Question

Let $\mathit{R}$ be a ring and let $\mathit{Z}\left(R\right)\mathbf{=}\left\{a\in RIar=raforeveryr\in R\right\}$. In other words,$\mathit{Z}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ consists of all elements of $\mathit{R}$that commute with every other element of $\mathit{R}$. Prove that $\mathit{Z}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ is a subring of $\mathit{R}$.$\mathit{Z}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ is called the center of the ring $\mathit{R}$. [Exercise 31 shows that the center of $\mathit{M}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ is the subring of scalar matrices.]

Short answer

It is proved that the $Z\left(R\right)$ is subring of $R$.

Step-by-step solution

Obtain, Z(R)which is a subring of  R

If we consider $z_1,z_2\in Z\left(R\right)$, then

1. Arbitrary $a$of $R$,

$$\begin{gathered} \left(z_1+z_2\right)a=z_{1}a+z_{2}a \\ =a{z}_1+a{z}_2 \\ =a\left(z_1+z_2\right) \end{gathered}$$

Here, the$z_1$ and $z_2$ commute with each element, therefore, $z_1+z_2\in Z\left(R\right)$.

2. Checking the multiplication property,

$$\begin{gathered} \left(z_1{z}_2\right)a=z_1\left(z_{2}a\right) \\ =z_1\left(a{z}_2\right) \\ =\left(z_{1}a\right)z_2 \\ =a\left(z_1{z}_2\right) \end{gathered}$$

Here, $z_1,z_2$are commute with each element, therefore, $z_1,z_2\in Z\left(R\right)$.

Determine that Z(R) is a subring of R

3. The additive property commutes with each element as$0a=a0=0$.

4. Now,

Let’s suppose that,$x\in R$ so that $x+z=0$then,

$\left(x+z\right)a=0=a\left(x+z\right)$

Further simplify,

$$\begin{gathered} xa+za=ax+az \\ ax=xa \\ x\in Z\left(R\right) \end{gathered}$$

The above equation shows that the additive inverse exists in $Z\left(R\right)$.

From above conditions, we can conclude that the $Z\left(R\right)$ is the subring of$R$.