Question

Question: Use the proof of Theorem 11 .18 to express each of these as simple extensions of $\mathrm{\mathbb{Q}}$:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\right) \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{\mathbb{Q}}\left(\sqrt{3},i\right)\mathbf{}\mathbf{} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)\mathbf{} \end{gathered}$$

Short answer

Answer:

1. The simple extension of$\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3}\right)is\mathrm{\mathbb{Q}}\left(\sqrt{2}+\sqrt{3}\right)$.
2. The simple extension of $\mathrm{\mathbb{Q}}\left(\sqrt{3},i\right)is\mathrm{\mathbb{Q}}\left(\sqrt{3}+i\right)$.
3. The simple extension of $\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)is\mathrm{\mathbb{Q}}\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)$.

Step-by-step solution

Describe the simple extension

A simple extension is a field extension which is generated by the adjunction of single element. Every finite field is a single extension of the prime field of the same characteristic.

Expressℚ(2, 3) as simple extensions of ℚ(a)

Consider $\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3}\right)$.

Let$v_{}=\sqrt{2},v_1=-\sqrt{2}and{w}_{}=\sqrt{3}and{w}_1=-\sqrt{3}$

Since $-\sqrt{2}-\sqrt{2}\ne \sqrt{3}+\sqrt{3}$.

Chose c = 1.

Therefore, $\left(c=1\right)\ne \frac{v_1-v}{w-w_1}$

Then as$u=v+cw$ . This gives $u=\sqrt{2}+\sqrt{3}$.

Therefore, the simple extension of $\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3}\right)is\mathrm{\mathbb{Q}}\left(\sqrt{2}+\sqrt{3}\right)$.

Express ℚ3, ias simple extensions of ℚ(b)

Consider$\mathrm{\mathbb{Q}}\left(\sqrt{3},i\right)$ .

Let ,$v_{}=\sqrt{3},v_1=-\sqrt{3}and{w}_{}=i,w_1=-i$

Since $-\sqrt{3}-\sqrt{3}\ne i-\left(-i\right)$.

Chose c =1 .

Therefore, $\left(c=1\right)\ne \frac{v_2-v}{w-w_2}$

Then as $u=v+cw$. This gives $u=\sqrt{3}+i$.

Therefore, the simple extension of $u=\sqrt{3},iisu=\sqrt{3}+i$.

Expressℚ2, 3, 5 as simple extensions of ℚ(c) 

Consider$\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$ .

Simplify as follows.

$$\begin{gathered} \left[\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right):\mathrm{\mathbb{Q}}\right]=\left[\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right):\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3}\right)\right]\left[\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3}\right):\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\right]\left[\mathrm{\mathbb{Q}}\left(\sqrt{2}\right):\mathrm{\mathbb{Q}}\right] \\ =2·2·2 \\ =8 \end{gathered}$$

This implies that $\left[\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right):\mathrm{\mathbb{Q}}\right]$is a simple extension.

Then explicit generator is$\alpha =\sqrt{2}+\sqrt{3}+\sqrt{5}$ because the minimal polynomial of Q over is$x^8-40x^6+352x^4-960x^2+576$ .

Then $\left[\mathrm{\mathbb{Q}}\left(\alpha \right):\mathrm{\mathbb{Q}}\right]=\left[\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right):\mathrm{\mathbb{Q}}\right]$and because $\mathrm{\mathbb{Q}}\left(\alpha \right)\subseteq \mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$. This gives $\mathrm{\mathbb{Q}}\left(\alpha \right)=\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$.

Therefore, the simple extension of $\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)is\mathrm{\mathbb{Q}}\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)$.