Question

What is the Galois group of ${\mathit{x}}^{\mathbf{5}}\mathbf{+}\mathbf{32}$over? [Hint: Show that$\mathbf{\mathbb{Q}}\mathbf{\left(}\mathit{\zeta }\mathbf{\right)}$is a splitting field, where$\mathit{\zeta }$ is a primitive fifth root of unity].

Short answer

The Galois group of is $x^5+32\mathrm{is}{\mathrm{\mathbb{Z}}}_4$.

Step-by-step solution

Determine solvable by Radicals

A radical extension overF is consider as a normal field and also $\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$. Then the solvable by radicals equation is considered as $\mathit{f}\left(x\right)\mathbf{=}\mathbf{0}$.

Find the Galois group

Suppose $\zeta$is the primitive fifth root of unity and a prime degree of splitting field $f\left(x\right)=x^5+32$be$p$.

The irreducible splitting field which can be defined as:

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

Suppose the root of the splitting field be $\alpha$, then the Galois group becomes .

$\alpha {\zeta }^0,\alpha {\zeta }^1,\alpha {\zeta }^2,\alpha {\zeta }^3,\alpha {\zeta }^4$

Now, the value of the root becomes:

$\begin{array}{l}{x}^5+32=0\\ x^5=-32\\ x=\sqrt[5]{-32}\\ x=\sqrt[5]{32i}\end{array}$

Since, the splitting root is $\alpha =\sqrt[5]{32}i$, then the Galois group over$\mathrm{\mathbb{Q}}$ is$\mathrm{\mathbb{Q}}(1+{\zeta }^1+{\zeta }^2+{\zeta }^3+{\zeta }^4)$ which is equivalent to ${\mathrm{\mathbb{Z}}}_4$.

Thus, the Galois group of$x^5+32=0$ is ${\mathrm{\mathbb{Z}}}_4$.