轉貼一篇爆文，這個題目出現在即將來臨的習題們中。 ----------------------------------------------------------- 發信人: [email protected] (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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