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In this lecture we develop a part of the theory
of polynomials over rings and fields.
Our main goal is to construct finite fields.
First part Presenter: Davidov Inna.
Second part Presenter: Vald Margarita.
Definition: 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:
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
Definition: A field is a commutative ring (R, +, •,0,1)
such that all elements of R except 0 have a
multiplicative inverse.
Example:
(Zm , m , m ,0,1) is a finite ring for each m  2,
it is a field  m is a prime number
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
n
f   aiXi   aiXi
i0
ai  R
i0
together with the following operations + and • ;
n
n
(a) f  g  (  aiX )  (  biX ) 
i
i0
(b) f  g  (
n
i
i0
m
 aiX )  ( b jX ) 
i0
for f ,g  R[X]
i
j0
j
n
 (a  b )X
i 0
i
mn
i
i
k
(
a

b
)X
  i j
k 0 i jk
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:
for every f  a0  a1X  ...adX d we have
X  f  a0X  a1X2  ... adXd1  1  0X  0X2  ... ( 1)
X f 1
X does not have a multiplicative inverse in R[X]
But, We will soon see how to use polynomials
to construct fields.
Proposition: Let p be a prime number. Then
(a) (f + g) p = f p + g P and (f • g) P = f p • g p , for f, g  ZP [X] ;
(b) for f  Z P [X] we have f P = f (X P ) and, more generally,
f
pk
pk
= f (X ) for all k  0.
Proof: The multiplication in
Z p [X]
is commutative
(f • g) P = (f • g) • ... • (f • g) = (f • ... • f ) • (g • ... • g)  f p • g p
 

 


p times
p times
p times
Proof: Continue…
The binomial theorem for the ring
p
p
(f + g) = f +
( )
1 j p -1
p
j
Z p [X]
says that:
• f j • g p- j  g P
! All factors in the sum are to be reduced modulo p
 p  p  (p -1)  (p - j +1)
 
j  (j-1)  2 1
j
The numerator is divisible by p; The denominator is not:
(f + g)p  f p  g p
Second part: On board.
Definition: The degree of a polynomial  R[X] is the
largest d such that the coefficient of X d is not zero.
In the case of zero polynomial the degree is defined
to be the −∞.
(-)  d  d  (-)  -  for d  N  {- },
and -  < d for all d  N.
Definition: An element a in a ring is called a unit
if it is invertible with respect to multiplication
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).
Example:
R = Z15
f = 4X4 + 5X2 + 6X + 1
h = X2 + 6
Solution:
f = (4X2 +11)  h + 6X + 10
f - 4X2  h = 4X4 + 5X2 + 6X + 1 - (4X4  9X2 ) =
11X2 + 6X + 1 = f1
f1  11  h = 11X2 + 6X + 1 - (11X2  6) = 6X + 10 = f2
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)
Example: In the ring
12 [X]
(6X2 +4) (6X2 +2)=(6X2 +4) (6X3 +8)=4
(6X2 +4) divides 4
The “quotient” is not uniquely determined
Question : Why?
This is due to the fact that 6 is not a unit in 12
on the contrary : (5X 2  4)(7X 2  4)  11X 4  4
Definition: 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*
!
The notion of irreducibility depends on the
Underlying field
Example: X2  1
F  Z3
The polynomial X2  1 is irreducible since has
no roots at Z3
F  Z2
The polynomial X2  1 is reducible
X2  1  (X  1)(X  1)
Lemma: 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 = g1 • g2 , then h divides g1 or h
divides g2 .
Theorem: Let F be a field. Then every nonzero
polynomial f  F[X] can be written as a product
a• h1 • • • hs , s  0, where a  F* and h1 ,..., hs are
monic irreducible polynomials in F[X] of
degree > 0.
This product representation is unique up to
the order of the factors.
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 Fp
We can factor f  Fq in O(n2  nlogq) operations in Fq
Under the ERH using randomized algorithm.
( deg(h) = n )
Theorem: Let F be a field, and let f  F[X] with
f  0. Then |{a  F | f(a) = 0}|  d = deg (f).
Proof: On 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 + h and • h;
f + h g= (f + g) mod h
for f,g  R[X]/(h).
and f
•h
g = (f • g) mod h,
Example:
R = Z12
h = X4 + 3X3 +1
f = 2X3
g = X2 + 5
Solution: f • g = 2X
5
 10X 3
Now we determine the reminder mod h

2X5  10X 3  2X5  10X 3  2X h 

6X 4  10X 3  10X  6X 4  10X 3  10X  6 h
4X 3  10X  6 (mod h)
2X3
X
h
2
 5  4X3 10X  6
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 g1  g2 (mod h), then f(g 1) mod h = f(g 2) mod h
for all f,g1 ,g 2 R[X]
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
2
performing d 2 multiplications and (d-1)
additions in R.
finally, we calculate (f·g) mod h by procedure for
polynomial division.
Overall O(d 2 ) multiplications and additions in R
Example: Z2[X]/(X 2  1)
Remark: The representation of a polynomial a+bX done
by it coefficients sequence ab
Example:
Z2[X]/(X 2  X  1)
Theorem: 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|deg(h) elements.
Proof: On board
2
Example: F  Z3[X]/(X  1)
!
all elements of F except 0 have a multiplicative inverse.
This is a field with 9 elements
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.
Note: if deg(h)  2 then  = X  F’ - F.
if deg(h) = 1, then h = X + a for some a  F
and  = - a.
r
Proposition: Let p and r be prime numbers
with p  r, and let h be a monic irreducible
r
factor of x  1 = Xr-1      X  1 .
x 1
r
Then in the field F’ = Fp [X]/(h) the element
 = X mod h satisfies ord ' ( ) = r.
F
Proof: On board
r
Proposition: Let p and r be prime numbers
with p  r, and q= Xr-1      X  1 .
Then q= h1 • • • hs
Where h1 ,…, hs  Fp [X] are monic irreducible
polynomials of degree ordr (p).
Proof: On board
Example: r  5
q  X4      X  1
p  11
In Z1 1 [x] q splits into linear factors
q  X4      X  1  (X  8)(X  7)(X  6)(X  2)
q  h1  h2  h3  h4
h1 ,h2 ,h3 ,h4 Z11[X]
ord5 (11) 1 = deg( h1) = deg(h ) = deg(h3 ) = deg(h )
2
4
p7
In Z [x] q is irreducible
7
ord5 (7)  4 = deg (q)