Prove that the center of a ring is a subring that contains the identity

The Mathematician — Mon, 13 Feb 2023 13:00:47 +0000

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

center of a ring is a subring that contains the identity Archives - Epsilonify