In this lecture we develop a part of the theory In this lecture we develop a part of the theory

of polynomials over rings and fields.of polynomials over rings and fields.

Our main goal is to construct finite fields. Our main goal is to construct finite fields.

First part Presenter: Davidov Inna.First part Presenter: Davidov Inna.

Second part Presenter: Vald Margarita.Second part Presenter: Vald Margarita.

A commutative ring (with 1) is a set R

together with two binary operations

+:R×R→R and •:R×R→R on R and two distinct

elements 0 and 1 of R with the following properties:

Definition:

• (a a + b) + c = a + (b + c c) (+ is associative))

• 0 + a = a (0 is the identity)

• a + b = b + a (+ is commutative)

• for each a in R there exists −a in R such that

a + (−a) = (−a) + a = 0 (exist inverse element)

for all a, b, c in R

• (a • b) • c = a • (b • c) (• is associative)

• 1 • a = a • 1 = a (1 is the identity)

• (a + b) • c = (a • c) + (b • c) (the distributive law)

Definition: Continue…

• a • b = b • a (• is commutative)

We write (R, +, •,0,1) for such a ring

Definition: A field is a commutative ring (R, +, •,0,1)

such that all elements of R except 0 have a

multiplicative inverse.

Example:

number prime a is field a is it

2, eachfor ring finite a is 0,1),,(Z ,mmm

m

m

Let (R ,+ ,• ,0 ,1 ) be a ring.

The set R[X] is defined to be the set of all

polynomials with coefficients in R

Definition:

R[X] g, ffor

ii

n

0ii

in

0ii

in

0ii )Xb (a Xb ( Xag f (a)

))(

nm

kj

kjii

jm

0jj

in

0ii )X b a ( Xb ( Xag f (b)

0

))(k

0i

ii

0i

ii XaXa f

n

Ria

together with the following operations + and • ;

If (R ,+ ,• ,0 ,1 ) is a ring

Then (R[X] ,+ ,• ,(0) ,(1) ) is also a ring.

Proposition:

Remark: For every field R, the ring R[X] is not a field:

But, We will soon see how to use polynomials

to construct fields.

X does not have a multiplicative inverse in R[X]

have weX...aXaa f everyfor dd10

1f X

1)( ...0X0XXa...XaXa f X 21dd

210 1

Let p be a prime number. Then Proposition:

; [X] Z gf,for , g f = g)(f and g + f = g) + (f (a) PppPPpp ••

0.k allfor )(X f = f

generally, more and, )(X f = f have we[X] Z ffor (b)kpkp

PPP

Proof: The multiplication in is commutative ]X[Zp

pp

timesp timesp timesp

P g f g)...(g) f...(f = g)(f ...g)(f = g)(f •••••••••••

Proof: Continue…

p p (p -1) (p - j+1)

j j (j -1) 2 1

The binomial theorem for the ring says that:]X[Zp

gg• f• + f = g) + (f Pj-pj

1-pj1

pp

( )j

p

All factors in the sum are to be reduced modulo p !

The numerator is divisible by p; The denominator is not:

p p p(f + g) f g

Second part: On board.

An element a in a ring is called a unit

if it is invertible with respect to multiplication

Definition:

Definition:

The degree of a polynomial R[X] is the

largest d such that the coefficient of is not zero.

In the case of zero polynomial the degree is defined

to be the −∞.

dX

N. d allfor d <-and

},{- N dfor - )(-dd)(-

Let R be a ring, and let h R[X] be a non zero

Polynomial whose leading coefficient is a unit on R.

Proposition:

Then for each f R[X] there are unique polynomials

q,r R[X] with f = h • q + r and deg(r) < deg(h).

Definition: if f = h • q (r=0) we say that h divides f.

For f,g R[X] we say that f and g are

congruent modulo h, if f - g is divisible by h.

Denoted by f g (mod h).

Definition:

Note: f r (mod h).

15R = ZExample:

4 2

2

f = 4X + 5X + 6X + 1

h = X + 6

Solution:

2f = (4X +11) h + 6X + 10

2 4 2 4 2

21

f - 4X h = 4X + 5X + 6X + 1 - (4X 9X ) =

11X + 6X + 1 = f

2 21 2f 11 h = 11X + 6X + 1 - (11X 6) = 6X + 10 = f

Division with Remainder -Time Analysis:

To obtain a degree smaller then d we need to

perform at most O(d’-d) iterations,

since on each iteration the degree is reduced by

at least 1.

If R, h, f are as in the preceding theorem with

On each iteration we perform O(d) operations

by multiplying a single element by the

polynomial h.

The total number of operations in R needed forthis procedure is O((d’ –d)d)

deg(f) = d’ and deg(h) = d Then:

12[X]

2 2 2 3(6X +4) (6X +2)=(6X +4) (6X +8)=4

Example: In the ring

2(6X +4) divides 4

The “quotient” is not uniquely determined

This is due to the fact that 6 is not a unit in 12

Question : Why?

411X 4)4)(7X(5X 422 on the contrary :

A polynomial f F[X] — {0} is called

irreducible if f does not have a proper divisor,

Or in other words,

if from f = g • h for g,h F[X] it follows that g F* or h F*

Definition:

The notion of irreducibility depends on the The notion of irreducibility depends on the

Underlying field Underlying field !

Example: 1X2

3Z F

The polynomial is irreducible since has

no roots at

1X2

3Z

2Z F

The polynomial is reducible1X2

1)1)(X(X 1X2

Let h F[X] be irreducible, and let f F[X]

be such that h does not divide f.

Then there are polynomials s and t such that:

1 = s • h + t • f.

Lemma:

Let h F[X] be irreducible. If f F[X] is

divisible by h and f = • , then h divides or h

divides .

Lemma: 1g 1g2g

2g

Let F be a field. Then every nonzero

polynomial f F[X] can be written as a product

a• • • • , s 0, where a F* and ,..., are monic

irreducible polynomials in F[X] of

degree > 0.

This product representation is unique up to

the order of the factors.

Theorem:

1h sh sh1h

There are efficient polynomial time randomized

algorithms for factoring f with coefficients in a

prime field

Algorithms for factoring polynomials :

!No Deterministic polynomial time algorithm is known

that can find the representation of a polynomial f as a

product of irreducible factors.

pF

We can factor f in operations in

Under the ERH using randomized algorithm.

( deg(h) = n )

qF qFnlogq)O(n2

Let F be a field, and let f F[X] with

f 0. Then |{a F | f(a) = 0}| d = deg (f).

Theorem:

Proof: On boardOn board

Definition: If (R, +, •, 0, 1) is a ring,

and h R[X], d = deg(h) 0,is a monic polynomial,

let R[X]/(h) be the set of all polynomials in R[X] of

degree strictly smaller than d, together with the

following operations hh and hh;

f hh g= (f + g) mod h and f hh g = (f g) mod h,

for f,g R[X]/(h).

•

+

•+ •

12R = ZExample:

4 3

3 2

h = X + 3X +1

f = 2X g = X + 5

Solution:

Now we determine the reminder mod h

35 10X2X f f • g =

h) (mod610X4X

h610X10X6X10X10X6X

h2X10X2X10X2X

3

3 43 4

3 53 5

610X4X 5X2X 3 2 3 hh

PropositionProposition: If R and h are as in the preceding

definition, then (R[X]/(h), +hh, ·hh ,0,1) is a ring with

1. Moreover, we have:

(a) f mod h = f if deg(f) < d;

(b) (f + g) mod h = ((f mod h) + (g mod h)) mod h

(f • g) mod h = ((f mod h) • (g mod h)) mod h

for all f,g R[Х];

(c) If g g (mod h), then f(g ) mod h = f(g ) mod h

for all f,g ,g R[X]11

11 22

22

11 22

The elements of R[X]/(h) are represented as arrays

of length d.

Adding two elements can be done by

performing d additions in R.

ImplementingImplementing R[X]/(h) & Time Analysis:R[X]/(h) & Time Analysis:

finally, we calculate (f·g) mod h by procedure for polynomial division.

Overall O( ) multiplications and additions in R2d

Multiplying two polynomials can be done by

performing multiplications and additions

in R.

2(d-1)2d

Example: 1)[X]/(XZ 22

Remark: The representation of a polynomial a+bX done by it coefficients sequence ab

Example: 1)X[X]/(XZ 22

Let F be a field, and let h F[X] be a

monic irreducible polynomial over F.

Then the structure F’= F[X]/(h) is a field.

If F is finite, this field has |F| elements.

Theorem:

)deg(h

Proof: On board

Example: 1)[X]/(XZ F 23

! all elements of F except 0 have a multiplicative inverse.

This is a field with 9 elements

Note: if deg(h) 2 then = X F’ - F.

if deg(h) = 1, then h = X + a for some a F

and = - a.

Proposition: Let F and h be as in the previous

theorem, and let F’ =F[X]/(h) be the corresponding

field.

Then the element = X mod h F’ is a root of h.

Let p and r be prime numbers

with p r, and let h be a monic irreducible

factor of = .

Then in the field F’ = F [X]/(h) the element

= X mod h satisfies ord ( ) = r.

r

Proposition:

1

1

rx

x

rr

'F

1 XX 1-r

Proof: On boardOn board

p

Let p and r be prime numbers

with p r, and q= .

Then q= • • •

Where ,…, are monic irreducible

polynomials of degree ord (p).r

Proposition:

rr

1 XX 1-r

Proof: On boardOn board

1h

sh

sh

1h

[X]Fp

In q splits into linear factors

Example: 1 XX q 4

1)11( 5ord

5r

11p

[x]Z11

2)6)(X7)(X8)(X(X XX q 4 1

= deg( ) = deg( ) = deg( ) = deg( )1h 2h 3h 4h

4321 hhhh q

In q is irreducible

4)7( 5ord

7p

[x]Z7

= deg (q)

[X]Zh,h,h,h 114321