NoteNextra-origin/content/CSE5313/CSE5313_L3.md

CSE5313 Coding and information theory for data science (Lecture 3)

Finite Fields

Why finite fields?

Most information systems are discrete.

• Use bits, byte etc.

Use bits/bytes to represent real numbers.

• Problems of overflow, accuracy, etc.

We wish to build "good" codes \mathcal{C} \subset \mathbb{F}^n:

• Large \frac{k}{n}
• Lage d_H(\mathcal{C})\implies error detection/correction, erasure correction.

Idea: Use linear algebraic operations to encode/decode.

• F=\mathbb{F}_q, a finite field with q elements.

Finite fields

Fields and field axioms

A field is a set \mathbb{F} with two operations + and \cdot that satisfy the following axioms:

• Associativity: (a+b)+c = a+(b+c) and (a\cdot b)\cdot c = a\cdot (b\cdot c)
• Commutativity: a+b = b+a and a\cdot b = b\cdot a
• Distributivity: a\cdot (b+c) = a\cdot b + a\cdot c
• Existence of Identity elements: a+0 = a and a\cdot 1 = a
• Existence of Inverse elements: a+(-a) = 0 and a\cdot a^{-1} = 1

Every set of elements which satisfies these axioms is a field.

We can "do algebra" over it (matrices, vector spaces, etc.).

Are there finite sets which satisfy the field axioms?

What are the possible sizes of such sets?

Background – Basic number theory

• For a, b \in \mathbb{N},
  • Greatest Common Denominator: \gcd(a, b) = the largest integer m such that m|a and m|b.
  • Lowest Common Multiplier: \operatorname{lcm}(a, b) = the smallest integer m such that a|m and b|m.
• a, b are coprime if \gcd(a, b) = 1.
• Fact: (Euclid’s lemma) Say a \geq b,
  • There exists a quotient q \geq 0 and a remainder 0 \leq r < b such that a = bq + r.
• Theorem (Euclid): If \gcd(a, b) = 1 then there exist m, n \in \mathbb{Z} such that am + bn = 1.
  • Proof by repeated application of Euclid’s lemma.
  • Example:
    • If a = 3, b = 8,
    • then m = -5, n = 2,
    • satisfy 3 \cdot -5 + 8 \cdot 2 = 1.

Modular arithmetic

Defined a set with addition \oplus and multiplication \odot that satisfy the field axioms.

\mathbb{Z}_p is a field if p is a prime number.

• Addition and multiplication are defined modulo p.

• a \oplus b = (a+b) \mod p

• a \odot b = (a\cdot b) \mod p

• 0 is the additive identity.

• 1 is the multiplicative identity.

• a has an additive inverse p-a.

• a has a multiplicative inverse a^{-1} such that a \odot a^{-1} = 1.

Proof for existence of multiplicative inverse for a\in \mathbb{Z}_p\setminus \{0\}:

Proof

Since p is prime, \gcd(a, p) = 1.

By euclid's theorem, there exist m, n \in \mathbb{Z} such that am + pn = 1.

Take mod p on both sides:

a_{\mod p}\odot m_{\mod p} \equiv 1_{\mod p}

Thus, m_{\mod p} is the multiplicative inverse of a_{\mod p}.

Polynomials over prime fields is also a field.

(\mathbb{Z}_2,\operatorname{XOR},\operatorname{AND}) is a field.

Polynomials over finite fields

A polynomial over a field \mathbb{Z}_p is a expression of the form:

a(x)=\sum_{i=0}^n a_i x^i

• Polynomial degree: largest index of a non-zero coefficient.
• Polynomial addition: a(x) \oplus b(x) = \sum_{i=0}^n (a_i \oplus b_i) x^i
• Polynomial multiplication: a(x)\odot b(x) = \sum_{i=0}^n \sum_{j=0}^n a_i \odot b_j x^{i+j}
• Polynomial equality: a(x) = b(x) if and only if a_i = b_i for all i.
• Polynomial division: suppose \deg(a(x)) \geq \deg(b(x)), then there exist unique polynomials q(x) and r(x) such that a(x) = b(x)q(x) \oplus r(x) and \deg(r(x)) < \deg(b(x)). (do long division for polynomials)

denoted as \mathbb{Z}_p[x].

Example

p(x) = x^2 + 6x+3\in \mathbb{Z}_7[x]

p(1) = 1^2 + 6\cdot 1 + 3 = 10 \equiv 3 \mod 7

p(2) = 2^2 + 6\cdot 2 + 3 = 4+5+3 = 12 \equiv 5 \mod 7

Irreducible polynomials

A polynomial p(x) is irreducible if it cannot be factored into two non-constant polynomials.

If \gcd(a(x),b(x))=1, then there exist m(x),n(x)\in \mathbb{Z}_p[x] such that a(x)m(x)\oplus b(x)n(x)=1.

Proved similar to euclid's theorem.

Tip

If a polynomial p(x) has a root, say r, then p(x) = (x-r)q(x) for some q(x)\in \mathbb{Z}_p[x].

Example in \mathbb{Z}_2[x]:

p(x) = x^2 \oplus 1

is reducible because p(x) = (x\oplus 1)(x\oplus 1).

p(x) = x^3 \oplus x \oplus 1

is irreducible.

Proof

We prove by contradiction.

Suppose p(x) is reducible, then p(x) = a(x)b(x) for some a(x),b(x)\in \mathbb{Z}_2[x].

Then \deg(p(x)) = \deg(a(x)) + \deg(b(x)).

Let \deg b(x)=1, then b(x) \in \{x, x\oplus 1\}.

If b(x) = x, then p(0)=0 but p(x) is 1.

If b(x) = x\oplus 1, then p(1)=0 but p(x) is 1.

It is not the case in \mathbb{Z}_2[x], that every polynomial with no root is irreducible. (e.g consider (x^3\oplus x\oplus 1)^2 has no root but is reducible.)

Polynomial modular arithmetic

There exist quotient q(x) and remainder r(x), \deg(r(x)) < \deg(b(x)) such that

a(x) = b(x)q(x) + r(x)

\implies a(x) \mod b(x) = r(x)

"$\mod b(x)$" is an operation on polynomials in \mathbb{Z}_p[x] that:

• Preserves polynomial addition:
  • a(x) \oplus c(x) \mod b(x) = a(x) \mod b(x) \oplus c(x) \mod b(x)
• Preserves polynomial multiplication:
  • a(x) \odot c(x) \mod b(x) = a(x) \mod b(x) \odot c(x) \mod b(x)

Extension fields

Let p be a prime number. then (\mathbb{Z}_p[x], \oplus, \odot) is a field.

Fix a polynomial f(x)\in \mathbb{Z}_p[x] of degree t.

Define a set

Elements: polynomials of degree at most t-1 in \mathbb{Z}_p[x]. (finite set, size is p^t.)

Define addition:

a(x) \oplus_f b(x) = (a(x) \oplus b(x)) \mod f(x)

Define multiplication:

a(x) \odot_f b(x) = (a(x) \odot b(x)) \mod f(x)

Denote this set as \mathbb{Z}_p[x] \mod f(x).

This is not a field because it does not have a multiplicative inverse for every element.

Proof

We prove by contradiction.

Suppose there exists a polynomial g(x)\in \mathbb{Z}_p[x] \mod f(x) such that a(x) \odot_f g(x) = 1.

Let p=2,f(x)=x^2\oplus 1.

The polynomials in \mathbb{Z}_2[x] \mod f(x) are \{0, 1, x, x\oplus 1\}.

Consider the modular inverse of (x\oplus 1).

• 0\odot_f (x\oplus 1) = 0
• 1\odot_f (x\oplus 1) = x\oplus 1
• x\odot_f (x\oplus 1) = (x^2\oplus x)\mod (x^2\oplus 1) = x\oplus 1
• (x\oplus 1)\odot_f (x\oplus 1) = (x^2\oplus 1)\mod (x^2\oplus 1) = 0

To make our field extension works, we need to find a polynomial f(x) that is irreducible.

Theorem: If f(x) is irreducible over \mathbb{Z}_p, then \mathbb{Z}_p[x] \mod f(x) is a field.

Proof

Let a(x)\in \mathbb{Z}_p[x] \mod f(x), a(x)\neq 0.

Existence of a(x)^{-1} in \mathbb{Z}_p[x] \mod f(x) can be done by Euclid's Theorem.

Since \gcd(a(x),f(x))=1, there exist m(x),n(x)\in \mathbb{Z}_p[x] such that a(x)m(x)\oplus f(x)n(x)=1.

Take mod f(x) on both sides:

a(x)m(x) \mod f(x) = 1 \mod f(x)

Thus, m(x) \mod f(x) is the multiplicative inverse of a(x) \mod f(x).

So a(x)^{-1} = m(x) \mod f(x).

Corollary:

We can extend a prime field \mathbb{Z}_p with irreducible polynomial

Intuitively, we add to \mathbb{Z}_p a new element x that satisfies f(x)=0.

Observation: – We only used the general field properties of \mathbb{Z}_p. – ⇒ any “base field” can be used instead of \mathbb{Z}_p. – ⇒ Any field can be “extended”.

Say we wish to build a field F with 2^8 elements.

• Option 1:

  • Take \mathbb{Z}_2 and f(x) irreducible of degree 8.
  • F = \mathbb{Z}_2[x] \mod f(x).

• Option 2:

  • Take \mathbb{Z}_2, and g_1(x) \in \mathbb{Z}_2[x] irreducible of degree 4,
  • F_1 = \mathbb{Z}_2[x] \mod g_1(x). Note |F_1| = 2^4 = 16.
  • Take g_2(x) \in F_1[x] irreducible of degree 2.
  • F_2 = F_1[x] \mod g_2(x).

Uniqueness of the finite field

Theorems:

• As long as it is irreducible, the choice of f(x) does not matter.
  • If f_1(x), f_2(x) are irreducible of the same degree, then \mathbb{Z}_p[x] \mod f_1(x) \cong \mathbb{Z}_p[x] \mod f_2(x).
• Over every \mathbb{Z}_p (𝑝 prime), there exists an irreducible polynomial of every degree.
• All finite fields of the same size are isomorphic.
• All finite fields are of size p^d for prime p and integer d.

Corollary: This is effectively the only way to construct finite fields!

Extension of fields

\mathbb{R}[x]\mod (x^2+1) is a field, \cong \mathbb{C}.

Terms	Finite field extension F_1\to F_2	\mathbb{R}\to \mathbb{C}
Base field	any field \mathbb{F}_1	\mathbb{R}
Irreducible polynomial	f(x)	x^2+1
New elements added	x	i
Add/mul	\mod f(x)	\mod (x^2+1)

You cannot do algebraic extension of \mathbb{Q} to \mathbb{R}.

Transcendental extension: