Question

1. Verify that $\mathbf{\mathbb{Q}}\left(\sqrt{3}\right)\mathbf{=}\left\{\left.r+s\sqrt{3}\right|r,s\in \mathbb{Q}\right\}$ is a subfield of $\mathbf{ℝ}$.
2. Show that$\mathbf{\mathbb{Q}}\left(\sqrt{3}\right)$ is isomorphic to $\mathbf{\mathbb{Q}}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathbf{}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{)}$.

Short answer

a.$\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)=\left\{\left.\mathbf{r}+\mathbf{s}\sqrt{3}\right|\mathbf{r}\mathbf{,}\mathbf{s}\in \mathrm{\mathbb{Q}}\right\}$ is a subfield of $\mathrm{ℝ}$.

Step-by-step solution

Verify that ℚ(3)={r+s3r,s∈ℚ} is a subfield of ℝ a)

$\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)\subset \mathrm{ℝ}$is a subfield because it has both 0 and 1, and also is closed under addition, multiplication, additive inverse, and multiplicative inverse.

Use computation to verify these conditions as follows:

$$\begin{gathered} \left(r_1\sqrt{3}+r_0\right)\left(s_1\sqrt{3}+s_0\right)=\left(r_1+s_1\right)\sqrt{3}+\left(r_0+s_0\right) \\ \left(r_1\sqrt{3}+r_0\right)\left(s_1\sqrt{3}+s_0\right)=\left(r_1{s}_0+r_0{s}_1\right)\sqrt{3}+\left(3r_1{s}_1+r_0{s}_0\right) \\ -\left(r_1\sqrt{2}+r_0\right)=\left(-r_1\right)\sqrt{3}+\left(-r_0\right) \\ -{\left(r_1\sqrt{3}+r_0\right)}^{-1}=\frac{-r_1}{-3r_1^0+r_0^2}\sqrt{3}+\frac{r_0}{-3r_1^2+r_0^2} \end{gathered}$$

Hence, $\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)=\left\{\left.\mathbf{r}+\mathbf{s}\sqrt{3}\right|\mathbf{r}\mathbf{,}\mathbf{s}\in \mathrm{\mathbb{Q}}\right\}$is a subfield of $\mathrm{ℝ}$.