Question

Let K is a normal extension of Q and $\left[K:Q\right]\mathbf{=}\mathit{P}$ with p prime show that $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{K}\mathbf{\cong }{\mathit{Z}}_{\mathbf{P}}$.

Short answer

$\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{K}\mathbf{\cong }{\mathit{Z}}_P$

Step-by-step solution

Definition of the splititing field.

A splititing field of a minimal polynomial with coefficient in an extension field is a smallest field extension of that field over which the polynomial splits into linear factors.

Step-2: Showing that GalQK≅ZP .

$\mathit{K}$If $\mathit{K}$is a normal field extension is contain the set of subgroup $\mathit{E}$ with $\mathit{K}\mathit{\subset }\mathit{E}$ and $\mathit{Q}$be an extension field. The minimal polynomial $\mathit{p}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{n}}\mathbf{-}\mathbf{1}$ and it is under the group of ${\mathit{Z}}_{\mathbf{p}}$. Now the derivative of the minimal polynomial,

$$\begin{gathered} \mathit{p}\left(x\right)\mathbf{=}{\mathit{p}}^{\mathbf{n}}{\mathit{x}}^{{\mathbf{p}}^{\mathbf{n}\mathbf{-}\mathbf{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}{\mathit{x}}^{{\mathbf{p}}^{\mathbf{n}\mathbf{-}\mathbf{1}}} \end{gathered}$$

Let subfield E of the normal field $\mathit{K}$consisted of the ordered ${\mathit{p}}^{\mathbf{n}}$ have a distinct root of the polynomial ${\mathit{x}}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{-}\mathbf{1}$. The set $\mathit{E}$ is the subfield over if the element if the subfield $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathbf{\in }\mathbf{}\mathit{E}$ using the addition property.

$$\begin{gathered} {\left(a+b\right)}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{=}{\mathit{a}}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{+}{\mathit{b}}^{{\mathbf{p}}^{\mathbf{n}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b} \end{gathered}$$

Now for multiplication property

$$\begin{gathered} {\left(ab\right)}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{=}{\mathit{a}}^{{\mathbf{p}}^{\mathbf{n}}}{\mathit{b}}^{{\mathbf{p}}^{\mathbf{n}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{a}\mathit{b} \end{gathered}$$

If a is nonzero normal field of the subfield E then $\mathbf{-}\mathit{a}\mathbf{,}{\mathit{a}}^{\mathbf{-}\mathbf{4}}$ are also in the subfield $\mathit{E}$since the normal field $\mathit{K}$ contain the smallest set of the normal subfield $\mathit{E}$and

the root of the set are same $\mathit{E}\mathbf{=}\mathit{K}$ here $\mathit{K}$ is an order of ${\mathit{p}}^{\mathbf{n}}$and the basis of the field is $\mathit{n}$.

Hence the Galois field $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{K}$over the extension field $\mathit{Q}$and the set is dependent on the order of the set that is ${\mathit{p}}^{\mathbf{n}}$so ${\mathit{p}}^{\mathbf{n}}\mathbf{\cong }{\mathit{Z}}_{\mathbf{p}}$${\mathit{p}}^{\mathbf{n}}\mathbf{\cong }{\mathit{Z}}_{\mathbf{p}}$

So Galois group $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{K}$ is the order of the ordered of the normal field $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{K}\mathbf{\cong }{\mathit{Z}}_{\mathbf{p}}$.