**Prove that the center of a ring is a subring that contains the identity.**
**Proof.** Recall that the center of a ring

R is defined as:

\begin{align*}
Z = \{z \in R \ | \ zr = rz \ \text{for all} \ r \in R\}.
\end{align*}

Firstly, note that

Z is not empty since it contains the identity

1 (1z = z1).
Secondly, we need to show that

Z is closed under subtraction. Let

z_1,z_2 \in Z, then

z_1r = rz_1 and

z_2r = rz_2 for all

r \in R. We get the following:

\begin{align*}
(z_1 - z_2)r &= z_1r - z_2r \\
&= rz_1 - rz_2 \\
&= r(z_1 - z_2),
\end{align*}

which implies that

z_1 - z_2 \in Z. So,

Z is closed under subtraction.
What is left is that

Z is closed under multiplication. Take again

z_1,z_2 \in Z. Then we need to show that

z_1z_2 \in Z. Notice that

\begin{align*}
z_1r = rz_1 \quad \text{and} \quad z_2r = rz_2.
\end{align*}

We need to show that

z_1z_2r = rz_1z_2 holds:

\begin{align*}
z_1z_2r = z_1rz_2 \iff z_1z_2r = rz_1z_2
\end{align*}

since

z_1r = rz_1 and

z_2r = rz_2.
Therefore, the center of a ring is a subring that contains the identity.H