Question

1. Verify that $\mathbf{\mathbb{Q}}\left(\sqrt{2}\right)\mathbf{=}\left\{\left.r+s\sqrt{2}\right|r,s\in \mathbb{Q}\right\}$is a subfield of $\mathbf{ℝ}$.
2. Show that$\mathbf{\mathbb{Q}}\left(\sqrt{2}\right)$ is isomorphic to $\mathbf{\mathbb{Q}}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{)}$. [Hint: Exercise 6 in Section 5.2 may be helpful.]

Short answer

a) $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{\left.\mathbf{r}+\mathbf{s}\sqrt{\mathbf{2}}\right|\mathbf{r}\mathbf{,}\mathbf{s}\in \mathrm{\mathbb{Q}}\right\}$is a subfield of $\mathrm{ℝ}$.

Step-by-step solution

Verify that ℚ(2)={r+s2r,s∈ℚ} is a subfield of ℝ a)

$\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\subset \mathrm{ℝ}$is a subfield because it has 0 and 1 and is closed under addition, multiplication, additive inverse, and multiplicative inverse.

Use computation to verify these conditions as follows:

$$\begin{gathered} \left(r_1\sqrt{2}+r_0\right)\left(s_1\sqrt{2}+s_0\right)=\left(r_1+s_1\right)\sqrt{2}+\left(r_0+s_0\right) \\ \left(r_1\sqrt{2}+r_0\right)\left(s_1\sqrt{2}+s_0\right)=\left(r_1{s}_0+r_0{s}_1\right)\sqrt{2}+\left(2r_1{s}_1+r_0{s}_0\right) \\ -\left(r_1\sqrt{2}+r_0\right)=\left(-r_1\right)\sqrt{2}+\left(-r_0\right) \\ {\left(r_1\sqrt{2}+r_0\right)}^{-1}=\frac{-r_1}{-2r_1^0+r_0^2}\sqrt{2}+\frac{r_0}{-2r_1^2+r_0^2} \end{gathered}$$

Hence, $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{\left.\mathbf{r}+\mathbf{s}\sqrt{\mathbf{2}}\right|\mathbf{r}\mathbf{,}\mathbf{s}\in \mathrm{\mathbb{Q}}\right\}$is a subfield of$\mathrm{ℝ}$.