Question

Show that $\mathrm{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{d}}\mathbf{\right]}$is a subring of$\mathrm{ℂ}$. If$\mathit{d}\mathbf{\ge }\mathbf{0}$, show that$\mathrm{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{d}}\mathbf{\right]}$is a subring of$\mathrm{ℝ}$.

Short answer

It has been proved that $\mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$is a subring of $\mathrm{ℂ}$and if $d\ge 0$, then it is a subring of$\mathrm{ℝ}$.

Step-by-step solution

Prove closure under addition

Clearly, $\mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$is a subset of $\mathrm{ℂ}$.

Let $p+q\sqrt{d},r+s\sqrt{d}\in \mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$.

$\left(p+q\sqrt{d}\right)+\left(r+s\sqrt{d}\right)=\left(p+q\right)+\left(r+s\right)\sqrt{d}\in \mathrm{\mathbb{Z}}\sqrt{d}$

Thus, it is closed under addition

Prove closure under multiplication

Let$p+q\sqrt{d},r+s\sqrt{d}\in \mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$.

$\left(p+q\sqrt{d}\right)\left(r+s\sqrt{d}\right)=\left(pr+qsd\right)+\left(ps+qr\right)\sqrt{d}\in \mathrm{\mathbb{Z}}\sqrt{d}$

Thus, it is closed under multiplication.

Hence, it is a subring.

If d≥0

If $d\ge 0$, then$\sqrt{d}$ is a real number.

Thus, $\mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$is a subset of $\mathrm{ℝ}$.

Hence,$\mathrm{\mathbb{Z}}\left[\sqrt{d}\right]$ is a subring of $\mathrm{ℝ}$.