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EE564/CSE554: Error Correcting Codes Spring 2018 Week 5: February 5-9, 2018 Lecturer: Viveck R. Cadambe Scribe: Songtao Feng

Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications. They may be distributed outside this class only with the permission of the Instructor.

5.1

Consider F[x] over field F.

Definition 5.1 () A polynomial is said to be monic if highest degree term has coeffi- cient 1.

2 2 2 2 Example 1 2x + 1 is not monic, x + 2 is monic. They are the same over F3 since 2 × (x + 2) = 2x + 1.

5.1.1

Definition 5.2 (Irreducible Polynomial) A polynomial p(x) is irreducible over F if the following state- ment is true: p(x) = a(x)b(x) ⇒ a(x) ∈ F or b(x) ∈ F.

Example 2 (Irreducible Polynomial over Z2) Notation: Z2 is the same as F2 and GF (2). Degree 1 → x, x + 1, are irreducible. Degree 2 → x2, x2 + 1, x2 + x, x2 + x + 1. The first three are reducible since x2 = x · x, x2 + 1 = (x + 1)2 and x2 + x = x(x + 1). The last polynomial x2 + x + 1 is irreducible. Degree3 → the irreducible of degree 3 must have the form p(x) = ax3 +bx2 +cx+d where a 6= 0. Then a = 1 (monic), d = 1 (since x|p(x) if d = 0) and a+b+c+d 6= 0 (since (x+1)|p(x) if a+b+c+d = 0). x3 + x2 + 1 and x3 + x + 1 are irreducible.

In the previous example, one can prove certain polynomial of degree d is irreducible by enumerate all reducible polynomials of d.

Theorem 5.3 For any finite field F, the of irreducible polynomials is infinite.

We state the following theorem without proof.

Theorem 5.4 For any finite field F and every positive d, there exists an irreducible polynomial p(x) of degree d in F[x].

5-1 5-2 Week 5: February 5-9, 2018

5.1.2 Polynomial GCD

Definition 5.5 (Polynomial GCD) For two polynomials a(x), b(x) ∈ F[x], gcd(a(x), b(x)) is defined as the highest degree monic polynomial that divides both a(x) and b(x).

The GCD of two polynomials is unique (since we consider monic polynomials).

Theorem 5.6 If d(x)|a(x) and d(x)|b(x), then d(x)|gcd(a(x), b(x)).

Theorem 5.7 For any polynomial a(x), b(x), there exists p(x) and q(x) such that p(x)a(x) + q(x)b(x) = gcd(a(x), b(x)).

The can be used to find the GCD of two polynomials.

Example 3 (An Example of Euclidean Algorithm) 4 2 3 Find gcd(x + x + x + 1, x + 1) over F2. Step1: x4 + x2 + x + 1 = x(x3 + 1) + x2 + 1 Step2: x3 + 1 = x(x2 + 1) + x + 1 Step3: x2 + 1 = (x + 1)(x + 1) + 0 4 2 3 Thus gcd(x + x + x + 1, x + 1) over F2 is (x + 1).

Reversing the above steps, we can write

x + 1 = x(x2 + 1) + x3 + 1 = x(x4 + x2 + x + 1 + x(x3 + 1)) + x3 + 1 = x(x4 + x2 + x + 1) + (x2 + 1)(x3 + 1)

5.2 Algebraic Extension Fields (Galois Fields)

5.2.1 Extension

Theorem 5.8 Let p(x) be an irreducible polynomial over field F, then the F[x]|p(x) is a field with operation modulo p(x).

Example 4 Consider field R: real , x2 + 1 is an irriducible polynomial over R (but reduccible over C). R[x]|(x2 + 1) has the form a + bx. (a + bx) + (c + dx) = (a + c) + (b + d)x. (a + bx)(c + dx) = ac + x(bc + ad + bdx2) = (ac − bd) + x(bc + ad)(mod(x2 + 1)). R[x]|(x2 + 1) → complex field.

Definition 5.9 (Extension Field) Let p(x) be an irreducible polynomial over field F, then F[x]|p(x) is a extension field of field F.

Consider F is a finite field. For every integer d, there exists irreducible p(x) such that deg[p(x)] = d. We d have the observation |F[x]|p(x)| = |F| . Week 5: February 5-9, 2018 5-3

3 Example 5 (Extension Field of Z2 with Irreducible Polynomial p(α) = α + α + 1) One can check every line of the table. We also have

α7 = 1 (α2 + 1)−1 = (α6)−1 = α (α + 1)−1 = α4 = α2 + α

Coefficients Field element Power of α 000 0 0 001 α α0 010 α α1 011 α + 1 α3 100 α2 α2 101 α2 + 1 α6 110 α2 + α α4 111 α2 + α + 1 α5

Consider polynomial g(x) = x3 + x + 1. It is easy to check g(α) = 0, thus α is a root of the polynomial g(x) = x3 + x + 1. The extension field can be represented as {0, 1, α, ..., α6} and α is called a primitive element of this field.

5.3 Properties of Generic Finite Fields

Theorem 5.10 Every finite field has pn elements for some prime p and some integer n.

The proof follows from Lemma 5.12 and Lemma 5.13 listed below.

Definition 5.11 (Field ) The characteristic of a finite field F is the smallest interger d such that 1 + 1 + ... + 1 (add d times) = 0

Lemma 5.12 The characteristic of a finite field is a .

Proof: Suppose not. The characteristic of finite field F is pq where p, q are intergers. Then pq p q X X X 0 = 1 = ( 1)( 1) i=1 i=1 i=1 Pp Pq Thus, either i=1 1 = 0 or i=1 1 = 0. If p 6= 1, q 6= 1, then p < pq and q < pq which implies pq is not the characteristic of field F.

For example, the characteristic of F8 is 2. Suppose the characteristic of field F is prime number p, then S = {0, 1, 1 + 1, ..., 1 + 1 + ... + 1} is isomorphic to Zp. S is a subfield of F . | {z } p−1 times

Lemma 5.13 Let prime number p be the characteristic of a finite field F, then F is a finite dimensional over Zp. 5-4 Week 5: February 5-9, 2018

Proof follows from checking the axioms of a vector space.

Theorem 5.14 Every finite field has a primitive element α. Equivalently, there exists α such that order(α) = |F | − 1.

To prove the theorem (next week), we need some preliminary definitions and results.

Definition 5.15 (Order of an Element in a Finite Field) Let β be an element of finite field F, i.e., β ∈ F. The order of β is the smallest interger d such that βd = 1.

Let S = {1, β, ..., βorder(β)−1} where β is a primitive number of field F. Then S is a subgroup of (F − {0}, ·).

Lemma 5.16 Let F be a finite field and ebery β ∈ F. Order(β) divides |F| − 1 for all β.

Proof:[Sketch] S = {1, β, ..., βorder(β)−1} is a subgroup of (F − {0}, ·). Apply Lagrange’s Theorem.

Lemma 5.17 Every element of finite field F is a root of x|F| − x = 0.

Proof: According to Lemma 5.16, for every β ∈ F, we have β|F|−1 = (βorder(β))m = 1m = 1. Fermat’s little theorem is a special case of Lemma 5.17, applied to prime-sized finite fields.

Corollary 5.18 (Fermat’s Little Theorem) For prime number p and for any integer a,

ap ≡ a (mod p)

.