COMMENTS ON CHAPTER 9 March 17, 2008 This chapter is all about the set U(R) of units of a <a href="/keyword/commutative-ring/" >commutative ring</a> R. The main focus is on the set U(Z/mZ). This set consists of the units (also called invertible elements ) of Z/mZ. Any general statement about U(R) applies, in particular, to U(Z/mZ), since Z/mZ is a <a href="/keyword/commutative-ring/" >commutative ring</a> . It is important to realize that U(R) is closed under multiplication. Also, the inverse of each element in U(R) is again in U(R). (In fact, U(R) is a group, but in this course we won t study groups per se, so we have no need for the additional terminology here.) Here is a summary of the main results we have proved: Theorem 1. Suppose R is a <a href="/keyword/commutative-ring/" >commutative ring</a> with exactly n units. Then un = 1 for each u U(R). How does this general result play out with respect to Z/mZ? We assume always that m 1. (The case m = 1 is exceptionally boring, but there s no need to exclude it.) First of all, a congruence class [a]m has an inverse (that is, [a]m is in U(Z/mZ)) if and only if GCD(a, m) = 1. For example, U(Z/15Z) = {[1]15 , [2]15, [4]15, , [7]15 , [8]15 , [11]15, [13]15 , [14]15 }. More succintly, U /15Z = { 1, 2, 4, 7}. Since Z/15Z has exactly 8 units, Theorem 1 guarantees that u8 = 1 for every u U(Z/15Z). The number of units of Z/mZ is equal to the number of positive integers a m such that GCD(a, m) = 1. This number is called Euler s phi function and is denoted by (m). We know, for example, that (p) = p 1 if p is a prime number. As a special case of Theorem 1 we have the following: Corollary 2. Let m be a positive integer, and let u U(Z/mZ). Then u (m) = [1]m . In terms of congruences, Corollary 2 says this: Euler s Theorem. Let m be a positive integer, and let a be any integer such that GCD(a, m) = 1. Then a (m) 1 (mod m). In the special case of a prime number, we get: Fermat s Theorem. Let p be a prime number, and let a be any integer such that p a. Then ap 1 1 (mod p). In class, we proved the following important corollary: 1 Corollary to Fermat s Theorem. Let p be a prime number, and let a be any integer. Then ap a (mod p). Proof. If p a, then ap 1 1 (mod p), by Fermat s Theorem; multiply boths sides by a to get the desired formula. If p | a, then ap 0 a (mod p). Thus it s true in either case. In Chapter 12 we will nd out how to evaluate (m) for large values of m. At this point, we can easily compute the number of units modulo a power of a prime. First, let s look at an example, say, 125 = 53 . The numbers on the list 1, 2, 3, . . . , 125 that are not relatively prime to 125 are the multiples of 5, that is 5, 10, 15, 20, . . . , 125. There are 25 multiples of 5 between 1 and 125. Therefore (125) = 125 25 = 100. Proposition 3. Suppose p is a prime number and m is a positive integer. Then (pm ) = pm 1 (p 1). Proof. On the list of integers 1, 2, . . . , pm , we have to throw out the ones that are not relatively prime to pm . These are exactly the integers that are multiples of p, and there are exactly pm 1 of them, namely, 1 p, 2 p, . . . , pm 1 p. Therefore the number of non-units is pm 1 , so the number of units is pm pm 1 , which equals pm 1 (p 1). The order of a unit. Let R be a <a href="/keyword/commutative-ring/" >commutative ring</a> , and let u U(R). Assume that U(R) is nite. Then we know, by Theorem 1, that there is a positive integer n such that un = 1. (For example, if R = Z/mZ we can take n = (m).) Of course, there may be a smaller positive integer q such that uq = 1. For example, in U(Z/15Z), ( 1)2 = 1 and 24 = 1. (In fact, it s not hard to see that u4 = 1 for every u U(Z/15Z).) The order of an element of U(R) is the smallest positive integer q such that uq = 1. In the special case (the main case of interest) where R = Z/mZ, we call this number q the order of u modulo m. Here are the orders of the elements of U(Z/15Z). u: 1 order of u : 1 2 4 4 2 7 8 4 4 11 13 14 2 4 2 The next two results are very useful in computing the order of a unit of Z/mZ. Proposition 4. Let R be a <a href="/keyword/commutative-ring/" >commutative ring</a> , let u U(R), and let n be a positive integer. If un = 1 , then the order of u is a divisor of n. We proved this in class, using the division algorithm, exactly as in the proof, in the book, of Proposition 2 on page 137. Corollary 5. Let m 1, and let u be any integer such that GCD(u, m) = 1. Then the order of u modulo m is a divisor of (m). For example, let s compute the order of 32 modulo 125. We know that the order has to be a divisor of (124) = 100. It makes sense to try exponents that are divisors of 100. Working in Z/125Z (bravely omitting brackets and foolishly refusing to use a calculator), we note that 322 = 1024 = 24, so 324 = 242 = 576 = 76 = 49. Next, working toward 3210 , we compute 328 = ( 49)2 = (1 50)2 = 1 100 + 502 = 99 = 26. Now 3210 = 322 328 = 24 26 = (25 1)(25 + 1) = 252 1 = 1. Finally, 2 3220 = ( 1)2 = 1. Now we know that the order of 32 is a divisor of 20, so it has to be on the list 1, 2, 4, 5, 10, 20. We have already observed that 324 = 1 and 3210 = 1, so the order of 32 cannot be a divisor of either 4 or 10. This rules out 1, 2, 4, 5 and 10, so the order of 32 is 20. Playing fast and loose with exponents (pardon the bad grammar). SupposeR is a <a href="/keyword/commutative-ring/" >commutative ring</a> , u U(R) and n is a positive integer. We de ne un = u u . . . u (n times). Also, by de nition, u n = (un ) 1 and u0 = 1. Thus um is de ned for every integer m, positive, negative or zero. The following laws of exponents, familiar in the context of real numbers, are valid: Let u U(R), and let m and n be arbitrary integers. Then the following equations are valid: um+n = um un um n = um (un ) 1 (um )n = u(mn) Let u, v U(R), and let m be any integer. Then um v m = (uv)m . The proofs of these are too boring to be done in public, but feel free to use these equations whenever they are useful. (The last equation, involving u and v, actually requires commutativity. The others do not, since an element always commutes with all of its powers.) The order of a power of a unit. In many situations we know the order of u, and we want to use this knowledge to gure out the order of ud . We use the following very important result (Proposition 3 on page 137 of the book): Theorem 6. Let u be a unit of a <a href="/keyword/commutative-ring/" >commutative ring</a> R, and suppose that u has order n n. Then, for every integer d, the order of ud is GCD(n,d) . In particular, the order of ud is always a divisor of the order of u. We proved this in class. Note that d can be a negative integer. In fact, it is easy to see that u and u 1 have the same order. For example, u 37 and u37 have the same order. Thus negative exponents don t really matter much in these considerations. Let s use this information to compute the orders of some units of Z/125Z, starting with the fact that 32 has order 20, as shown above. 20 The order of 322 is GCD(20,2) = 20 = 10. 2 The order of 323 is The order of 325 is We can work backwards too, to gure out, for example, the order of 2. Let n denote the order of 2. By Theorem 6, the order of 25 is a divisor of the order of 2. Since 25 = 32, and since we know that 32 has order 20, we see that 20 | n. Of course n | 100, by Corollary 5. Therefore the order of 2 has to be either 20 or 100. If it were 20, then 324 = (25 )4 = 220 = 1, which is false. Therefore the order of 2 is 100. Here s another little tidbit that we proved in class: 3 20 20 GCD(20,3) = 1 = 20. 20 = 20 = 4. GCD(20,5) 5 Proposition 7. Let R be a <a href="/keyword/commutative-ring/" >commutative ring</a> with exactly n units. If u is a unit of order n, then the list 1, u, u2 , . . . , un 1 is a list of all units of R (with no repetitions). Conversely, if v is any element such that the list 1, v, v 2 , . . . , v n 1 is a list of all units of R, with no repetitions, then v has order n. Well, maybe we didn t prove the Conversely statement, so let s do that now. Since there are no repetitions, and since 1 is on the list, we know that v 1 , v 2 . . . , v n 1 are all di erent from 1. Since v n = 1 by Theorem 1, we see that the order of v is exactly n. For example, since we showed that 2 has order 100 modulo 125, we know that 1, 2, 22 , . . . , 299 is a list of all 100 units of Z/125Z. Then, using Theorem 5, we can easily read o the order of every element of U(Z/125Z). Suppose, for example, that we want an element u of order 5. We can take u = 220 = 324 = 49. Warning. You can t always nd an element like the element u in U(Z/mZ), whose powers run through all of the units. For example, there is no such element in Z/15Z, since every element has order 1, 2 or 4. Comments on the exercises. Be sure that you know how to do all of the problems in Sections 9A and 9B. Most of them are quite straightforward adaptations of examples we have worked in class. Here are comments on a few: 9A: The correct answer to E4 (v) is something like 2 is not a unit modulo 1, so it does not have an order. In E4, since you know, from E3 (i), that 2 has order 10, every unit of Z/11Z occurs on the list 1, 2, 22 , 23 , . . . , 29 . Then use Theorem 6 to get the order of each of the 10 units. E8: Of course you must assume that GCD(a, r) = 1 and GCD(a, s) = 1, or else the problem does not make sense. Let f be the least common multiple of d and e. You have to show (i) af 1 (mod m); and (ii) if 1 t &lt; f then at 1 (mod m). Showing (i) is easy (and it s something you have done several times already). To verify (ii), suppose that at 1 (mod m) and look for a contradiction. Then at 1 (mod r) (Why?), so by Proposition 3 d | t. Similarly, e | t. Now use the fact, proved in class, that the least common multiple of two integers is a divisor of every common multiple of the two integers. Finish the problem. E12: Proposition 3 (in the book) gives you a formula for the order of af , and you know that this order is 1 (since the order of 1 is 1 (duh!)). Now what? E16: You are trying to show that n 1 (mod 41). Go back to basics. What does it mean to say that n2 1 (mod 41)? By the way, 41 is prime; this should be helpful. E18: 97 is prime. (Trust me.) What is (97)? 9B: E1 and E2 are silly problems. The smart-alecky (and perfectly correct) answer is: They are true because Fermat s Theorem is true. There s no reason to run through the list to see that Fermat s Theorem gives the correct answer in a speci c situation, since it has been proved in complete generality. These are not problems; they are suggestions for you to work on your own, in order to see how Fermat s Theorem plays out in a speci c situation. This is, in fact, a useful thing to do. E3: I think what is intended is something like this (in the case p = 11): 210 = 1, so 2 1 = 210 2 1 = 210 1 = 29 = 25 24 = ( 1) 16 = 5 = 6. The veri cation is 4 that 2 6 = 12 = 1. I leave it to you to decide whether or not Fermat s Theorem is actually useful in nding the inverse of a unit, when the modulus is reasonably small. My opinion is no . E7: This is the Corollary to Fermat s Theorem. Feel free to use this result on the other exercises. E8: We did this in class, using the Corollary to Fermat s Theorem. 5 3 E12: The expression equals 3n +5n +7n , and your mission is to show that this is an 15 integer. That is, you must show that 3n5 + 5n3 + 7n 0 (mod 15). E17: In addition to the hint in the back of the book, use the fact in the penultimate sentence of the comment on E8. E18: The commas before hence in (ii) and (iii) should be semicolons (at least according to the rules of grammar and punctuation that I learned in the 1950 s). 5