Question

Question: Prove that the Galois group of an irreducible quadraic polynomial is solvable.

Short answer

Answer:

Irreducible cubic polynomial is solvable.

Step-by-step solution

Defintion of Galois extension.

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 split into linear factors.

Step-2: Showing that irreducible quadraic polynomial is solvable.

Consider the splitting field$f\left(x\right)$ here the field is the irreducible quadraic polynomial since

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

Let is the root of the splititing field $f\left(x\right)$ over F then $Q\left(\alpha ,\sqrt{a}\right)$ contains all the root of $f\left(x\right)$. And Galois group of $Q\left(\alpha ,\right)$ over the Q is the degree of 1,2 and 4 so, $Q\left(\alpha ,\sqrt{a}\right)$ over $Q\left(\alpha ,\right)$ is a degree of 1 and 2. Since

$$\begin{gathered} \raisebox{1ex}{$Q\left(\alpha ,\sqrt{a}\right)$}\!\left/ \!\raisebox{-1ex}{$Q\left(\alpha ,\right)=\left(Q\left(\alpha /Q\right)\right)Q\left(\alpha ,\sqrt{a}/Q\left(\alpha ,\right)\right)$}\right. \\ =1,2,4,8 \end{gathered}$$

There 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 isrole="math" localid="1659341907535" $Z_2$and if it is the order 4 so the group is$Z_2\times Z_2$or$Z_4$and the overall group is isomorphic to degree 4. Since the of order 4 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 he Galois group also the solvable.