in this lecture we develop a part of the theory of polynomials over rings and fields. our main goal...
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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