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.

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)

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

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

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 • ;

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.

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

( ) ! 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.

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:

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).

Division with remainder Example: Solution:

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)

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 :

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:

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

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:

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

! 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 )

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

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). • +

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

(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

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

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

Example:

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

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

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.

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

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

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