A ring {R, +, .} is called Boolean if r2 = r for all r ∈ R. We present four proofs that a Boolean ring is commutative

MacHale, Desmond

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UCC Library and UCC researchers have made this item openly available.Please let us know how this has helped you. Thanks!Title Boolean rings are definitely commutative!Author(s) MacHale, DesmondPublication date 2015Original citation MacHale, D. (2015) 'Boolean Rings are Definitely Commutative!',Bulletin of the Irish Mathematical Society, 76, pp. 77-78.Type of publication Article (peer-reviewed)Link to publisher'sversionhttp://banach.ucd.ie/bull76/index.phpAccess to the full text of the published version may require asubscription.Rights © 2015 Irish Mathematical SocietyItem downloadedfromhttp://hdl.handle.net/10468/9603Downloaded on 2021-11-27T09:34:08ZIrish Math. Soc. BulletinNumber 76, Winter 2015, 77–78ISSN 0791-5578Boolean Rings are Definitely Commutative!DESMOND MACHALEAbstract. A ring {R,+, .} is called Boolean if r2 = r for allr ∈ R. We present four proofs that a Boolean ring is commutative.A ring {R,+, .} is called Boolean if r2 = r for all r ∈ R. Inthis bicentenary year of Boole’s birth we present four proofs that aBoolean ring is commutative. Our first proof is the standard onefound in many textbooks.Proof 1. For all r ∈ R we have r = r2 = (−r)2 = −r, so r + r = 0.Next, for all x and y in R, x+y = (x+y)2 = x2+xy+yx+y2, so bycancellation in the group {R,+}, we have xy+yx = 0 = xy+xy, bythe above. Again by cancellation we have xy = yx, as required. Proof 2. As in Proof 1, xy + yx = 0, for all x and y in R. Sincefor all r ∈ R, 0.r = 0 = r.0 we have (xy + yx)x = x(xy + yx) orxyx + y.x2 = x2.y + xyx. Cancelling xyx and remembering thatx2 = x, we get xy = yx, as required. Proof 3. Since for all r, r2 = r it follows that if r2 = 0 then r = 0.Now for all x and y in R we have (xy − xyx)2 = xyxy + xyxxyx−xyxyx−xyxxy = xyxy+xyxyx−xyxyx−xyxy = 0. So xy−xyx = 0and xy = xyx. Then (yx − xyx)2 = yxyx + xyxxyx − yxxyx −xyxyx = yxyx + xyxyx− yxyx− xyxyx = 0. So yx− xyx = 0 andyx = xyx. Thus xy = yx as required. Proof 4. For a, b ∈ R if ab = 0, then ba = (ba)2 = b(ab)a = 0. Now,0 = xy − xy = xy − x2y = x(y − xy), so 0 = (y − xy)x = yx− xyx.Also, 0 = yx − yx = yx − yx2 = (y − yx)x, so 0 = x(y − yx) =xy − xyx. Thus xy = yx for all x and y in R. We note it is immediate in all four proofs that xy = yx = xyx =yxy, for all x and y.2010 Mathematics Subject Classification. 19E50.Key words and phrases. Boolean Rings.Received on 8-6-2015.c©2015 Irish Mathematical Society7778 MACHALEDesmond MacHale received his Ph.D. from the University of Keele and isEmeritus Professor of Mathematics at University College Cork where he taughtfor nearly forty years. His research interests include commutativity in groupsand rings, automorphisms of groups, Euclidean geometry, number theory, andmathematical humour.School of Mathematical Sciences, University College Cork.E-mail address : d.machale@ucc.ieCopyright of Bulletin of the Irish Mathematical Society is the property of IMS Bulletin and itscontent may not be copied or emailed to multiple sites or posted to a listserv without thecopyright holder's express written permission. However, users may print, download, or emailarticles for individual use.

Boolean rings are definitely commutative!

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Boolean rings are definitely commutative!

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