The Structure of Z p when p is Prime

Transcription

1 LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z p (3) Wheever [b] p = [0] p i Z p, the = [0] p or [b] p = [0] p Proof (1) (2) Suppose p is a positive prime ad [0] p i Z p We wat to show that the equatio X = [1] p has a solutio i Z p Now sice [0] p, a 0 kp so a is ot divisible by p Sice the oly divisors of p are ±1 ad ±p ad because p a, we must have GCD(a, p) = 1 But the by Theorem 13, there exists itegers u ad v such that This equatio, however, is equivalet to ua + vp = 1 ua 1 = vp which implies that ua 1 (mod p), or [ua] p = [1] p Settig X = [u] p we have so X = [u] p is a solutio [x] p = [u] p = [au] p = [1] p, (2) (3) Suppose [b] p = [0] p i Z p If = [0] p there is othig to prove, If [0] p the by (2) there exists a solutio [u] p Z p such that But the [u] p = [1] p [0] p = [u] p [0] p = [u] p ( [b] p ) = ([u] p ) [b] p = [1] p [b] p Hece, i every case we have either = [0] p or [b] p = [0] p (3) (1) Let a be ay divisor of p; say p = ab I order to show that p is prime we must show a = ±1, ±p Now p = ab ab 0 = p [ab] p = [0] p [b] p = [0] p i Z p By (3) the either = [0] p or [b] p = [0] p Now = [0] p implies a 0 = kp which implies p a, or that a = sp But the p = ab = spb Dividig both sides by p shows that sb = 1 Sice s ad b are itegers the oly possibilities are that s = ±1 ad b = ±1 Hece b = ±1 ad so a = ±p O the other had, a similar argumet shows that whe [b] p = 0, we must have a = ±1 ad b = ±p Hece if (3) holds, the the oly factors of p are ±1 ad ±p, so p is prime 47

2 13 THE STRUCTURE OF Z p WHEN p IS PRIME 48 We ll ow prove three easy corollaries to this theorem Corollary 132 Let p be a positive prime For ay 0 ad ay [b] p Z p, the equatio X has a uique solutio i Z p Proof We eed to prove that two thigs, that X has a solutio i Z p ad that that solutio is uique Existece: Sice p is prime, by (2) of the precedig theorem, X = [1] p has a solutio i Z p Let [c] p be that solutio Multiplyig both sides of this equatio by [b] p, we get [b] p [c] p [1] p = ( [bc] p ) Thus, [bc] p will be a solutio of x Uiqueess: Suppose both Subtractig these two equatios we have [c 1 ] p [c 2 ] p ) [a p ] ([c 1 ] p [c 2 ] p = [0] p Sice p is prime ad [0] p by hypothesis, statemet (3) of the precedig theorems says [c 1 ] p [c 2 ] p = [0] p = [c 1 ] p = [c 2 ] p Corollary 133 Let a ad be itegers with > 1 The GCD(a, ) = 1 if ad oly if the equatio X = [1] i Z has a solutio Proof Suppose GCD(a, ) = 1 The by Theorem 13, there exist itegers u ad v such that But the so au is cogruet to 1 modulo Hece Thus, [u] is a solutio of X = [1] i Z 1 = au + v au 1 = v [1] = [au] = [u] Suppose [x] = [1] has a solutio [u] i Z The au is cogruet to modulo But this implies au 1 = q or au q = 1 It follows from this equatio that ay commo divisor of a ad must divide 1 Therefore, GCD(a, ) = 1 Defiitio 134 Wheever there is solutio i Z to the equatio X = [1] we say that is a uit i Z Wheever there is a o-trivial solutio (ie, a solutio other tha the obvious oe X ) of X we say that is a zero divisor i Z

3 13 THE STRUCTURE OF Z p WHEN p IS PRIME 49 Lemma 135 Let be a positive iteger If Z, the is either a uit or a zero divisor Proof From the fact that GCD (a, ) 1 always, we have two distict cases: GCD (a, ) = 1 I this case, we kow from Corollary 133 that is a uit i Z We will show that caot also be a zero divisor Suppose we had a elemet [b] [0] such that [b] Let [a] 1 be the solutio of X = [1] guarateed by Corollary 133The we would have [1] = [a] 1 ( ) [b] [1] = [b ] [a] 1 [b] = ([b] ) [a] 1 [a] 1 which cotradicts our hypothesis that [b] [0] Therefore whe GCD (a, ) = 1, is a uit but ot a zero divisor Suppose GCD (a, ) = d > 1 I this case, the if ad oly if part of Corollary 133 tells us that ca ot be a uit i Z To see that is a zero divisor, we ote GCD (a, ) = d meas d divides both a ad, ad moreover, 1 < d Now if d =, the this meas that divides a ad so, ad hece will be a zero divisor (as ay [k] time [0] produces [0] ) So ow we suppose 1 < d < Write We the have a = qd = sd with 1 < s, d < [s] = [as] = [(qd) s] = [q (ds)] = [q] Sice 1 < s < we have [s] [0] ad yet [s] Thus, whe GCD(a, ) > 1 is a zero divisor but ot a uit Corollary 136 Let a, b, be itegers with > 1 ad GCD (a, ) = 1 The the equatio has a uique solutio i Z x Proof Suppose GCD (a, ) = 1, the as above we have itegers u, v Z such that Now multiply both sides by [b] ad we get So [bu] [u] is a solutio of x au + v = 1 = [au v] = [1] = [au] [v] = [1] = [au] [0] = [1] = [u] = [1] ([b] [u] ) = [1] [b] To see that this solutio is uique argue as i Corollary 132 Suppose we had two solutios Subtractig oe equatio from the other we get [c 1 ] [c 2 ] ([c 1 ] [c 2 ] )

4 13 THE STRUCTURE OF Z p WHEN p IS PRIME 50 Because [a ] has o zero divisors (by Corollary 133 ad Lemma 135), we must coclude that [c 1 ] [c 2 ] [c 1 ] = [c 2 ] ad so the two solutios i fact must coicide Theorem 137 Let a, b, be itegers with > 1, ad let d = GCD (a, ) The (i) The equatio x has a solutio i Z if ad oly if d b (ii) If d b, the the equatio x has d distict solutios i Z p Proof (i) = Suppose x has a solutio i Z ad let [c] be that solutio We have [c] = [ac] = ac = b (mod ) = ac b = k for some k Z But the (*) b = ac k So aythig that divides both a ad, will divide the right had side of (*) ad hece, b (the left had side of (*)) I particular, the greatest commo divisor of a ad divides the right had side of (*), so d = GCD(a, ) divides b (i) = Suppose d = GCD(a, ) ad d b Sice d = GCD (a, ) there exists itegers u, v such that (**) d = au + v Sice d b, there exists a iteger k such that b = kd Now multiply both sides of (**) by k The we have b = kd = a (ku) + (kv) = b a (ku) (mod ) = [b] = [aku] = [ku] Hece [ku] is a solutio of x (ii) Suppose d = GCD (a, ) ad d b I fact, sice d = GCD (a, ), d a ad d Write = rd a = sd I claim (ar) Ideed, ar = (sd) r = s (rd) = s = (ar) Now suppose [c] is a solutio of x I claim [c + r] is also a solutio Ideed, if the if we replace c by c + r, we get [c] [c + r] = [c] + [ar] + [ar], sice [c] is a solutio of x + [0], sice ar is divisible by But if [c + r] is a solutio so is [c + r + r] = [c + 2r], as well as [c + 3r], etc Clearly we ca geerate lots of solutios this way The questio is, whe do stop gettig ew solutios this way (recall that Z oly has elemets, so we ca t get a ifiite umber of solutios) Well, we will keep gettig ew cogruece

5 13 THE STRUCTURE OF Z p WHEN p IS PRIME 51 classes util [c + kr] = [c + ] I other words util kr = But r was defied as the solutio of dr = Therefore, we ll get the followig cogruece classes as solutios [c], [c + r], [c + 2r],, [c + (d 1) r] It is easy to see that these are all distict sice 0 kr < for k {0, 1,, d 1}