Question

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

Short answer

It is proved that $\mathbb{Z}\left[i\right]$ is a subring of C.

Step-by-step solution

Prove that S is a ring  

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 $\mathbb{Z}\left[i\right]=\left\{a+bi|a,b\in \mathbb{Z}\right\}$.

Show the properties as:

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

(b) Consider $x=a_1+b_{1}i$and $y=a_2+bi$where $a_1,a_2,b_1,b_2$are integers. This implies that,

$x+y=(a_1+a_2)+(b_1+b_2)i$

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

(c) Assume that $a+bi+x=0$, that is, $x=-a-bi$. This implies that every element has an additive inverse.

(d) Let $a+bi$and $c+di$are two elements of $\mathbb{Z}\left[i\right]$, this implies that

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

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

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