Question

Prove that the group $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathbf{K}$in Theorem 12.18 is cyclic. [Hint: Define a mapf from$\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathbf{K}$ to additive group${\mathbf{\mathbb{Z}}}_{\mathbf{n}}$by $\mathit{f}\left(\sigma \right)\mathbf{=}\mathit{k}$, where $\mathit{\sigma }\left(u\right)\mathbf{=}{\mathit{\zeta }}^{\mathbf{k}}\mathit{u}$. Show thatf is a well-defined injective homomorphism and use theorem 7.17].

Short answer

It is proved that group$\mathrm{Ga}{l}_{F}K$ is cyclic.

Step-by-step solution

Determine the splitting field

The shortest field extension of the extension field over which a minimum polynomial break or dissociates into linear components is known as the splitting field.

Determine the proof

Suppose the primitive$n^{th}$ root of unity in the splitting field of characteristic 0,F . Letu is the root of polynomial$x^n-c$ in field F.

Now, if$\zeta$is the primitive$n^{th}$ root of unity over F, then

$$\begin{gathered} {\left({\zeta }^{k}u\right)}^n={\left({\zeta }^{k}u\right)}^n{u}^n \\ ={\left({\zeta }^{k}u\right)}^n{u}^n \\ =1_{F}c \\ =c \end{gathered}$$

Now, the value of uis $u=\zeta u,{\zeta }^{2}u,{\zeta }^{3}u,...,{\zeta }^{n}u$. The reason behind this is $1_F=\zeta ,{\zeta }^2,{\zeta }^3,...,{\zeta }^n$ and this has an n distinct root of the polynomial $x^n-c$.

Therefore, we can say that the splitting field overF is the subgroup $K=F\left(u\right)$.

Here, the subgroup is normal overF and Galois group of the field is $\sigma ,\tau \in Ga{l}_{F}K$. Then the automorphism of the root for all$k,t$ are$\sigma \left(u\right)={\zeta }^{k}u$ and $\tau \left(u\right)={\zeta }^{t}u$.

Therefore, it can be interpreted that the Galois group of the splitting field overF is a cyclic group because the polynomial of the form$x^n-c=0$ is solvable by radicals and the Galois group of the polynomial$x^n-1_F$ is abelian and each abelian is a cyclic group.