# Prove that the center of a ring is a subring.

The center of a ring $R$ is $\{c\in R : cr=rc$ for every $r \in R\}$. Prove that the center of a ring is a subring. What is the center of a commutative ring?

Is my solution right?

solution

You just need to prove that the centre is a ring within itself.

1. Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.

2. Show that it is a group under addition: Take a and b in the centre and r in R. Then $(a + b)r = ar + br = ra + rb = r(a + b)$

hence it is closed under addition. Show that $0$ (additive identity) is in the centre.

$0 = 0.r = r.0$ so $0$ is in the centre.

For a in the centre, there exists $-a$ in R such that $a + (-a) = 1 = (-a) + a$

Show that $-a$ is in the centre. Let r in R $0 = 0.r = (a + (-a))r = ar + (-a)r$ and $0 = r.0 = r(a + (-a)) = ra + r(-a)$

as $ra = ar$, it follows that $(-a)r = r(-a)$ Hence inverses exist in the centre. So the centre is a group under +.

1. Show that the centre is closed under multiplication: For a,b in the centre and r in R,# $(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab)$

thus multiplication is closed. And show that 1 (the multiplicative identity) is in the centre. $1.r = r = r.1$
hence 1 is in the centre.

So the center is a subring of R.

The center of a commutative ring is the ring itself. (By definition the centre is the ring of all commutative elements.)

• It is correct but showing the identity is superflous, as rings do not necciserily have an identity. – Zelos Malum Nov 27 '16 at 17:47
• @ZelosMalum some authors impose that rings must have an identity. – Xam Nov 27 '16 at 18:31
• Possible duplicate of On the center of a ring – user2371916 Jul 26 '17 at 19:42

Your proof is right, but unnecessarily long. When you want to prove that some nonempty set is a subring you have to use the subring test. Denote the center of your ring by $Z(R)$, you only have to prove that $1\in Z(R)$ and if $x,y\in Z(R)$, then $x-y, x\cdot y\in Z(R)$. Since you have proved all that, then $Z(R)$ is a subring of $R$.