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The Center of a Ring and Subrings! 
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#1
Feb2808, 02:12 PM

P: 12

[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? 


#2
Feb2808, 03:07 PM

P: 206

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).



#3
Feb2808, 03:07 PM

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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)?



#4
Feb2808, 07:00 PM

P: 12

The Center of a Ring and Subrings!
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: xy=yx yy=xx 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? 


#5
Feb2808, 07:19 PM

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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.



#6
Nov1710, 05:06 PM

P: 1

By criterion theorem, S is a subring of R iff xy 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 xy E Z(R), you have to look at r(xy). by definition, rx=xr and ry=yr, so r(xy)= rxry = xryr= (xy)r so xyE 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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