More generlly, we define ring to be non-empty set R hving two binry opertions (we ll think of these s ddition nd multipliction) which is n Abelin group under + (we ll denote the dditive identity by 0), nd stisfies the following dditionl properties: (v) (bc) = (b)c, b, c R i.e., ssocitivity for multipliction (vi) 1 R such tht 1 = = 1 R i.e., multiplictive identity * (vii) (b + c) = (b) + (c) nd ( + b)c = (c) + (bc), b, c R i.e., distributivity ** Nturl exmples of rings re the ring of integers, ring of polynomils in one vrible, the ring Zn of integers mod n, the Boolen ring of P(A) for some set A under + (symmetric difference) nd (s multipliction), nd the ring of 3 3 mtrices of rel numbers. Since + is lwys commuttive in ring, we sy tht ring is commuttive if its multipliction is commuttive (notice tht the mtrix ring fils to be commuttive). Notice lso tht there s no requirement for ring to hve multiplictive inverses. If ring is commuttive nd hs multiplictive inverses for everything other tht 0, then it s clled field. For exmple, the field of rtionl numbers, or of rel numbers, or integers modulo prime, or non-singulr 3 3 mtrices together with the zero mtrix. The fct tht the integers re commuttive ring, but fil to be field, nd tht they re very useful(!!), is why ttention is focussed on rings. In this course we ll mostly confine our ttention to commuttive rings becuse of the more immedite ppliction to the integers. * Some uthors prefer not to include the requirement of generl ring hving multiplictive identity, preferring to tlk bout rings nd rings with identity s very distinct nimls. There re some benefits to this, however the community is divided on this, so for simplicity we ll go with the side tht includes 1. 29 ** There re bunch of useful properties common to ll rings. For exmple, 0x = 0 lwys, since 0x = (0 + 0)x = 0x + 0x. Also ( 1)x = x lwys, since x + ( 1)x = 1x + ( 1)x = (1 + ( 1))x = 0x = 0. Similrly, ( )( b) = b lwys.
Returning to where we left off two slides go, we observed tht in the field Zp ny non-zero element hs the property tht p 1 = 1. Wht bout the commuttive ring Zn when n isn t prime, for exmple when n = p r? First some definitions. Let s define unit in commuttive * ring to be ny element u R such tht v R with uv = 1. The set of ll units in ring (clerly non=empty since 1 is unit) is clled the group of units of the ring R, nd is often denoted R. We define p R to be prime if it s not unit nd if p b p or p b. ** Similrly, we cn define zero divisor to be ny non-zero element r R such tht s R with s 0 nd rs = 0. Notice tht zero divisors cn t hve multiplictive inverses. *** If t R is such tht t k = 0 for some k 1, then t is sid to be nilpotent. Notice tht if ring hs no zero divisors ****, then even if it doesn t hve multiplictive inverses, we cn still solve equtions like x = b if 0, for then 0 = x b = (x b), nd the lck of zero divisors then implies tht either = 0 or x b = 0, hence x b = 0 nd so x = b. So cncelltion is still possible even without inverses! So wht does the group of units of R = Zn look like? We could list the elements, so if n = p r : \ \ \ \ \ R = { 0, 1, 2,..., p 1, p, p + 1,..., 2p 1, 2p, 2p + 1,..., 3p 1, 3p, 3p + 1,....., p r 1 1, p r 1, p r 1 + 1,..., p r 1 } So there were totl of p r elements in R, from which we ve deleted p r 1 of them (since multiples of p cn t hve multiplictive inverses), leving us with R = p r p r 1. * If defining this for non-commuttive ring, then we d hve to require both tht uv = 1 nd vu = 1. *** Otherwise if m k = 1 nd r m = 0, then r = r1 = r (m k) = (r m) k = 0 k = 0. 30 ** Recll tht the nottion x y mens tht x divides y exctly. Notice lso tht our definition specificlly excludes 1 from being prime. **** Such ring is clled n integrl domin.
More generlly, if n is some composite number, then n = rm, whereupon m is zero divisor nd so cn t hve multiplictive inverse. So R comprises numbers between 1 nd n which re co-prime to n. This quntity is function, nmely φ(n), nd is clled the Euler phi function. Notice tht φ(n) = 1 for ny non zero-divisor R when R = Zn. How cn we clculte φ(n)? We ve seen tht φ(n) = p 1 if n = p ( prime), nd lso tht φ(n) = p r p r 1 if n = p r. Indeed, it s not hrd to show * tht if nd b re co-prime, then φ(b) = φ() φ(b), hence for n ny composite number, by fctorising n completely we cn get φ(n) = φ(p r.... q z ) = (p r p r 1 ).... (q z q z 1 ) = n(1 p 1 ).... (1 q 1 ). Let s ply with wht we hve for bit. We ll clculte 2 35 (mod 7). Since 35 = (6)(5) + 5 nd φ(7) = 6, we cn see tht 2 35 = (2 6 ) 5 2 5 = 1 32 = 4 (mod 7). Similrly, we cn clculte 11 81050696835 (mod 1176). Firstly we fctorise 1176 = (2 3 )(3)(7 2 ), so then φ(1176) = (8 4)(3 1)(49 7) = 336, nd then compute 81050696835 (mod 336), nmely 3. Hence 11 81050696835 = (11 336 ) 241222312 11 3 = 1331 (mod 1176) = 155 (mod 1176). In order to be ble to hndle multiple simultneous modulr equtions, we ll need to be ble to relte multiple rings. Essentilly, ech ring will define the effect of ech modulr fctor, but there s nturl isomorphism which will mke life much esier. Let s strt by supposing tht we hve two rings R nd S, then we cn mke the set R = R S into ring in nturl wy by defining (r, s) + (r, s ) = (r + r, s + s ) nd (r, s)(r, s ) = (rr, ss ). * you cn check tht this is true simply by compring the number of zero divisors of, b, nd b. We cn lso get this s corollry of the Chinese reminder theorem we ll prove shortly. 31
Define the subset A R to be bsorptive * if it s subgroup of R under + nd tht ra A nd Ar A ** for ll r R (hence the use of the word bsorptive ). As nturl exmple of such set in the ring Z consider A = nz, since for ny r R, ny t ra is utomticlly multiple of n nd so lives in A. It s not hrd to show tht if f : R S is ring homomorphism, then ker f is n bsorptive set (the kernel for rings will be the set of ll elements of R which get mpped to 0S). Notice tht if A is bsorptive nd 1 A, then A = R, so typiclly we re more interested in the cses where A doesn t contin 1. We define A + B = { + b A, b B } nd AB = { Σ ibi i A, bi B }, where the summtion in AB is only ever llowed to be over finitely mny things. It s esy to show tht both of these re bsorptive if A nd B re bsorptive. If A nd B re bsorptive in R, then we cn define f : R R/A R/B by f(r) = (r + A, r + B). Then f is (ring) homomorphism, *** with ker f = A B. If moreover A + B = R, then f is onto nd A B = AB, **** hence R/AB is isomorphic to R/A R/B. In the context of R = Z nd A = mz nd B = nz, then if m nd n re coprime, then AB = mnz. However, why is it obvious tht A + B = R in this cse? It s ll down to the fct tht the gcd (gretest common divisor) of m nd n is 1. * This is not stndrd nottion. The stndrd lnguge is to cll these sets idels. However, the word bsorptive seems to chrcterise how they behve, so for now we ll use this more descriptive lnguge. ** Notice tht in fct, ra = A = Ar. Tht s esy to see once you spot tht 1 R mens tht 1A A. 32 *** Showing f(r + s) = f(r) + f(s) is strightforwrd. However, the bsorptive nture of A nd B is needed to show tht f(rs) = f(r) f(s) since, for exmple, (r + A) (s + B) = rs + ra + Ar + AA = rs + A + A + A = (rs) + A **** Since A is bsorptive, ibi A, nd since B is bsorptive, ibi B, nd since bsorptive sets re subgroups under +, Σ ibi A B! AB A B. Now if A + B = R then 1 A + B so 1 = + b for some A nd b B, but then c A B! c = c1 = c( + b) = c + cb AB. Hence A B AB.
So we need n effective wy to compute the gcd of pir of numbers. Clim: if d = gcd(m,n) then the eqution mx + ny = kd hs solutions x, y Z k Z. We ll pproch this by ctully constructing solution to the eqution 13155x + 2367y = 3 using the Eucliden subtrction lgorithm. Divide 2367 into 13155 to get 13155 = 5 2367 + 1320 Divide 1320 into 2367 to get 2367 = 1 1320 + 1047 Divide 1047 into 1320 to get 1320 = 1 1047 + 273 Divide 273 into 1047 to get 1047 = 3 273 + 228 Divide 228 into 273 to get 273 = 1 228 + 45 Divide 45 into 228 to get 228 = 5 45 + 3 Divide 3 into 45 to get 45 = 15 3 + 0 STOP. Hence this lst dividing vlue is the gcd of 2367 nd 13155. This is ctully process to find lrgest common mesurement unit for two lengths; t lest, tht ws how it ws formulted in ncient Greek times. It ws used to get pproximtions for the rtios of side to the digonl of squre, for side to the digonl Hence if we let units = 3, then working from the bottom up, 45 = 15 units 228 = 5(45) + 3 = 5(15 units) + 1 unit = 76 units 273 = 1(228) + 45 = 1(76) + 15 units = 91 units 1047 = 3(273) + 228 = 3(91) + 76 units = 349 units 1320 = 1(1047) + 273 = 1(349) + 91 units = 440 units 2367 = 1(1320) + 1047 = 1(440) + 349 units = 789 units 13155 = 5(2367) + 1320 = 5(789) + 440 units = 4385 units We cn lso use this informtion to get solution to the eqution by focussing on the reminders, letting = 13155 nd b = 2367, vi 1320 = 5b 1047 = b 1320 = b ( 5b) = + 6b 273 = 1320 1047 = ( 5b) ( + 6b) = 2 11b 228 = 1047 3(273) = ( + 6b) 3(2 11b) = 7 + 39b 45 = 273 228 = (2 11b) ( 7 + 39b) = 9 50b 3 = 228 5(45) = ( 7 + 39b) 5(9 50b) = 52 + 289b of regulr pentgon, nd for the rtio of the circumference to the dimeter of circle, nd is strongly relted to the construction of continued frctions. 33
Notice tht if we tke successive trunctions of the previous continued frction, we get sequence of rtionl pproximtions to the vlue of r = 13155 2367, nd tht they lternte round bove nd below, nmely: r 5, 6, 5+ 1/2, 5+ 4/7, 5+ 5/9, 5+ 29/52, 5+ 440/789, or in deciml form: r 5, 6, 5.5, 5.57143, 5.55556, 5.55769231, =5.55766793. We ll pply this to find n sequence of pproximtions for the dimeter : side rtio for regulr pentgon. It ll be reminder of eucliden geometry... We ll lbel the length of side by s nd digonl by d. So in our picture, d = 2 + b. Since AB is prllel to EC, AEF is isoceles, nd AF = s. So d = s + nd s = + b Since EA is prllel to GF nd BD, EFG is isoceles, nd GF =. But GF is the digonl of nother regulr pentgon with side b. So d : s is the sme rtio s b :. So in the spirit of the eucliden subtrction lgorithm... D d = 1s +, s = 1 + b, b = 1s1 + 1, s1 = 11 + b1, b1 = 1s2 + 2, s2 = 12 + b2, b2 = 1s3 + 3,.... where the bn, sn, nd n re repeting the sme pttern of sides nd digonls within ever smller pentgons. This gives continued frction expnsion: d / s = 1 + 1/ (1 + 1/ (1 + 1/ (1 + 1/ (1 +... ) ) ) ) which gives successive rtionl pproximtions: 1, 1+ 1/2, 1+ 2/3, 1+ 3/5, 1+ 5/8, 1+ 8/13, 1+ 13/21, 1+ 21/34,..... Not only does this method give surprisingly fst rtionl pproximtion, but it lso musingly uses the Fiboncci sequence! Applied to estimting π this process yields: 3, 22/7, 333/106, 355/113, 103993/33102,...... 34 E s A b G F b C B
This pproch cn be generlised. Suppose ring R cn dmit function f : R * N such tht, b R * q, r R with = bq + r nd f(r) < f(b) or r = 0. Certinly the ring of integers stisfies this (with for exmple, f(n) = n being the function). Cn we do this for Zn? Any ring which hs this property is sid to be Eucliden domin. Notice tht we could extend this to the ring of polynomils hving integer (or rtionl, or rel, or complex) coefficients by defining f to yield the degree of the polynomil. The results of pplying the Eucliden subtrction lgorithm to such polynomils will of course be different depending on the rnge of coefficients vilble. Before our detour on finding gcds, we rised the question of solving multiple simultneous modulr equtions, nd observed tht for A nd B bsorptive in R with A + B = R, then R/AB is isomorphic to R/A R/B, nd so if m nd n re coprime, then Zmn is isomorphic to Zm Zn. This is known s the Chinese reminder theorem. A consequence of this is tht if we re given pir of equtions x = mod m, x = b mod n with m, n coprime, then (, b) Zm Zn corresponds to unique vlue in Zmn, hence we cn solve in the trditionl simultneous eqution style, substituting successively. * We cn even do this for mny equtions, s long s the moduli re pirwise coprime. Suppose x = 3 mod 11, x = 6 mod 8, x = 1 mod 15. The first eqution gives us x = 3 + 11, so from the second eqution we get x = 6 + 8b = 3 + 11. This gives 3 = 11 8b = 1 = b x = 14 + 88t s generl solution for the first two equtions. The third eqution gives us x = 1 + 15c = 14 + 88t 15 = 15c 88t c = 1, t = 0 x = 14 + 1320s is the generl solution for the three equtions. 35 * This pproch ctully feels bit like using the Eucliden lgorithm -- we ll work through more efficient wy to formulte this in the homework.