Question

If p and q are distinct primes then prove that $\mathbf{Q}\mathbf{(}\sqrt{\mathbf{p}}\mathbf{,}\sqrt{\mathbf{q}}\mathbf{)}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{p}}\mathbf{+}\sqrt{\mathbf{q}}\mathbf{)}$ .

Short answer

It is proved that $\mathrm{Q}(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}})=\mathrm{Q}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})$.

Step-by-step solution

Definition of field extension

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

Showing that Q(p+q)=Q(p,q) 

Let$\mathrm{Q}(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}})$. As$\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}$is contained in the setrole="math" localid="1659080776592" $\mathrm{Q}(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}})$. Hence,

role="math" localid="1659080789007" $\mathrm{Q}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})\subseteq \mathrm{Q}(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}})\left(1\right)$

For the element$(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}})\in Q(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}})$. Now consider the difference of square of the two element.

${\left(\sqrt{\mathrm{p}}\right)}^2-{\left(\sqrt{\mathrm{q}}\right)}^2=\mathrm{p}-\mathrm{q}\in \mathrm{Q}\subseteq \mathrm{Q}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}})$

This implies that:

$$\begin{gathered} {\left(\sqrt{\mathrm{p}}\right)}^2-{\left(\sqrt{q}\right)}^2\in \mathrm{Q}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}) \\ \frac{{\left(\sqrt{\mathrm{p}}\right)}^2-{\left(\sqrt{\mathrm{q}}\right)}^2}{\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}}=\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\in \mathrm{Q}(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}) \end{gathered}$$

The element$\sqrt{q}$can be written as:

$\begin{array}{rcl}\frac{1}{2}\left(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\right)+\left(\sqrt{\mathrm{p}}-\sqrt{\mathrm{q}}\right)& =& \sqrt{\mathrm{q}}\\ \sqrt{\mathrm{q}}& \in & \mathrm{Q}\left(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\right)\\ \sqrt{\mathrm{p}}& \in & \mathrm{Q}\left(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\right)\\ \sqrt{\mathrm{p}},\sqrt{\mathrm{q}}& \in & \mathrm{Q}\left(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\right)\end{array}$

This gives:

$Q\left(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}}\right)\subseteq Q\left(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}}\right)$ (2)

From 1 and 2.

$Q\left(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\right)=Q\left(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}}\right)$

It is proved that $Q\left(\sqrt{\mathrm{p}},\sqrt{\mathrm{q}}\right)=Q\left(\sqrt{\mathrm{p}}+\sqrt{\mathrm{q}}\right)$.