Is the center of a ring an ideal?

Let $Z(R) = \{ a \in R : ax = xa,\text{ for all $x \in R$}\}$

Is $Z(R)$ an ideal of $R$?

Attempt: I already proved that $Z(R)$ is a subring of $R$. I would say yes, since if $x \in R$, then $xa$ is an element in $Z(R)$ and if $a\in Z(R)$ then we have $ax\in Z(G)$. So by definition it is an ideal for $R$.

Please can anyone please give some feedback? Thank you.

Solution 1:

Hint: If $I$ is an ideal in a ring $R$ and $1\in I$, then $I=R$. Moreover, one may show that $1\in Z(R)$. Hence $Z(R)$ is an ideal if and only if $Z(R)=R$. Can you use this to find a counterexample?

The hint above tells us that $Z(R)$ is an ideal of $R$ if and only if $R$ is a commutative ring. So letting $R$ be your favorite noncommutative ring (say the ring of $n\times n$ matrices over $\Bbb C$) gives a counterexample.

Solution 2:

To show $I = Z(R)$ is an ideal, you need to show two things: (1) $I$ is an additive subgroup of $R$. (2) $I$ absorbs $R$ on both sides. You showed (2) already. The quickest way to (1) is showing that $I$ is closed under the binary map $f(x,y) = x - y$. Then from that you can prove that you indeed have an additive group.

Let $x, y \in I$. Then for any $a \in R$, $a(x-y) = ax - ay = xa - ya = (x - y) a$ from the ring axioms, so $I$ is closed under subtraction. QED

$a \in R , x \in Z(R) \implies (ax)b = a(xb) = abx \neq b(ax)$ necc. So you were wrong about the first part. It is an additive subgroup though!