Question

Let$\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$ be separable of degreerole="math" localid="1657966853457" $\mathit{n}$ and$\mathit{k}$ a splitting field of$\mathit{f}\left(x\right)$ . Show that the order of$\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{k}$ divides$\mathit{n}\mathbf{!}$ .

Short answer

It is proved that the order of $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{k}$divides$\mathit{n}\mathbf{!}$.

Step-by-step solution

Spiting field of a polynomial.

A splitting group of a polynomial with coefficients in a field is a smallest field of that field over which the polynomial splits or decomposes into linear factors.

Finding the root of F(x) .

If$\mathit{k}$is the splitting field of a separable polynomial$\mathit{f}\left(x\right)$of degree$\mathit{n}$in$\mathit{f}\left(x\right)$.

Since,$\mathit{f}\left(x\right)\mathbf{\in }\mathit{f}\left[x\right]$is separable of degree$\mathit{n}$, and the set of polynomial function is .

$\mathit{R}\mathbf{=}\left\{u_1,u_2,....,u_n\right\}$

If the Galois of the extension$\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$, then$\mathit{\sigma }\left(U_1\right)\mathbf{,}\mathit{\sigma }\left(U_2\right)\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathit{\sigma }\left(U_n\right)$are the roots of minimal polynomial.

Since$\mathit{f}\left(x\right)$ is a maximal ideal. Hence,$\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$ is a field.

Assume$\mathit{\sigma }\mathbf{:}\mathit{k}\mathbf{\to }\mathit{k}\mathbf{\in }\mathit{F}$ be the natural projection of the ring then,

$$\begin{gathered} \mathit{f}\left(\sigma \left(x\right)\right)\mathbf{=}\mathit{\sigma }\left(f\left(x\right)\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{f}\left(x\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left|f\left(x\right)\right| \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

So,$\mathit{\sigma }\left(x\right)$ is a root of$\mathit{f}\left(x\right)$ and of$\mathit{p}\left(x\right)$ .

Sowing that the Galois group of field is divides .

Now, the degree of a single extension$\left[K_{i+1}:k_i\right]$ is equal to the degree of the irreducible factor of$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ . Since, the degree$\left[K,F\right]$ is$\left[k_r:k_{r-1}\right]\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left[k_2:k_1\right]\left[k_1:F\right]$ .

This is almost equal to the factorial $\mathit{n}$. Therefore, by the definition of splitting field $\mathit{\sigma }\left|R=\tau \right|\mathit{R}\mathbf{,}$then $\mathit{\sigma }\left(u_i\right)\mathbf{=}\mathit{\tau }\left(u_i\right)$for all $\mathit{i}$elements. Hence,$\mathit{\sigma }\mathbf{=}\mathit{\tau }$.

Thus,$\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$ and is divisible by order of polynomial. So,$\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$ is also divisible by the factorial of degree of polynomial.

Hence, the order of the Galois group of field is divisible by$\mathit{n}\mathbf{!}$ .