Question

If r and s are nonzero prove that $\mathit{Q}\sqrt{\mathbf{r}}\mathbf{=}\mathit{Q}\sqrt{\mathbf{s}}$if and only if $\mathit{r}\mathbf{=}{\mathit{t}}^{\mathbf{2}}\mathbf{}\mathit{s}$for some $\mathit{t}\mathbf{\in }\mathit{Q}$.

Short answer

$Q\sqrt{r}=Q\sqrt{s}$

Step-by-step solution

Definition of a simple extension.

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

Showing that Qr=Qs .

Let r and s are nonzero and $r=t^{2}s$.

Consider

$$\begin{gathered} Q\left(\sqrt{r}\right)=Q\left(\sqrt{t^{2}s}\right) \\ =Q\left(\sqrt[t]{s}\right) \end{gathered}$$

As.$t\in Q$ Therefore

$Q\sqrt{r}=Q\sqrt{s}$

Then $\sqrt{r}\in Q\sqrt{s}$

This implies

$\sqrt{r}=x+y\sqrt{s}x,y\in Q$

Square both the side

$$\begin{gathered} {\left(\sqrt{r}\right)}^2={\left(x-y\sqrt{s}\right)}^2 \\ r=x^2+r{y}^2+2xy\sqrt{s} \\ \mathrm{If}2\mathrm{xy}\ne 0 \\ \sqrt{\mathrm{s}}=\frac{\mathrm{r}-\left({\mathrm{x}}^2+{\mathrm{ry}}^2\right)}{2\mathrm{xy}}\in \mathrm{Q} \end{gathered}$$

$$\begin{gathered} \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{contradicitionas}\sqrt{\mathrm{s}}\notin \mathrm{Q}.\mathrm{So},2\mathrm{xy}=0.\mathrm{This}\mathrm{implies}\mathrm{either}\mathrm{x}=0\mathrm{or}\mathrm{y}=0. \\ \mathrm{Now}\mathrm{if}\mathrm{x}=0\mathrm{then} \\ \mathrm{r}={\mathrm{sy}}^2,\mathrm{y}\in \mathrm{Q} \\ \mathrm{And}\mathrm{if}\mathrm{y}=0\mathrm{then} \\ \mathrm{r}={\mathrm{x}}^2,\sqrt{\mathrm{r}}=\mathrm{x}\notin \mathrm{Q} \\ \mathrm{This}\mathrm{gives} \\ \mathrm{r}={\mathrm{y}}^2\mathrm{s}orr=t^{2}s.t\in Q \end{gathered}$$