Hi ! i have this exercise , i need help

For a fixed element a we have . Prove that is a subring of R containing a . Prove that the center of R is the intersection of the subrings for all s R

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- August 18th 2012, 12:12 PMpeter1991subring
Hi ! i have this exercise , i need help

For a fixed element a we have . Prove that is a subring of R containing a . Prove that the center of R is the intersection of the subrings for all s R - August 18th 2012, 04:22 PMDevenoRe: 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:

,

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)?