Question

Let $\mathbb{Z}\left[\sqrt{2}\right]$denote the set$\mathbf{\left\{}\mathbf{a}\mathbf{+}\mathbf{b}\sqrt{\mathbf{2}}\mathbf{|}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{\right\}}$ . Show that$\mathit{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{2}}\mathbf{\right]}$ is a subring of$\mathit{R}$ .

Short answer

It is proved that$\mathbb{Z}\left[\sqrt{2}\right]$is a subring of R.

Step-by-step solution

Subring properties

Assume that R is a ring and S is a subset of R. If the following properties are satisfied, then S is a subring of R.

(i) S is closed under addition.

(ii) S is closed under multiplication

(iii)$0_R\in S$

(iv) If $a\in S$, then the solution of the equation$a+x=0_R$ lies in S.

Show the required result

It is given that$\mathbb{Z}\left[\sqrt{2}\right]=\left\{a+b\sqrt{2}|a,b\in \mathbb{Z}\right\}$.

Show the properties as:

(a) For $a=0,b=0$in $a+b\sqrt{2}$, we have $0\in \mathbb{Z}\left[\sqrt{2}\right]$.

(b) Consider $x=a_1+b_1\sqrt{2}$ and$y=a_2+b\sqrt{2}$ , where $a_1,a_2,b_1,b_2$ are integers. This implies that,

$x+y=(a_1+a_2)+(b_1+b_2)\sqrt{2}$

This implies that $x+y\in \mathbb{Z}\left[\sqrt{2}\right]$.

(c) Assume that$a+b\sqrt{2}+x=0$ , that is$x=-a-b\sqrt{2}$, . This implies that every element has an additive inverse.

(d) Let$a+b\sqrt{2}$ and$c+d\sqrt{2}$ are two elements of $\mathbb{Z}\left[\sqrt{2}\right]$. This implies that:

$\begin{array}{c}(a+b\sqrt{2})(c+d\sqrt{2})=(a^2+2b^2)+\sqrt{2}(ad+bc)\\ \in \mathbb{Z}\left[\sqrt{2}\right]\end{array}$

Therefore,$\mathbb{Z}\left[\sqrt{2}\right]$is multiplicatively closed.

Since all the properties are satisfied, this implies that$\mathbb{Z}\left[\sqrt{2}\right]$is a subring of R.