1. Introduction Polynomials
In this lecture we develop a part of the theory of polynomials over rings and fields.
Our main goal is to construct finite fields.
Introduction
First part Presenter: Davidov Inna.
Second part Presenter: Vald Margarita.

2. rings A commutative ring (with 1) is a set R Definition:
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:
for all a, b, c in R
(a + b) + c = a + (b + 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)

3. Definition: Continue…
(a • b) • c = a • (b • c) (• is associative)
1 • a = a • 1 = a (1 is the identity)
a • b = b • a (• is commutative)
(a + b) • c = (a • c) + (b • c) (the distributive law)
We write (R, +, •,0,1) for such a ring

4.  A field is a commutative ring (R, +, •,0,1) Definition:
such that all elements of R except 0 have a
multiplicative inverse.
Example: 

5. polynomials over rings
Definition:
Let (R ,+ ,• ,0 ,1 ) be a ring.
The set R[X] is defined to be the set of all
polynomials with coefficients in R
together with the following operations + and • ;

6. Then (R[X] ,+ ,• ,(0) ,(1) ) is also a ring.
Proposition:
If (R ,+ ,• ,0 ,1 ) is a ring
Then (R[X] ,+ ,• ,(0) ,(1) ) is also a ring.
Remark:
For every field R, the ring R[X] is not a field:
X does not have a multiplicative inverse in R[X]
But, We will soon see how to use polynomials
to construct fields.

7. substitution Proposition: Let p be a prime number. Then
Proof: The multiplication in is commutative

8. ( ) ! Proof: Continue… The binomial theorem for the ring says that:
( ) !
All factors in the sum are to be reduced modulo p
The numerator is divisible by p; The denominator is not:
Second part: On board.

9. 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 −∞.
An element a in a ring is called a unit
if it is invertible with respect to multiplication
Definition:

10. Division with remainder
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.
Definition:
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).
Note:
f r (mod h).

11. Division with remainder
Example:
Solution:

12. Division with Remainder -Time Analysis:
If R, h, f are as in the preceding theorem with
deg(f) = d’ and deg(h) = d Then:
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.
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 for
this procedure is O((d’ –d)d)

13. The “quotient” is not uniquely determined
Example:
In the ring
divides 4
The “quotient” is not uniquely determined
Question : Why?
This is due to the fact that 6 is not a unit in
on the contrary :

14. Irreducible Polynomials & Factorization
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:

15. ! The notion of irreducibility depends on the Underlying field
Example:
The polynomial is irreducible since has
no roots at
The polynomial is reducible

16. 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:

17. Unique Factorization for Polynomials
Theorem:
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. s h s h

18. ! 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. !
There are efficient polynomial time randomized algorithms for factoring f with coefficients in a prime field
We can factor f in operations in
Under the ERH using randomized algorithm.
( deg(h) = n )

19. Roots of Polynomials Let F be a field, and let f F[X] with Theorem:
f Then |{a F | f(a) = 0}| d = deg (f).
Theorem:
Proof: On board

20. Quotients of Polynomial Rings
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 h and h;
f h g= (f + g) mod h and f h g = (f g) mod h,
for f,g R[X]/(h). • +

21. Now we determine the reminder mod h
Example:
Solution:
f • g =
Now we determine the reminder mod h h

22. (c) If g g (mod h), then f(g ) mod h = f(g ) mod h
Proposition: If R and h are as in the preceding definition, then (R[X]/(h), +h, ·h ,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] 1 2 1 2 1 2

23. Implementing R[X]/(h) & Time Analysis:
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.
Multiplying two polynomials can be done by performing multiplications and additions in R.
finally, we calculate (f·g) mod h by procedure for polynomial division.
Overall O( ) multiplications and additions in R

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

25. Example:

26. Finite Fields
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:
Proof: On board

27. ! Finite Fields Example:
all elements of F except 0 have a multiplicative inverse.
This is a field with 9 elements

28. 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.
Proposition:
Note: if deg(h) 2 then = X F’ - F.
if deg(h) = 1, then h = X + a for some a F
and = - a.

29. Roots of the Polynomial X -1
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.
Proposition:
Proof: On board

30. Roots of the Polynomial X -1
Let p and r be prime numbers with p r, and q=
Then q= • • •
Where ,…, are monic irreducible polynomials of degree ord (p).
Proposition: s h s h
Proof: On board

31. Example: In q splits into linear factors
= deg( ) = deg( ) = deg( ) = deg( )
In q is irreducible
= deg (q)