Question

$$\begin{gathered} \cancel{\mathbf{c}}\left[\sqrt{2}\right]\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathbf{}\mathit{r}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{b}\mathit{y}\mathbf{}\mathit{E}\mathit{x}\mathit{e}\mathit{r}\mathit{c}\mathit{i}\mathit{s}\mathit{e}\mathbf{}\mathbf{13}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{.}\mathbf{}\mathit{L}\mathit{e}\mathit{t}\mathbf{}\mathit{f}\mathbf{:}\cancel{\mathbf{c}}\mathbf{\left[}\sqrt{\mathbf{2}}\mathbf{\right]}\mathbf{\to }\cancel{\mathbf{c}}\mathbf{\left[}\sqrt{\mathbf{2}}\mathbf{\right]}\mathbf{} \\ \mathit{b}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{f}\mathit{u}\mathit{n}\mathit{c}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{d}\mathit{e}\mathit{f}\mathit{i}\mathit{n}\mathit{e}\mathit{d}\mathbf{}\mathit{b}\mathit{y}\mathbf{}\mathit{f}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\sqrt{\mathbf{2}}\mathbf{)}\mathbf{=}\mathit{a}\mathbf{-}\mathit{b}\sqrt{\mathbf{2}} \end{gathered}$$.

$$\begin{gathered} \left(a\right)\mathit{S}\mathit{h}\mathit{o}\mathit{w}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathbf{}\mathit{s}\mathit{u}\mathit{r}\mathit{j}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathbf{}\mathit{h}\mathit{o}\mathit{m}\mathit{o}\mathit{m}\mathit{o}\mathit{r}\mathit{p}\mathit{h}\mathit{i}\mathit{s}\mathit{m}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{r}\mathit{i}\mathit{n}\mathit{g}\mathit{s}\mathbf{.}\mathbf{}\mathbf{} \\ \left(b\right)\mathbf{}\mathit{U}\mathit{s}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathit{o}\mathit{r}\mathit{e}\mathit{m}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{11}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathit{s}\mathit{h}\mathit{o}\mathit{w}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathit{l}\mathit{s}\mathit{o}\mathbf{}\mathit{i}\mathit{n}\mathit{j}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{h}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathit{n}\mathbf{}\mathit{i}\mathit{s}\mathit{o}\mathit{m}\mathit{o}\mathit{r}\mathit{p}\mathit{h}\mathit{i}\mathit{s}\mathit{m}\mathbf{.}\mathbf{} \\ \mathbf{[}\mathit{Y}\mathit{o}\mathit{u}\mathbf{}\mathit{m}\mathit{a}\mathit{y}\mathbf{}\mathit{a}\mathit{s}\mathit{s}\mathit{u}\mathit{m}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{i}\mathit{r}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{a}\mathit{l}\mathbf{.}\mathbf{]} \end{gathered}$$

Short answer

(a) and (b) are proved.

Step-by-step solution

Solution to Part (a)

To show that$f$ is a surjective homomorphism of rings compute as follows:

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

On further solving, we get,

$$\begin{gathered} f\left(\left(a+b\sqrt{2}\right)\right)+\left(c+d\sqrt{2}\right)=f\left(\left(ac+2bd\right)+\left(ad+bc\right)\sqrt{2}\right) \\ =\left(ac+2bd\right)-\left(ad+bc\right)\sqrt{2} \\ =\left(a-b\sqrt{2}\right)+\left(c-d\sqrt{2}\right) \\ =f\left(a+b\sqrt{2}\right)+f\left(c+d\sqrt{2}\right) \end{gathered}$$

Thus, $f$is a homomorphism. It is surjective since for all$a+\sqrt{2}\in \cancel{c}\left[\sqrt{2}\right]$ then we have,$f\left(a-b\sqrt{2}\right)=a+\sqrt{2}$.

Hence, the statement is proved.

Solution to Part (b)

Assume$a-b\sqrt{2}\in ker\left(f\right)$then,$f\left(a+b\sqrt{2}\right)=0$ which is equal to $a-b\sqrt{2}=0$.

If$b\ne 0$ then the equation$a-b\sqrt{2}=0$ becomes$\sqrt{2}=\frac{a}{b}\in$ which is a contradiction$\sqrt{2}$ being irrational. Thus,$b=0$ which also implies that $a=0$.

By using theorem 6.11 which states that assume$f:R\to S$ be a homomorphism of rings with kernel $K$. Then,$K=\left(0_R\right)$ if and only if$f$ is injective.

Therefore, the kernel of$f$ is trivial then by applying theorem 6.11$f$ is injective.

Hence, the statement is proved.