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

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

Step-by-step solution

Statement of Exercise 5.2.6

Exercise 5.2.6states each element of the congruence-class ring $\mathrm{\mathbb{Q}}\left[x\right]/(x^2-2)$can be expressed in the form [ax + b].

Every equivalent classes contain a certain representative of the form [a₁x + a_o].

The addition is given from the formula as follows:

$\left[a_0+a_{1}x\right]·\left[b_0+b_{1}x\right]=\left[\left(a_0+b_0\right)+\left(a_1+b_1\right)x\right]$

The multiplication is given from the formula as follows:

$$\begin{gathered} \left[a_0+a_{1}x\right]·\left[b_0+b_{1}x\right]=\left[a_0{b}_0+\left(a_0{b}_1+a_1{b}_0\right)x+a_1{b}_1{x}^2\right] \\ =\left[a_0{b}_0+\left(a_0{b}_1+a_1{b}_0\right)x+a_1{b}_{1}2\right] \\ =\left[\left(a_0{b}_0+2a_1{b}_1\right)+\left(a_0{b}_1+a_1{b}_0\right)x\right] \end{gathered}$$

Show that ℚ(2) is isomorphic to ℚ[x]/(x2-2)b

Let the map $\varphi :\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\to \mathrm{\mathbb{Q}}\left[x\right]/(x^2-2)$with $\varphi \left(r_1\sqrt{2}+r_0\right)=\left[r_{1}x+r_0\right]$. According to the description of the field operations in $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$and$\mathrm{\mathbb{Q}}\left[x\right]/(x^2-2)$from Exercise 5.2.6,$\phi$is a homomorphism.

With every localid="1649066614302" $\left[r_{1}x+r_0\right]\in \mathrm{\mathbb{Q}}\left[x\right]/(x^2-2)$, there is $\varphi \left(r_1\sqrt{2}+r_0\right)=\left[r_{1}x+r_0\right]$. As a result, $\varphi$is surjective.

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

Therefore, $r_1\sqrt{2}+r_0=s_1\sqrt{2}+s_0$. As a result, $\phi$is injective.

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