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轉貼一篇爆文，這個題目出現在即將來臨的習題們中。
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發信人: [e-mail] (Apostol), 看板: math
標  題: 分享...(代數環論)
發信站: 清華資訊(楓橋驛站) (Thu May 27 07:45:11 1999)
轉信站: ntnumath!news.ntnu!ctu-gate!news.nctu!news.ntust!news.civil.ncku!netn


  Let a^3 = a for all a in a ring R, show that R is commutative.

  (pf):

  Claim that if y^2 = y, and note that (yx -yxy)^2 = 0 = (xy - yxy)^2
  for all x in R.  (證:我們僅需要展開即可)

  And a^3 = a ==> a^4 = a^2.
  Now we have ( (a^2) x - (a^2) x (a^2) )^2 = 0
             =( x (a^2) - (a^2) x (a^2) )^2  by the claim.
  So, ( (a^2) x - (a^2) x (a^2) )^3 = ( (a^2) x - (a^2) x (a^2) ) =
                                      ~~~~~~~~~~~~~~~~~~~~~~~~~~~
      ( (a^2) x - (a^2) x (a^2) )^2 * ( (a^2) x - (a^2) x (a^2) ) = 0.
                                                                    ~~
  and ( x (a^2) - (a^2) x (a^2) )^3 = ( x (a^2) - (a^2) x (a^2) ) =
                                      ~~~~~~~~~~~~~~~~~~~~~~~~~~~
      ( x (a^2) - (a^2) x (a^2) )^2 * ( x (a^2) - (a^2) x (a^2) ) = 0.
                                                                   ~~
 畫線部分是我們要的, So we have

  ( (a^2) x - (a^2) x (a^2) ) = 0 = ( x (a^2) - (a^2) x (a^2) )

 That's (a^2) x = x (a^2) for all x in R.
 i.e. a^2 in Z(R) for all a in R. ------------------(A)

 Using the fact: if a^2 + a in Z(R) for all a in R, then R is commutative.
 Z(R) = { z: rz = zr for all r in R } is a sub-ring of R.
 (這個事實一會再證;先拿來用)

 We want to show that a^2 + a in Z(R) for all a in R.  Let's consider that
 ( a^2 + a )^2 = a^4 + 2 * a^3 + a^2 = a^2 + 2 * a + a^2 = 2( a^2 + a )
 And note that (a^2 + a)^2 in Z(R) and a^2 in Z(R), => 2a in Z(R)
 by close property of group.
 (a^2 + a )= ( a^2 + a )^3 = a^6 + 3 * a^5 + 3 * a^4 + a^3
 = a^2 + 3 * a + 3 * a^2 + a => 3 ( a^2 + a ) = 0 in Z(R).
 And note that 3 (a^2 + a) in Z(R) and a^2 in Z(R) => 3a in R.
 by close property of group.
 Therefore, we get a in Z(R).-----------------------(B)
 So, by (A) and (B), we can use the fact to Q.E.D.

  Now we come to show the fact:
  Let R be a ring.
  Show that if a^2 + a in Z(R) for all a in R.
  Then R is a commutative ring.

  Consider (a+b)^2 + (a+b) = a^2 + ab + ba + b^2 + a + b. in Z(R)
  Therefore, ab + ba in Z(R)
  that's for every x in R, we have x(ab+ba)=(ab+ba)x.
  Choose x = a , we can get aab = baa => aa in Z(R) since b is arbitrary.
  So, by a^2 + a in Z(R) and a^2 in Z(R) => a in Z(R) since Z(R) is still a
  ring under the reduced operations.
  Since a is arbitary chosen , that is Z(R) = R.
  We are done!

  Now we give it a try for another statement:

  Let a^4 = a for all a in a ring R , show that R is commutative.

  (Pf):

  Still want to show that if a^2 + a in Z(R) for all a in R, then R is
  Commutative.
  Consider that r^4 = r = (-r)^4 = -r for all r in R => 2r = 0 for all r in R.
  So, (a^2 + a)^2 = a^4 + 2 * a^3 + a^2 = a + a^2 .
  Let y = a^2 + a. we have y^2 = y.
  So we have (yx -yxy)^2 = 0 = (xy - yxy)^2 for all x in R.
  Implies (yx - yxy)^4 = 0 = yx - yxy  &
          (xy - yxy)^4 = 0 = xy - yxy.
  => for all x in R, we have (a^2 + a)x = x(a^2 + a)
  Therefore, a^2 + a in Z(R) for all a in R.          Q.E.D.


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