# subring

• August 18th 2012, 01:12 PM
peter1991
subring
Hi ! i have this exercise , i need help

For a fixed element a $\in{} \mathbb{R}$ we have $C(a)=r \in{} R\ /\ ra=ar$ . Prove that $C(a)$ is a subring of R containing a . Prove that the center of R is the intersection of the subrings $C (a)$ for all s $\in{}$ R
• August 18th 2012, 05:22 PM
Deveno
Re: subring
in this case, i believe that you want to show this for an arbitrary ring (possibly non-commutative) R, *not* the ring of real numbers (because every real number centralizes every other real number, because multiplication is commutative on the reals).

you need to show 3 things:

1) if x,y are in C(a), then x - y is in C(a).
2) if x,y are in C(a) , so is xy.
3) C(a) is non-empty (so we can actually use (1) & (2)).

proving a is in C(a) will show (3). does aa = aa?

the distributive law should figure heavily in your proof of (1). the associative law of multiplication should figure heavily in your proof of (2).

to prove that:

$C(R) = \bigcap_{a \in R} C(a)$,

is equivalent to saying that for r in C(R), ra = ar for EVERY a in R. is that not the definition of C(R)?