Question

Let $\mathbf{\zeta }$be a primitive fifth root of unity. Prove that ${\mathbf{Gal}}_{\mathbf{Q}}\mathbf{Q}\left(\mathbf{\zeta }\right)$ the Galois group of ${\mathit{x}}^{\mathbf{5}}\mathbf{-}\mathbf{1}$ is cyclic of order 4.

Short answer

Answer:

$Ga{l}_{Q}Q\left(\zeta \right)$is Galois group of $x^5-1$ is cyclic of order 4.

Step-by-step solution

Step 1:Defintion of splititing field. 

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

Step-2: Showing that GalQQζ  is Galois group ofx5-1   is cyclic of order 4.

Letis the primitive fifth root of unity. Andis a prime degree of the splititingfield.

Assume the function$f\left(x\right)=x^5-1$splititing field as irreducible is$Q\left[x\right]$then the splititing field is define as

$f\left(x\right)=x^{p-1}+x^{p-2}+.........x+1$

Since$\left(x-1\right)f\left(x\right)=x^p-1$here the degree of the field is fifth orderthen

$$\begin{gathered} \left(x-1\right)f\left(x\right)=x^5-1 \\ {x}^5-1=\left(x-1\right)\left(x^4+x^3+x^2+x+1\right) \end{gathered}$$

After the factorization of splititing field the first termis not irreducible and$\left(\zeta -1\right)\ne 0$since the factor does not exist now for second term$\left(x^4+x^3+x^2+x+1\right)$is equal to zero and it is a irreducible field so the Galois group of the fourth order splititing field is the group of order 1,2 or 4 and every quadric order of the splititing field is a cyclic and Galois group is the cyclic group and every cyclic group is abelian.

Hence the order of the Galois group of fifth order splititing field 4.