Question

Find a radical extension of $\mathit{\mathbb{Q}}$containing the given number:

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{}\sqrt[\mathbf{4}]{\mathbf{1}\mathbf{+}\sqrt{\mathbf{7}}}\mathbf{-}\sqrt[\mathbf{5}]{\mathbf{2}\mathbf{+}\sqrt{\mathbf{5}}} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{\left(}\sqrt[\mathbf{5}]{\sqrt{\mathbf{2}}\mathbf{+}\mathbf{i}}\mathbf{\right)}\mathit{l}\mathbf{\left(}\sqrt[\mathbf{3}]{\mathbf{5}}\mathbf{\right)} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{\left(}\sqrt[\mathbf{3}]{\mathbf{3}\mathbf{-}\sqrt{\mathbf{2}}}\mathbf{\right)}\mathit{l}\mathbf{(}\mathbf{4}\mathbf{+}\sqrt{\mathbf{2}}\mathbf{)} \end{gathered}$$

Short answer

1. The radical extension is $\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{5}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7})\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[4]{2+\sqrt{5}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[5]{2+\sqrt{5}},\sqrt[4]{1+\sqrt{7}})$.
2. The radical extension is $\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}},\sqrt[3]{5})$.
3. The radical extension is $\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}},4+\sqrt{2})$.

Step-by-step solution

Solvable Group

Suppose K is a normal radical extension field F and there is an intermediate field E that has all characteristic 0. For the condition that E normal over F, then this implies Gal_FE is a solvable group.

(a)Step 2: Find the Radical Extensions

Here, we have to find the radical extension of $\mathrm{\mathbb{Q}}$ such that it contains $\sqrt[4]{1+\sqrt{7}}-\sqrt[5]{2+\sqrt{5}}$. So, recall the fact that $\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{5}\right)$ and $\mathrm{\mathbb{Q}}\left(\sqrt{5}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7})$.

In the same way, we can say that $\mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7})\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[4]{2+\sqrt{5}})$and $\mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[4]{2+\sqrt{5}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[5]{2+\sqrt{5}},\sqrt[4]{1+\sqrt{7}})$.

Thus, the radical extension becomes .

$\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{5}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7})\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[4]{2+\sqrt{5}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{5},\sqrt{7},\sqrt[5]{2+\sqrt{5}},\sqrt[4]{1+\sqrt{7}})$.

(b)Step 3: Find the Radical Extensions

Here, we have to find the radical extension of$\mathrm{\mathbb{Q}}$ such that it contains$\left(\sqrt[5]{\sqrt{2}+\mathrm{i}}\right)/\left(\sqrt[3]{5}\right)$ . So, recall the fact that$\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$ and $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5})$.

In the same way, we can say that $\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}})$and$\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}},\sqrt[3]{5})$ .

Thus, the radical extension becomes $\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{5},\sqrt[5]{\sqrt{2}+\mathrm{i}},\sqrt[3]{5})$.

(c)Step 4: Find the Radical Extensions

Here, we have to find the radical extension of$\mathrm{\mathbb{Q}}$ such that it contains$\left(\sqrt[3]{3-\sqrt{2}}\right)/(4+\sqrt{2})$ . So, recall the fact that$\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$ and $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})$.

In the same way, we can say that$\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}})$ and $\mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}},4+\sqrt{2})$.

Thus, the radical extension becomes $\mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}})\subseteq \mathrm{\mathbb{Q}}(\sqrt{2},\sqrt{3},\sqrt[3]{3-\sqrt{2}},4+\sqrt{2})$.