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 $\mathbf{\mathbb{Q}}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{}\mathbf{/}\mathbf{}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{3}\mathbf{)}$.

Short answer

b. It is proved $\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$is isomorphic to $\mathrm{\mathbb{Q}}\left[x\right]/(x^2-3)$.

Step-by-step solution

Show that ℚ(3) is isomorphic to ℚ[x] / (x2 - 3) b)

Let the map $\varphi :\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)\to \mathrm{\mathbb{Q}}\left[x\right]/(x^2-3)$with $\varphi \left(r_1\sqrt{3}+r_0\right)=\left[r_{1}x+r_0\right]$. It is observed that in$\left.\mathrm{\mathbb{Q}}\left[x\right]\right|\left(x^2-3\right)$as follows:

$$\begin{gathered} \left[r_{1}x+r_0\right]+\left[s_{1}x+s_0\right]=\left[\left(r_1+s_1\right)x+\left(r_0+s_0\right)\right] \\ \left[r_1\sqrt{3}+r_0\right]+\left[s_1\sqrt{3}+s_0\right]=\left[\left(r_1{s}_0+r_0{s}_1\right)x+\left(3r_1{s}_1+r_0{s}_0\right)\right] \end{gathered}$$

As a result, according to the description of the field operations in$\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$above and the ones with $\mathrm{\mathbb{Q}}\left[x\right]/(x^2-3)$, Ф is a homomorphism.

With every localid="1649069790535" $\left[r_{1}x+r_0\right]\in \mathrm{\mathbb{Q}}\left[x\right]/(x^2-3)$, there is $\varphi \left(r_1\sqrt{3}+r_0\right)=\left[r_{1}x+r_0\right]$. As a result,$\phi$is surjective.

When $\varphi \left(r_1\sqrt{3}+r_0\right)=\varphi \left(s_1\sqrt{3}+s_0\right)$,then$\left.x^2-3\right|\left(r_1-s_1\right)x+\left(r_0-s_0\right)$because the degrees involved indicates$\left(r_1-s_1\right)x+\left(r_0-s_0\right)=0$ .

Therefore, $r_1\sqrt{3}+r_0=s_1\sqrt{3}+s_0$. As a result,$\phi$ is injective.

Hence, it is proved$\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$ is isomorphic to $\mathrm{\mathbb{Q}}\left[x\right]/(x^2-3)$.