## Saturday, March 01, 2008

### The set of congruence classes modulo n: Z/nZ

In considering modular arithmetic (see here for review if needed), we can divide all integers into congruence classes modulo n.

Definition 1: Congruence class modulo n: [i]

Let [i] represent the set of all integers such that [i] = { ..., i-2n, i-n, i, i+n, i+2n, ... }

To make this definition even clearer, let's consider the following lemma.

Lemma 1: x ∈ [i] if and only if x ≡ i (mod n)

Proof:

(1) Assume x ≡ i (mod n)

(2) Then n divides x - i

(3) So, there exists an integer a such that an = x - i

(4) So that x = an + i

(5) Assume that x ∈ [i]

(6) Then, there exists a such that x = i + an

(7) This shows that an = x - i and further that n divides x - i.

(8) Therefore, we have x ≡ i (mod n)

QED

Lemma 2: There are only n distinct congruence classes modulo n

Proof:

(1) To prove this, I will show that for all x ∈ Z, there exists i such that:

x ≡ i (mod n)

and

0 ≤ i ≤ n-1

(2) For x ∈ Z, there exists k such that x ≡ k (mod n)

(3) We can assume that k is positive since if k is negative, k ≡ -k (mod n) since n divides k + -k.

(4) We can further assume that k ≤ n-1, since if k ≥ n, it follows that k ≡ (k-n) (mod n) since n divides (k - [k-n]) = k -k + n = n.

QED

We can now consider the set of congruence classes modulo n as the set Z/nZ.

Definition 2: set of congruence classes modulo n: Z/nZ

Z/nZ = { [0], ..., [n-1]}

Example 1: Z/3Z

Z/3Z = { [0], [1], [2] }

For purposes of showing the relationship between Z and Z/nZ, I will from this point on view Z/nZ = {0, ..., n-1}.

So that Z/3Z will also be represented as {0, 1, 2}

Lemma 3: Z/nZ is a commutative ring

Proof:

(1) Z/nZ has commutative rule for addition

For all a,b ∈ Z/nZ: a+b ≡ b + a (mod n)

(2) Z/nZ has associative rule for addition

For all a,b,c ∈ Z/nZ: (a + b) + c ≡ a + (b + c) (mod n)

(3) Z/nZ has additive identity rule

0 ∈ Z/nZ and for all a ∈ Z/nZ: a ≡ a + 0 (mod n)

(4) Z/nZ has additive inverse rule:

(a) Let a be any congruence class modulo n such that a ∈ Z/nZ

(b) Assume a ≠ 0 for if a = 0, then a + a ≡ 0 (mod n) and a is its own additive inverse.

(c) Let b = n - a

(d) b ∈ Z/nZ since 1 ≤ b ≤ n-1

(e) a + b ≡ n ≡ 0 (mod n)

(5) Z/nZ has an associative rule for multiplication:

For all a,b,c ∈ Z/nZ: (ab)c ≡ a(bc) (mod n)

(6) Z/nZ has a distributive rule:

For all a,b,c ∈ Z/nZ: a(b+c) ≡ ab + ac ≡ (b + c)a (mod n)

(7) Z/nZ has a commutative rule for multiplication

For all a,b ∈ Z/nZ: ab ≡ ba (mod n)

(8) Thus, Z/nZ is a commutative ring. [See Definition 2, here]

QED

Lemma 4: if p is a prime, then Z/pZ is a field

Proof:

(1) Z/pZ has a multiplicative identity rule.

1 ∈ Z/pZ and for all a ∈ Z/pZ 1*a ≡ a*1 ≡ a (mod p)

(2) Z/pZ has a multiplicative inverse rule since:

(a) Let a be any nonzero element of Z/pZ

(b) If p = 2, then a is its own inverse and a*a ≡ 1*1 ≡ 1 (mod 2).

(c) If p ≥ 3, then let b = ap-2

(d) Then a*b ≡ a*(ap-2) ≡ ap-1 ≡ 1 (mod p). [See Fermat's Little Theorem, here]

(3) Thus, Z/pZ is a field. [See Definition 3, here]

QED

Lemma 5:

if p is prime and ab ≡ 0 (mod p), then a ≡ 0 (mod p) or b ≡ 0 (mod p)

Proof:

(1) Assume a is not ≡ 0 (mod p)

(2) Since ab ≡ 0 (mod p), it follows that p divides ab.

(3) If a is not ≡ 0 (mod p), then it follows that p does not divide a.

(4) So, using Euclid's Lemma (see Lemma 2, here), it follows that p divides b.

(5) And equivalently, b ≡ 0 (mod p)

QED

References
• Gareth A. Jones, J. Mary Jones, , Springer, 2002.