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A Principal Ideal Ring That Is Not a Euclidean Ring Author(S): Jack C A Principal Ideal Ring That Is Not a Euclidean Ring Author(s): Jack C. Wilson Reviewed work(s): Source: Mathematics Magazine, Vol. 46, No. 1 (Jan., 1973), pp. 34-38 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2688577 . Accessed: 08/11/2011 11:53 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to Mathematics Magazine. http://www.jstor.org 34 MATHEMATICS MAGAZINE [Jan.-Feb. terminalpositions. The firstselection of a turningarc must be at the largestacute angle in orderto minimizethe arc, and hence,to minimizethe path. B a'~~~~B FIG. 7. Crossedjoins, one pivotpoint. 6. The restrictedproblem with crossed joins and close terminalpositions. In each of the foregoingcases, the motionis composed of a combinationof rotationsand translations.Figure 8, however,shows a case in whichAA' crosses BB', and the positionsAB and A'B' are so close that tangentscannot be drawn to the turning arcs as in Figures6 and 7. Then, it is necessaryto use pure rotationof the line AB about a centerP whichis at the intersectionof the perpendicularbisectors of AA' and BB'. B AA FIG. 8. Crossedjoins, pure rotation. A PRINCIPAL IDEAL RING T HAT IS NOT A EUCLIDEAN RING JACKC. WILSON, Universityof NorthCarolina at Asheville. 1. Introduction.In introductoryalgebra textsit is commonlyproved that every Euclidean ringis a principalideal ring.It is also usuallystated that the converseis false,and the studentis oftenreferred to a paper by T. Motzkin[1]. Unfortunately, this referencedoes not contain all of the details of the counterexample,and it is not easy to findthe remainingdetails from the referencesgiven in Motzkin'spaper. The object of this articleis to presentthe counterexamplein completedetail and in a formthat is accessibleto studentsin an undergraduatealgebra class. Not all authorsuse preciselythe same definitionsfor these two typesof rings. Throughoutthis paper the followingdefinitions will hold. 1973] A PRINCIPAL IDEAL RING THAT IS NOT A EUCLIDEAN RING 35 DEFINITION 1. An integraldomain R is said to be a Euclidean ring ifforevery x : 0 in R thereis defineda nonnegativeinteger d(x) such that: (i) For all x and y in R, bothnonzero, d(x) < d(xy). (ii) For any x and y in R, both nonzero,there exist z and w in R such that x = zy + w whereeither w = 0 or d(w) < d(y). DEFINITION 2. An integral domain R with unit elementis a principal ideal ring if everyideal in R is a principal ideal; i.e., if everyideal A is of theform A = (x) for some x in R. The ring,R, to be consideredis a subsetof the complexnumbers with the usual operationsof addition and multiplication: R = {a + b(t + 1-9)/21 a and b are integers}. It is elementaryto showthat R is an integraldomain withunit element. The purpose of thisarticle then is to showthat R is a principalideal ring,but thatit is impossible to definea Euclideannorm on R so thatwith respect to thatnorm R is a Euclidean ring. 2. The ringis a principalideal ring. In R thereis theusual norm,N(a + bi) = a2 + b2, which has the propertythat N(xy) = N(x) N (y) for all complex numbers x and y. In R thisnorm is alwaysa nonnegativeinteger. The essentialtheorem for this part of the exampleis due to Dedekind and Hasse, and the proof is taken from[2, p. 100]. THEOREM 1. Iffor all pairs of nonzer-oelements x and y in R withN(x) ? N(y), eitheryI x or thereexist z and w in R with0 < N(xz - yw) < N(y), thenR is a principal ideal ring. Proof. Let A : (0) be an ideal in R. Let y be an elementof A withminimal nonzeronorm, and let x be any otherelement of A. For all z and w in R, xz - yw is in A so that eitherxz - yw = 0 or N(xz - yw) > N(y). Hence the assumed conditionson R requirethat yI x; i.e., A = (y). The ring R under considerationwill now be shown to satisfythe hypo- theses of Theorem t. Observe that 0 < N(xz - yw) < N(y) if and only if 0 < N[(x/y)z - w] < 1. Given x and y in R, both nonzero and ytx, writex/y in the form (a + bV-19)/c where a, b,c are integers,(a, b,c) = 1, and c > 1. First of all, assume that c > 5. Choose integersd, e,f, q, r such that ae + bd + cf = 1, ad- 19 be = cq + r,and I rl ? c/2. Set z=d + ej-19 and w=q-fJ -19. Thus, (x/y)z-w = (a + bJ-19)(d + e 19)/c--(q-f /-l9) = r/c+ -19/c. This complexnumber is not zero and has norm(r2 + 19)/c2,which is less than I since I r _? c/2and c ? 5. The onlycase thatis not immediatelyobvious is c = 5, but then Ir 2 so that r2 + t9 < 23 < c2. 36 MATHEMATICS MAGAZINE [Jan.-Feb. The remainingpossibilities are c = 2, 3, or 4. Consider these in order: (i) If c = 2, y, x and (a, b,c) = 1 implythat a and b are of opposite parity. Set z = 1 and w = [(a -1) + bVI-19]/2 which are elements of R. Thus, (x/y)z - w = 1/2 # 0 and has norm less than 1. (ii) If c = 3, (a, b,c) = 1 impliesthat a2 + 19b2 _ a2 + b2 0 0 (mod 3). Let z = a - b-19 and w = q where a2 + 19b2 = 3q + r with r = 1 or 2. Thus, (x/y)z - w = r/3: 0 and has norm less than 1. (iii) If c = 4, a and b are not both even. If they are of opposite pa- rity,a2 + 19b2 _ a2 -b2 O (mod 4). Let z = a-b /-19 and w =q, where a2 + 19b2 = 4q + r with 0 < r < 4. Thus, (x/y)z-w = r/4:# 0 and has norm less than 1. If a and b are both odd, a2 + 19b2 -a2 + 3b2 X 0 (mod 8). Let z = (a - bV-19)/2 and w = q, wherea2 + 19b2 = 8q + r with0 < r < 8. Thus, (x/y)z - w = r/8: 0 and has norm less than 1. This completesthe proof that R is a principalideal ring. 3. The ringis nota Euclideanring. This part of thecounterexample is takenfrom [1]. The materialis repeatedand slightlyelaborated here in order to give a self- containedresult accessible to an undergraduateclass. As withthe previoussection the resultsare statedwithin the contextof the ringR underconsideration, but the theoremapplies to more general integraldomains. Throughoutthis section Ro will denote the set of nonzero elementsof R. DEFINITION 3. A subsetP ofRo withthe property PRO(P; i.e., xy is an element of P for all x in P and y in Ro, is called a productideal of R. (Notice that Ro is a product ideal.) DEFINITION 4. If S is a subsetof R, thederived set ofJS,denoted by S', is defined by S' = {xeSj y + xR c S, for some y in R}. LEMMA 1. IfS is a productideal, thenS' is a productideal. Proof. If x is in S', thenx is in S and thereexists y in R such thaty + xR c S. Let z be in Ro. Since S is a productideal and x is in S, xz is in S. Further, y + (xz)R e y + xR e S. This shows that S'Ro e S'; i.e., S' is a product ideal. LEMMA 2. If S c T, then S' c T'. Proof.If x is in S', thenx is in S and hencein T, and thereexists a y in R such that y + xR c S c T. Therefore,x is in T', and S' c T'. THEOREM 2. If R is a Euclidean ring, then thereexists a sequence, {P,}, of product ideals with thefollowing properties: (i) Ro = PO D P1 DP2 D' D P D , (ii) P, = 0, (iii) PI c , for each n7, and (iv) For each n , dnerivedthe nth set of Ro, is a subsetof P,, Proof.Let the Euclidean normin R be symbolizedby d(x) forx in Ro. For each 1973] A PRINCIPAL IDEAL RING THAT IS NOT A EUCLIDEAN RING 37 nonnegativeinteger n, defineP, = {x eRoR d(x) > n} . This definesthe sequence which obviouslyhas properties(i) and (ii). Suppose thatx is in Pn and y is in Ro. d(xy) ? d(x) > n whichimplies that xy is in P,. This shows that P, Ro ' P,,; i.e., for each n, P,, is a product ideal. For property(iii) let x be in Pn; i.e., x is in P, and thereexists a y in R such that y + xR P,n.Applying the Euclidean algorithm,there exist elements q and r in R with y =xq + r and r = 0 or d(r) < d(x). Hence, r = y + x(-q) is in v + xR c P, whichimplies that d(r) _ n, and in turn,d(x) > d(r) _ n1, so that d(x) > n + 1 and x is in P,1 . This proves property(iii) P Pn+1 . For property(iv), clearlyRo = P0 and applicationof (ii) gives Ro = P' c P1.
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