Question

Let$\mathrm{\mathbb{Z}}\left(i\right)$ denote the set$\left\{a+\overline{)bi}a,b\in \mathrm{\mathbb{Z}}\right\}$ . Show that$\mathrm{\mathbb{Z}}\left(i\right)$ is a subring of $C$.

Short answer

It is proved that $\mathrm{\mathbb{Z}}\left(i\right)$ is a subring ofC.

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.

1. S is closed under addition.
2. S is closed under multiplication
3. $0_R\in S$
   
4. If $a\in S$, then the solution of the equation localid="1659328738941" $a+x=0_R$ lies in S.

Show the required result

It is given that$\mathrm{\mathbb{Z}}\left[i\right]=\left\{a+\overline{)bi}a,b\in \mathrm{\mathbb{Z}}\right\}$ .

Show the properties as:

1. For $a=0,b=0$ in $a+bi$ , we have role="math" localid="1646373429218" $0\in \mathrm{\mathbb{Z}}\left[i\right]$ .
2. Consider $x=a_1+b_{1}i$ and role="math" localid="1646373260710" $y=a_2+b_{2}i$ where $a_1,a_2,b_1,b_2$ are integers. This implies that,$x+y=\left(a_1+a_2\right)+\left(b_1+b_2\right)i$

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

1. Assume that $a+bi+x=0$ , that is, $x=-a-bi$. This implies that every element has an additive inverse.
2. Let $a+bi$and $c+di$ are two elements of $\mathrm{\mathbb{Z}}\left[i\right]$ , this implies that $\left(a+bi\right)\left(c+di\right)=\left(a^2-b^2\right)+i\left(ad+bc\right)$

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

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