Question

Prove that$\mathbf{\mathbb{Q}}\mathbf{(}\mathbf{3}\mathbf{+}\mathit{i}\mathbf{)}\mathbf{=}\mathbf{\mathbb{Q}}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{i}\mathbf{)}\mathbf{.}$

Short answer

It is proved that $\mathrm{\mathbb{Q}}(3+i)=\mathrm{\mathbb{Q}}(i-1).$

Step-by-step solution

Describe the concept of simple exyension

A simple extension is s field extension that is generated by the adjunction of a single element. Every finite field is a simple extension of the prime field of the same characteristics.

Prove that ℚ(3+i)=ℚ(1-i)

Consider$\mathrm{\mathbb{Q}}(3+i)=\mathrm{\mathbb{Q}}(1-i).$ .

First,

Asrole="math" localid="1659164900268" $4\in \mathrm{\mathbb{Q}}\subseteq \mathrm{\mathbb{Q}}(1-i)$ and $(1-i)\in \mathrm{\mathbb{Q}}(1-i)$. So, their difference will also belong to role="math" localid="1659164984121" $\mathrm{\mathbb{Q}}(1-i)$as follows $4-(1-i)\in \mathrm{\mathbb{Q}}(1-i)$.

This implies that,

$(3+i)\subseteq \mathrm{\mathbb{Q}}(1-i)$

Also,

$(3+i)\subseteq \mathrm{\mathbb{Q}}(3+i)$

Therefore, this gives,

$\mathrm{\mathbb{Q}}(3+i)\subseteq \mathrm{\mathbb{Q}}(1-i)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}......\left(1\right)$

Now, get reverse containment of the above.

As $4\in Q\subseteq Q(3+i)$and $(3+i)\in Q(3+i).$So, their difference will also belong to$Q(3+i)$ as follows $4-(3+i)\in Q(3+i)$.

This implies that,

$(1-i)\subseteq \mathrm{\mathbb{Q}}(3+i)$

Also,

$(1-i)\subseteq \mathrm{\mathbb{Q}}(1-i)$

Therefore, this gives,

$\mathrm{\mathbb{Q}}(1-i)\subseteq \mathrm{\mathbb{Q}}(3+i)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}......\left(2\right)$

From equations (1) and (2),

role="math" localid="1659164420860" $\mathrm{\mathbb{Q}}(3+i)=\mathrm{\mathbb{Q}}(i-1)$

Hence, it is proved that$\mathrm{\mathbb{Q}}(3+i)=\mathrm{\mathbb{Q}}(i-1).$