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The Center of a Ring and Subrings!

  1. Feb 28, 2008 #1
    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?
     
  2. jcsd
  3. Feb 28, 2008 #2
    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).
     
  4. Feb 28, 2008 #3

    Dick

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    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)?
     
  5. Feb 28, 2008 #4
    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?
     
  6. Feb 28, 2008 #5

    Dick

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    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.
     
  7. Nov 17, 2010 #6
    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.
     
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