The Center of a Ring and Subrings!

lostinmath08

[b]1. The problem statement

Let R be a ring. The center of R is defines as follows:

Z(R)= {x E R where xy = yx for all y E R}

Show that Z(R) is a subring of R



3. The attempt at a solution

I know that rings have to follow 4 axioms
a) its an abelian group under addition
b) Closure (ab E R)
c) Associativity ((ab)c =a(bc)
d) Distributivity a(b+c)=ab+ac and (b+c)a= ba+ca

Do the axioms apply to sub rings as well? and how would u go about solving it?

Mystic998

In general, a subring of R is a subset of R which is a ring with structure comparable to R. So you don't actually have to show all the axioms because multiplication being associative and distributive is inherited just by being a subset of R. Similarly, some of the additive group structure is inherited from R (associativity, commutativity of addition in particular). In short, you just have to show that whatever subset that you're claiming is a subring is closed under addition, multiplication, and taking of additive inverses (or more compactly, closed under subtraction and multiplication).

Dick

Exactly right. The axioms have to apply to the subring as well. Start proving them one by one. E.g. if a and b are in Z(R), is a+b in Z(R)?

lostinmath08

i) (for all or any) x,y E R implies x+(-y) E R
ii) (for all or any) x,y E R implies xy E R ( R is closed under mulitplication)

so using the requirements of a subring...this is what i came up with:

x-y=y-x
-y-y=-x-x
-2y=-2x
y=x

and vice versa.

The above is just to satisfy the first axiom.
am i on the right track or completely off?

Dick

Off. You know x+y is in R. R is already a ring. You just want to show for x and y in Z(R), x+y is in Z(R). To do that you have to show that for any w in R, (x+y)*w=w*(x+y). Remember you can use that x and y are in Z(R). So x*w=w*x and y*w=w*y.

frnkarich

By criterion theorem, S is a subring of R iff x-y E R and xy E R for all x, y E S so you must show these two are true.

So let x,y E Z(R) and r E R,

To show x-y E Z(R), you have to look at r(x-y).

by definition, rx=xr and ry=yr, so

r(x-y)= rx-ry = xr-yr= (x-y)r so x-yE Z(R). (This proves that Z(R) is an abelian subgroup of R under addition).

Next, you need to show xy E Z(R) by looking at r(xy).

r(xy)= (rx)y=x(ry)= (xy)r. (Proving Z(R) is closed under multiplication)

Thus by criterion theorem, Z(R) is a subgroup of R.