Question

Let $\mathbf{\mathbb{Q}}\left(\sqrt{2}\right)$be as in Exercise 39 of Section 3.1. Prove that the function $\mathit{f}\mathbf{:}\mathbf{\mathbb{Q}}\mathbf{\left(}\sqrt{\mathbf{2}}\mathbf{\right)}\mathbf{\to }\mathbf{\mathbb{Q}}\mathbf{\left(}\sqrt{\mathbf{2}}\mathbf{\right)}$given by $\mathit{f}\left(a+b\sqrt{2}\right)\mathbf{=}\mathit{a}\mathbf{-}\mathit{b}\sqrt{\mathbf{2}}$ is an isomorphism.

Short answer

It is proved that $f$is an isomorphism.

Step-by-step solution

Property of Rings

If any ring$R$has elements such that,$a,b\in R$, then, addition and multiplication of the function of its elements is respectively given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Isomorphism

From exercise 39, we already have a ring as:$\mathrm{\mathbb{Q}}\sqrt{2}=\left\{a+b\sqrt{2}|a,b\in \mathrm{\mathbb{Q}}\right\}$

Let us consider a ring the function$f$such that:

$f:\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)\to \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$

According to question, $f\left(a+b\sqrt{2}\right)=a-b\sqrt{2}$. This implies that:

$$\begin{gathered} a+b\sqrt{2}=a-\left(-b\right)\sqrt{2} \\ =f\left(a-b\sqrt{2}\right) \\ f\to surjective \end{gathered}$$

Now, for $a_1,b_1,a_2,b_2\in \mathrm{\mathbb{Q}}$, we have:

$$\begin{gathered} f\left(a_1+b_1\sqrt{2}\right)=f\left(a_2+b_2\sqrt{2}\right) \\ {a}_1-b_1\sqrt{2}=a_2+b_2\sqrt{2} \\ {a}_1-a_2=\left(b_2-b_1\right)\sqrt{2} \end{gathered}$$

Now, for $b_2=b_1$, we have: $a_1=a_2$. This clearly implies:

$$\begin{gathered} f\left(a_1+b_1\sqrt{2}\right)=f\left(a_2+b_2\sqrt{2}\right) \\ {a}_1+b_1\sqrt{2}=a_2+b_2\sqrt{2} \\ f\to injective \end{gathered}$$

Checking for addition and multiplication property, we have:

$$\begin{gathered} f\left\{\left(a_1+b_1\sqrt{2}\right)+\left(a_2+b_2\sqrt{2}\right)\right\}=\left(a_1+a_2\right)-\left(b_1\sqrt{2}+b_2\sqrt{2}\right) \\ =\left(a_1-b_1\sqrt{2}\right)+\left(a_2-b_2\sqrt{2}\right) \\ =f\left(a_1+b_1\sqrt{2}\right)+f\left(a_2+b_2\sqrt{2}\right) \end{gathered}$$

$$\begin{gathered} f\left\{\left(a_1+b_1\sqrt{2}\right)\left(a_2+b_2\sqrt{2}\right)\right\}=\left(a_1{a}_2+2b_1{b}_2\right)-\left(a_2{b}_1\sqrt{2}+a_1{b}_2\sqrt{2}\right) \\ =\left(a_1-b_1\sqrt{2}\right)\left(a_2-b_2\sqrt{2}\right) \\ =f\left(a_1+b_1\sqrt{2}\right)f\left(a_2+b_2\sqrt{2}\right) \end{gathered}$$

Hence proved, $f$is an isomorphism.