Question

Question: Use Galois Criterion to prove that every polynomial of degree $⩽\mathbf{4}$is solvable by radicals.

Short answer

Answer:

Every polynomial of degree$⩽4$is solvable by radicals.

Step-by-step solution

Defintion of Galois extension

The Galois group of the polynomial f (x ) over F is${\mathbf{Gal}}_{\mathbf{F}}\mathbf{K}$ .

Step-2: Showing that every polynomial of degree ⩽4  is solvable by radicals. 

Consider the Galois group of the polynomialhere the field is the irreducible quadratic polynomial or less than degree 4 since the degree of the polynomial is degree. So let the polynomial is

$f\left(x\right)=x^4-a$

Let is the root of splititing filed F (X) over F. Then $Q\left(\zeta ,\sqrt{a}\right)$ is contain all the root of . And Galois group of over the is the degree of 1,2 and 4. So, $Q\left(\zeta ,\sqrt{a}\right)$ over $Q\left(\zeta ,\right)$ is a degree of 1 or 2.

$$\begin{gathered} Q\left(\zeta ,\sqrt{a}\right)/Q\left(\zeta ,\right)=Q\left(\zeta /Q\right)\left(Q\left(\zeta ,\sqrt{a}\right)/Q\left(\zeta ,\right)\right) \\ =1,2,4,8 \end{gathered}$$

Since the derivatives of the polynomial is non zero so the fieldhas characteristic 0 and relative prime to$x^n-1_F$with. It$\zeta$andis nth root od unity in the subfield K . Then

$$\begin{gathered} \left(\zeta \tau \right)={\zeta }^n{\tau }^n \\ =1_F{1}_F \\ =1_F \end{gathered}$$

Since the product of root in degree first$\zeta \tau$is also an nth root of umity. Hence the set of root is closed under multiplication.

These are the order of the splititing field. Hence the order of the Galos group is the same as the degree of the splititing field. If it is order 2 so the group is $Z_2\times Z_2$ and if it is the order 4 so the group is or and the overall group is isomorphic to degree 4. Since the Galois group of the irreducible quartic polynomial is consist the splititing field of order 4 and all four order polynomial is always solvable hence the order 4 is solvable so the Galois group of degree also the solvable.