The structure of finite rings and finite exponentiation
The multiplicative residues • We have seen that the finite ring Zp is a field, that is, every non-zero element of Zp has a multiplicative inverse. • It is a convention to write Zp* for the non-zero elements {1, 2, 3, ..., p-1}. • Zp* is the set of multiplicative residues modulo p.
Modular exponentiation • Public key cryptography explores the properties of the exponentiation function in Zp* • Defined as repeated multiplication: • g5mod p := g * g * g * g * g mod p. • To exponentiate by negative values, exponentiate the inverse: • g-3 := g-1 * g-1 * g-1mod p.
Exponent rules • Addition/subtraction rules: • gk gj = gk+jin Zn* • gk g-j = gk-jin Zn* • Multiplication rule: • (gk)j = gkj in Zn*
Non-prime modulus • If n is not prime, then not all non-zero elements are invertible. • In this case, we write Zn* for the invertible elements only. • Examples: • Z14* = {1, 3, 5, 9, 11, 13} • Z15* = {1, 2, 4, 7, 8, 11, 13, 14}
Generators • Consider the following: • In Z14* = {1, 3, 5, 9, 11, 13}; • 32 =9 mod 14; 33 =13 mod 14; 34 = 11 mod 14; 35 = 5 mod 14; 36 = 1 mod 14. • In Z14* every element is a power of 3. We say that 3 is a generator. • Do generators always exist?
Prime modulus • If n is a prime, or twice a prime, then Zn* always has a generator. • We have already seen this for n = 14 = 2*7. • Otherwise, generators do not exist. • An important case is when n = pq, where both p and q are odd and prime. In this case, there is an element that generates 1/2 of Zn* .
Example • Z15* = {1, 2, 4, 7, 8, 11, 13, 14} • 21 =2 mod 15; 22 =4 mod 15; 23 =8 mod 15; 24 =1 mod 15 • 41 = 4 mod 15; 42 = 1 mod 15; • 71 =7 mod 15; 72 =4 mod 15; 73 =13 mod 15; 74 =1 mod 15; • 81 =8 mod 15; 82 =4 mod 15; 83 =2 mod 15; 84 =1 mod 15; • 111 =11 mod 15; 112= 1 mod 15; • 131 =13 mod 15; 132 =4 mod 15; 133 =7 mod 15; 134 =1 mod 15; • 141 = 14 mod 15; 142 =1 mod 15; • No element is a generator, as predicted
Order of an element • Take g in Zn* . The list • g1, g2, ..., gk, k = 1, 2, ... must eventually repeat. • Otherwise get infinite sequence of elements from a finite set, a contradiction. • Let gj = gk, j < k. k = j + t. • gj = gk = g j+t; • gj = g j+t = gj gt; • gt = 1 • Cancellation rule applies because g is invertible
Order (continued) • We have shown that: • g is invertible if and only if there is t > 1 such that gt = 1 mod Zn* . • Indeed, if g is invertible we have shown that t exists. On the other hand, if t exists, then g has an inverse, equal to gt-1. • g g t-1 = gt = 1 in Zn* . • The smallest such t is the order of g.
Order of Zn* • The order of an element can also be defined as the size of the set generated by it: • t = order(g) = #{g, g2, g3, ..., gt = 1} • The order of the group Zn* is simply its cardinality | Zn* |. The function • (n) = | Zn* | is called the Euler totient function.
Euler totient • We know that all non-zero residues modulo a prime p are invertible. In other words: • (p) = p - 1, if p is a prime. • It is easy to see that, if n = p q is a product of two primes, then • (n) = (p - 1)(q - 1) = (p)(q) • In general: • (n)(m) = (nm)if n, m are relatively prime.
Relations between orders • Fact: If g is a residue in Zn* , then • order(g) divides (n) = order(Zn* ). • An important special case is when p is a prime. In that case, • order(g) divides p-1 • gp-1 = (gt)k = 1k = 1 mod p; t = order(g)
Fermat’s Little Theorem • The previous result is called Fermat’s Little Theorem. • (FLT) For every non-zero g in Zp* , where p is a prime: • gp-1 = 1 mod p • This can be generalized for all g in Zp* , • gp = gmod p
Generalizing FLT • For any finite ring Zn* : • g(n) = 1 mod n,g in Zn* . • Proof will not be given. • The special case n = pq is important. • Claim: If n is a product of two primes: • g(n)+1 = g mod n,g in Zn= {0, 1, ..., n-1}
The Remainder Theorem • In order to appreciate the structure of finite rings when the modulus is composite, the remainder theorem applies: • Given n = s t, where GCD(s, t) = 1 • For each element a mod n, there corresponds a unique pair • (b mod s, c mod t).
Example (CRT) • n = 15 = 3*5 • a = 7 mod 15 corresponds to • (1 mod 3, 2 mod 5) • To go from “a mod n” to (b mod s, c mod t): • Just compute b = a mod s, c = a mod t. • How to go backwards? • Let represent s-1 mod t, • represents t-1 mod s.
CRT backwards • Given (b mod s, c mod t), compute • a = c s + b t mod n • In other words a = c s + b t + k n • Consider ”a mod s” (similar for a mod t) • a mod s = • c s + b t + k s t mod s = • b t mod s = • b mod s
CRT backwards example • given b = 1 mod 3, c = 5 mod 7 • Compute 3-1 mod 7 = 5, as 3*5 = 1 mod 7 • Compute 7-1 mod 3 = 1, as 7 = 1 mod 3 • a =1 * 7 * 1 + 5 * 3 * 5 = 82 mod 21 = 19 mod 21
Returning to FLT for n = pq • To prove: • g(n)+1 = g mod n,g in Zn= {0, 1, ..., n-1}, when n = pq, and p, q are primes. • For invertible elements, i.e., GCD(g, n) = 1, it is the previous claim • For g=0 mod n, i.e.,GCD(g, n) = n it is clear. • Consider now the case GCD(g, n) = p.
FLT (continued) • By the CRT, g is defined by • g is invertible mod q • g = 0 mod p • We get that • gq = g mod q • gq = 0 = g mod p • By backwards CRT, we get • gq = g mod pq; g (n)+1 = gpq- p - q +2 = = g-p+2 (gq)p-1 = g mod pq
