$|Gal(f(x),\mathbb Q )|$?

Question

Let $f(x) = x^5 -7$ then i need to find $|Gal(f(x),\mathbb Q )|$, I can use Eisenstein to see that $f(x)$ is irreductible but the order isn't $5$, what i'm doing wrong?

Answer 1

The $f(x)$ roots are: $$\sqrt[5]{7},\;\omega\sqrt[5]{7},\;\omega^2\sqrt[5]{7},\;\omega^3\sqrt[5]{7},\;\omega^4\sqrt[5]{7},$$ where $\omega=e^{\frac{2\pi}{5}i}$.

Let $\Sigma$ the splitting field of $f$ over $\mathbb{Q}$, then \begin{equation} \Sigma=\mathbb{Q}(\sqrt[5]{7},\omega). \end{equation}

From the tower theorem we have \begin{equation} [\Sigma:\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{7},\omega):\mathbb{Q}(\sqrt[5]{7})]\cdot[\mathbb{Q}(\sqrt[5]{7}):\mathbb{Q}]. \end{equation} We observe that $\text{irr}(\sqrt[5]{7},\mathbb{Q})=x^5-7$, then $[\mathbb{Q}(\sqrt[5]{7}):\mathbb{Q}]=5$.

Then \begin{equation} [\Sigma:\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{7},\omega):\mathbb{Q}(\sqrt[5]{7})]\cdot 5. \end{equation} We are looking for $\text{irr}(\omega,\mathbb{Q}(\sqrt[5]{7}))$, we observe that $\mathbb{Q}(\sqrt[5]{7},\omega)\ne\mathbb{Q}(\sqrt[5]{7})$, because every element in $\mathbb{Q}(\sqrt[5]{7})$ is real, while $\omega$ is a complex number.

Now, $\text{irr}(\omega,\mathbb{Q}(\sqrt[5]{7}))=x^4+x^3+x^2+x+1$, then \begin{equation} |Gal(\Sigma:\mathbb{Q})|=[\Sigma:\mathbb{Q}]=20. \end{equation}

Answer 2

The splitting field contains all fifth roots of $7$, so all of $\exp(2\pi ik/5)\sqrt[5]7$. The $\exp(2\pi ik/5)$ generate a cyclotomic field of degree $4$. It follows that the splitting field has degree $20$ and so the Galois group has order $20$. By studying the action of the elements of the Galois group on $\sqrt[5]7$ and $\exp(2\pi i/5)$ it's possible to elucidate its structure.