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Chapter 2, Congruence in Z and modular arithmetic.
This leads us to an understanding of the kernels and images of functions between rings (ideals, quotient rings, ring homomorphisms). It will also give us more examples of rings to think about.
Definition. An equivalence relation is a binary relation which is reflexive, symmetric and transitive.
Note that an equivalence relation on a set S partitions the set into subsets; these are called the equivalence classes. The application of this to congruence of integers is Theorem 2.3. (See Appendix D.)
Examples: =, congruence and similarity of triangles, congruence and similarity of ma- trices and our real interest:
Definition, p. 24. Let a, b, n ∈ Z with n > 0. We say a is congruent to b modulo n (written a ≡ b (mod n)) if n divides b − a.
Theorem 2.1. Congruence of integers is an equivalence relation.
Definition, p. 26. The equivalence class of an integer a under the relation of congruence modulo n is called the congruence class of a modulo n and denoted by [a].
Example. [a] = { b ∈ Z | b ≡ a (mod n) } = { a + kn | k ∈ Z } Modulo 2, there are two classes: [0], the set of even numbers and [1], the set of odd numbers.
Corollary 2.5. Fix n > 1.
(1) If r is the remainder when a is divided by n, then [a] = [r] (2) There are n distinct congruence classes, [0], [1],... , [n − 1].
Proof. (1) If a = qn + r, then n divides a − r, so a ≡ r (mod n).
(2) 0, 1 ,... , n − 1 are the only possible remainders. Show that the n given classes are all distinct.
Definition, p. 28. The set of congruence classes modulo n is denoted by Zn.
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Theorem 2.2. If a ≡ b (mod n) and c ≡ d (mod n), then
(1) a + c ≡ b + d (mod n) (2) ac ≡ bd (mod n)
Proof. Use the definition of congruence.
Theorem 2.6. If [a] = [b] and [c] = [d] in Zn, then [a + c] = [b + d] and [ac] = [bd].
Proof. This is a translation of Theorem 2.2 to new notation.
This says that the set of equivalence classes, Zn, can have addition and multiplication defined by: [a] ⊕ [c] = [a + c] and [a] [c] = [ac]
Theorem 2.6 says these operations are “well-defined”; that is, it does not matter which representative we pick from the congruence class to do our addition or multiplication with. We use ⊕ and only temporarily to emphasize that they are not the same operations that we have in Z. In fact, we will usually write a rather than [a] as long as it is clear that we are talking about an element of Zn rather than an element of Z.
Do an example of arithmetic in Z 3. We don’t have to think about the fact that we are working with remainders after division by 3.
Example. Z 3 = { [0], [1], [−1] }. Discuss arithmetic in Z 3 , solving equations such as x^2 ≡ 1 (mod 3), x^2 ≡ −1 (mod 3), x^2 ≡ −1 (mod 5).
All the usual rules for arithmetic (distributive, commutative laws, etc.) are inherited from the integers: this is the content of Theorem 2.7, page 34.
Some new things happen:
Example. 2 · 3 = 0 in Z 6 , but 2 6 = 0 and 3 6 = 0. x^2 = −1 has a solution in Z 5 , but not in Z 3. Adding 1 to itself n times gives 0 in Zn. If a 6 = 0 in Z 5 , then a has a multiplicative inverse. Check. Every a 6 = 0 of Z 5 satisfies a^4 = 1. In Z 5 , (a + b)^5 = a + b. Compute.
Good things happened in Zn when n = 5, but bad things happened when n = 6. What is different? Section 2.3 concentrates on Zp when p is a prime.
Theorem 2.11. Let a, b, n ∈ Z with n > 1 and let d = gcd(a, n). Then
(1) [a]x = [b] has a solution in Zn iff d|b. (2) If d|b, the equation [a]x = [b] has d distinct solutions.
Partial proof. The proof is outlined in the exercises, numbers 8–10, page 40. We will see where the different solutions come from. If there is any solution [r], so that ar − b = nq, then d|a and d|n, so d divides b = ar − nq. Thus we can write au + nv = d by Theorem 1.3, a = da 1 , b = db 1 , n = dn 1. Check that ub 1 + kn 1 gives a solution for k = 0,... , d − 1:
a(ub 1 + kn 1 ) = da 1 (ub 1 + kn 1 ) = b(a 1 u) + n(a 1 k) ≡ ba 1 u = b(1 − n 1 v) = b − b 1 vn ≡ b (mod n)
What we have not shown is that these solutions are all different and include all possible solutions.
