}; \tag{18}$
then if $\eta_{min}$ is an element of $L$ corresponding to $k_{min}$, since $\eta_{min}^{k_{min}} = 1$ we have
$\eta_{min}^r = (\eta_{min}^{k_{min}})^q \eta_{min}^r = \eta_{min}^{qk_{min} + r} = \eta_{min}^n = 1, \tag{19}$
which contradicts the minimality of $k_{min}$ amongst the $k_i$ such that 
$\eta_i^{k_i} = 1$; thus $\Upsilon$ must contain a primitive $n$-th root of unity $\omega \ne 1$.  Now if
$f(a) = a^n - \theta = 0, \tag{20}$ 
then for $0 \le k \le n -1$,
$f(\omega^k a) = (\omega^k a)^n - \theta = \omega^{kn}a^n - \theta = (\omega^n)^k a^n - \theta = 1^k a^n - \theta = a^n - \theta = 0; \tag{21}$
furthermore, we see these zeroes $\omega^k a$ of $f(x)$ are distinct, for if
there were $0 \le i < j \le n - 1$ with
$\omega^i a = \omega^j a, \tag{22}$
then
$\omega^{j - i} = 1,\; 0 < j -i \le n -1, \tag{23}$
precluded by the primitivity of $\omega$.  Thus the $\omega^k a$, $0 \le k \le n - 1$, form a collection of $n$ distinct roots of $f(x)$ in $L$; since $\deg f(x) = n$ there can be no other zeroes of $f(x)$; hence we have presented all roots of $f(x)$ in $L$.  It is at this point clear that
$f(x) = x^n - \theta = \displaystyle \prod_0^n (x - \omega^k a) \in L,  \tag{24}$
the splitting field of $f(x)$ over $K$.
Now we also have
$\omega, a \in K(\omega, a) \subset L, \tag{25}$
so also then
$\omega^k a \in K(\omega, a), \; 0 \le k \le n - 1; \tag{26}$
if follows that (24) binds in $K(\omega, a)$ and thus $f(x)$ splits in $K(\omega, a)$; but since $L$ is the smallest field in which $f(x)$ splits, we must indeed have
$K(\omega, a) = L. \tag{27}$
(d.)  Now let
$\sigma \in \text{Aut}(L/K(\omega)); \tag{28}$
that is, $\sigma$ is a field automorphism of $L$ which fixes $K(\omega)$; then, since $\sigma$ fixes $\theta \in K$,
$(\sigma(a))^n = \sigma(a^n) = \sigma(\theta) = \theta; \tag{29}$
it thus follows that $\sigma(a)$ is a zero of $x^n - \theta$; therefore
$\sigma(a) = \omega^k a, \; \text{for some} \; k, \;0 \le k \le n - 1; \tag{30}$
it further follows that $k$ is the sole integer in the range $0 \le k \le n - 1$ for which (30) holds, since $\sigma(a)$ is uniquely determined for $\sigma \in \text{Aut}(L/K(\omega))$; since $k$ is uniquely determined by $\sigma$ we may denote this affiliation by writing
$k = j(\sigma), \tag{31}$
so that (30) is now written
$\sigma(a) = \omega^{j(\sigma)}a, \; 0 \le j(\sigma) \le n - 1; \tag{32}$
now because any $\sigma$ also fixes $\omega \in K(\omega)$, for $\sigma_1, \sigma_2 \in \text{Aut}(L/K(\omega))$, 
$\sigma_1 \sigma_2(a) = \sigma_1(\sigma_2(a)) = \sigma_1(\omega^{j(\sigma_2)}a) = \sigma_1(\omega^{j(\sigma_2)})\sigma_1(a)$
$= (\sigma_1(\omega))^{j(\sigma_2)}\sigma_1(a) = \omega^{j(\sigma_2)} \omega^{j(\sigma_1)}a = \omega^{j(\sigma_1) + j(\sigma_2)}a, \tag{33}$
where we understand $j(\sigma_1) + j(\sigma_2) \mod n$ since $\omega^n = 1$; thus we take $j(\sigma_1) + j(\sigma_2)$ as a coset representative $\mod n$ and hence $0 \le j(\sigma_1) + j(\sigma_2) \le n - 1$ is the unique value of $j(\sigma_1 \sigma_2)$:
$j(\sigma_1 \sigma_2) = j(\sigma_1) + j(\sigma_2); \tag{34}$
we now see we may define the homomorphism
$\phi: \text{Aut}(L/K(\omega)) \to \Bbb Z / n \Bbb Z, \; \phi(\sigma) = j(\sigma) + n \Bbb Z \tag{35}$
since
$\phi(\sigma_1 \sigma_2) = j(\sigma_1 \sigma_2) + n \Bbb Z = (j(\sigma_1) + j(\sigma_2)) + n \Bbb Z = (j(\sigma_1) + \Bbb Z) + (j(\sigma_2) + \Bbb Z); \tag{36}$
if
$\phi(\sigma) = j(\sigma) + n \Bbb Z = 0 + n \Bbb Z = n \Bbb Z, \tag{37}$
then
$\sigma(a) = \omega^{j(\sigma)}a = a, \tag{38}$
and by virtue of the fact that $L = K(\omega, a)$ we see that
$\sigma = e \in \text{Aut}(L/K(\omega)), \tag{39}$
$e$ being the identity automorphism; this shows that
$\ker \phi = \{ e \} \subset \text{Aut}(L/K(\omega)), \tag{40}$
and therefore $\phi$ is an injective homomorphism into $\Bbb Z / n \Bbb Z$.
(e.)  By virtue of what we have just proved in (d), we immediately see that $\phi$ is an isomorphism 'twixt $\text{Aut}(L/K(\omega))$ and a subgroup of $\Bbb Z/n \Bbb Z$; since this group is cyclic or order $n$, every subgroup is cyclic with order dividing $n$; thus 
the isomorphic image $\phi(\text{Aut}(L/K(\omega))) \subset \Bbb Z / n \Bbb Z$ must itself be cyclic with order dividing $n$, as must $\text{Aut}(L/K(\omega))$ be itself.
As a final observation, we see that if $n$ is prime we must have
$\text{Aut}(L/K(\omega, a)) \cong \Bbb Z / n \Bbb Z, \tag{41}$
since in this case $\Bbb Z / n \Bbb Z$ has no proper subgroups.
