Transcript Polynomials over finite fields

Probabilistic verification
Mario Szegedy, Rutgers
www/cs.rutgers.edu/~szegedy/07540
Lecture 3
Fields
• A set F with two operations:
+ (addition), x (multiplication)
• (F, +) is an Abelian group with unit element 0.
• (F\{0}, x) is an Abelian group with unit element 1.
• For all x, y, z Є F: (x+y)z = xz + yz. (Distributivity)
• (We get the same definition if the multiplicative
part is not restricted to Abelian.)
Characteristic
Let F be a field (finite or infinite).
Let U = < 1>+
= {0, 1, 1+1, 1+1+1,…}, if |F| is finite
|U| is the characteristic of F. If |U| is infinite then the characteristic is 0.
(1 + 1)(1+1+1) = (1+1+1) + (1+ 1 + 1).
Similarly, product of any two elements from U is also from U by distributivity.
Let |U| = p, finite. Then U is isomorphic with Z/pZ with respect to. addition and
multiplication.
In this case p is a prime, otherwise F would have a zero divisor, so U= Fp.
And Fp is also called the prime subfield of F.
LEMMA:
If a field (finite or infinite) has finite characteristic p, then p is a prime.
A finite field F has positive characteristic p for some prime p.
Size of a finite field
Theorem 1.1 The cardinality of F is pn where n = [F : Fp] and Fp
denotes the prime subfield of F.
Proof. The prime subfield Fp of F is isomorphic to the field Z/pZ of
integers mod p. Since the field F is an n-dimensional vector space
over Fp for some finite n, it is set-isomorphic to Fpn and thus has
cardinality pn.
(Uni-variate) Polynomials
P(x) = xn + an-1 xn-1 + … + a1 x + a0
(deg P = n)
ais are the coefficients.
Roots:
P(c) = 0 → P(x) = (x-c) Q(x)
(deg Q = n-1)
→ P(x) can have at most n roots
Reducibility:
P(x) = Q(x)S(x)
(deg Q, deg S < n)
If there are no factors Q,S as above, then P is irreducible.
Theorem: the multiplicative group
of every finite field is cyclic
Let |F| = q.
The theorem says that there is g Є F such that F = { 0, g, g2,…, gq-1 }
We need to prove that there is a g with order q-1 (smallest power that is 1).
Let ORD(a) = { z | ord(z) = a}. ORD(a) is empty unless a|q-1.
LEMMA:
| ORD(a) | = φ(a) , where φ(a) is the number of those residue classes mod a
That are relatively prime to a.
REMARK:
The lemma immediately gives the theorem, since φ(q-1) ≥ 1.
Proof of the lemma:
We proceed by induction on a. ORD(1) = {1}. Consider a > 1.
z Є F is a root of xa -1 ↔ for some a’|a it holds that z Є ORD(a’).
→ xa -1 = Πa’|a Πf Є ORD(a’) (x-f) .
→ ∑a’|a |ORD(a’)| = a.
From the inductional hypothesis: |ORD(a)| = a - ∑a’|a; a’<a φ(a) .
To prove |ORD(a)| = φ(a) it is sufficient to show that:
LEMMA: ∑a’|a φ(a) = a.
Proof of the lemma:
Classify all numbers in {1,2,…,a} according to its greatest
common divisor with a, and for every b|a let
β(b) = | { z | (a,z) = b}|.
Clearly, ∑b|a β(b) = a.
We claim β(b) = φ(a/b). (If so, we are done, since {a/b | b|a} = {b | b|a}).
Indeed, (a,z) = b ↔ z = λb, where (λ, a/b) = 1.
Field extensions
Transcendental extension:
F(x) = { q(x)/r(x), where q,r are polynomials}
Algebraic extension (with a root of some irreducible polynomial, s(x)):
F(α) = {q(x) | q is a polynomial over F such that deg q < deg s}
q(α) ↔ q(x) mod s(x)
Alternative notation: F(α) ↔ F[x]/(s(x))
Inverse of r(x) for an algebraic extension:
If xists r’(x) such that
r’(x) r(x) + s’(x)s(x) = 1
→
r’(x) r(x) = 1 (mod s(x))
→ r’ = r-1
Splitting field
F’ is the splitting field of a polynomial r(x) in F
1. if r(x) decomposes into linear factors in F’.
2. F’ is the smallest field with this property
Remark: if (r’(x),r(x)) = 1, then all linear factors are different.
Linear spaces (classical approach)
S = Fn (dimension =n)
S = {(x1,x2,…,xn) | xi Є F }
Subspace:
S’ ≤ S, iff S’ is closed under linear combinations:
x,y Є S → λx + μy Є S
Affine subspaces
1 dimensional affine subspaces = lines
Lx,y = { x+λy | λ Є F }
2 dimensional affine subspaces = planes
Px,y,z = { x+λy+μz | λ,μ Є F }
n-1 dimensional affine subspaces = hyperplanes
S = { a1x1 + a2x2 + … + anxn =b}