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.
-
Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.
-
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 +.
- 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.)
Best Answer
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 $R$.
The answer to the other question is right too; the center of a commutative ring is the ring itself.