Question

Let $\mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$be the set of all real numbers of the form

${\mathbf{r}}_{\mathbf{0}}\mathbf{+}{\mathbf{r}}_{\mathbf{1}}\sqrt{\mathbf{2}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}{\left(\sqrt{\mathbf{2}}\right)}^{\mathbf{2}}\mathbf{+}\cdots \mathbf{+}{\mathbf{r}}_{\mathbf{n}}{\left(\sqrt{\mathbf{2}}\right)}^{\mathbf{n}}$, with $n\ge 0$and${\mathbf{r}}_{\mathbf{i}}\in \mathrm{\mathbb{Q}}$ .

1. Show that $\mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$is a subring of $\mathrm{ℝ}$.
2. Show that the function$\theta :\mathrm{\mathbb{Q}}\left[x\right]\to \mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$defined by $\theta \left(f\left(x\right)\right)=f\left(\sqrt{2}\right)$is a surjective homomorphism, but not an isomorphism. Thus Theorem 4.1 is not true with$\mathbf{R}=\mathrm{\mathbb{Q}}$ and$\sqrt{2}$ in place of x. Compare this with Exercise 25.

Short answer

1. It is proved that $\mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$is a subring of $\mathrm{ℝ}$.
2. It is proved that the function $\theta :\mathrm{\mathbb{Q}}\left[x\right]\to \mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$defined by $\theta \left(f\left(x\right)\right)=f\left(\sqrt{2}\right)$is a surjective homomorphism.

Step-by-step solution

Statement of corollary 3.11 

Corollary 3.11states that when $f:R\to S$would be a homomorphism of rings, then the image of$f$ would be a subring of S.

Show that  ℚ2 is a subring of ℝ

a)

Construct $\theta :\mathrm{\mathbb{Q}}\left[x\right]\to \mathrm{ℝ}$as$\theta \left(f\left(x\right)\right)=f\left(\sqrt{2}\right)$ . It is known that 0 is a homomorphism using the same approach as in Exercise 23. Therefore, according to corollary 3.11, its image is a subring of$\mathrm{ℝ}$ . From the definition, $\mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$would be the image of$\theta$ because ${\sum }_{i=0}^n{r_i\left(\sqrt{2}\right)}^i=\theta \left({{\sum }_{i=0}^n{r}_{i}x}^i\right)$.

Hence, it is proved that$\mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$ is a subring of $\mathrm{ℝ}$.

Show that function θ:ℚx→ℚ2  defined by θfx=f2is a surjective homomorphism, but not an isomorphism.  

b)

It is known that $\theta$would be a surjective homomorphism, according to the solution to the preceding item. The fact that it is not injective could be seen.

$\begin{array}{c}\theta \left(x^2-2\right)={\left(\sqrt{2}\right)}^2\\ =0\\ =\theta \left(0\right)\end{array}$

However,$x^2-2\ne 0$ .

Hence, it is proved that the function $\theta :\mathrm{\mathbb{Q}}\left[x\right]\to \mathrm{\mathbb{Q}}\left[\sqrt{2}\right]$defined by$\theta \left(f\left(x\right)\right)=f\left(\sqrt{2}\right)$ is a surjective homomorphism.