Section10Math623

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Section 10: Finite Fields
HW p. 8 # 1-3 at the end of the notes
Finite Fields are fields with a finite number of elements. We have already examined
finite fields. Recall that given a prime p that the set of integers modulo p given by
Z p  {0, 1, 2, 3,  , p  1}
is a finite field of p elements.
In this section, we want to consider finite fields constructed from the extension field
Z p [ x] /( p ( x)) containing Z p , where p (x ) is an irreducible polynomial in Z p [x] .
We would like an “easy” way to generate the elements in Z p [ x] /( p ( x)) . We start with
the following fact:
Fact: The non-zero elements of a finite field, K * , form a cyclic group, that is, if number
of elements in K * is q  1 , then there exists an element a  K where
K *  {a, a 2 , a 3 , , a q2 , a q1  1} .
Recall that for an irreducible polynomial p (x ) in Z p [x] . Then recall that the congruence
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class x in the extension field Z p [ x] /( p ( x)) is a root of the polynomial. If we set a  x ,
then p (a )  0 . Generating other elements in Z p [ x] /( p ( x)) can be done by taking
powers of the root a , which is demonstrated by the following example.
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Example 1: Consider the polynomial p( x)  x 2  x  2  Z 3 [ x] . Noting the fact that
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p (x ) is irreducible and that a  x in Z 3 [ x] /( x 2  x  2) is a root of p (x ) , use the
powers of a to generate the non-zero elements of the finite field Z 3 [ x] /( x 2  x  2) .
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For a field F, let a  x  K  F[ x] /( p( x)) , where p (x ) is an irreducible polynomial in
F [x ] . Then a is a root of p (x ) , that is , p (a )  0 . We want a to be a cyclic generator of
K * , the non-zero elements of the finite field K  F [ x] /( p( x)) . However, p (x ) must be
chosen “correctly”. The polynomial p (x ) being irreducible does not guarantee that the root
a can be used to generate K * .
Fact: There always exists and irreducible polynomial p (x ) where the root a generates all
of K * . In this case, the root a is called a primitive element of K  F [ x] /( p( x)) and the
polynomial p (x ) is called a primitive polynomial.
This leads to another important fact concerning finite fields.
Theorem 10.1: All finite fields have q  p n elements for some integer n and prime p.
Although to give a general proof of Theorem 10.1, we need information concerning vector
spaces, we can sketch the important ideas in the proof using the concepts in this section. If
F is a field that is finite, its characteristic (the smallest integer p where p  1  0 ) be prime
(see Exercise 10 in Section 3). Hence, F  Z p . If we consider an irreducible polynomial
p (x ) of degree n, that is, if
p( x)  an x n  an1 x n1    a1  a0 ,
then the division algorithm says that each congruence class of the finite field
Z p [ x] /( p ( x)) can be represented by polynomial of the form
r ( x)  bn1 x n1  bn2 x n2    b1  b0
since deg r ( x)  deg( p( x)) . Since the coefficients (the bi ' s ) of r (x ) are contained in field
Z p  {0, 1, 2, 3,  , p  1} , each coefficient has p choices giving p n possible polynomials.
Notes
1. We denote the finite field Z p [ x] /( p ( x)) containing the field Z p as the finite field of
characteristic p .
2. A primitive polynomial p (x ) of degree n defined over Z p can be used to generated
the finite field Z p [ x] /( p ( x)) of p n elements.
Procedure for Constructing Finite Fields of p n Elements
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1. Consider the field Z p , where p is a prime.
2. Choose a primitive polynomial p( x)  Z p [ x] where deg( p( x))  n .
3. Construct the finite field Z p [ x] /( p ( x)) by taking the root a of primitive polynomial
p (x ) and constructing powers of a up to a p 1 and obtaining the non-zero elements of
Z p [ x] /( p ( x)) . The field is completed by adding the 0 element.
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Example 2: Construct the finite field Z 3 [ x] /( x 2  x  2) .
Solution: As we noted in Example 1, p( x)  x 2  x  2 is irreducible. Since deg p( x)  2
and for Z 3 , p  3 , we should obtain a finite field of p n  32  9 elements. Let a be a
zero of p( x)  x 2  x  2 . Then, as we saw in Example 1,
a2  a  2  0
or
a 2   a  2  2a  1 .
In Example 1, we demonstrate how by taking increasing powers of a , we can generate
the non-zero elements of the finite field. The results are given by
Powers of a
a
a2
a3
Finite Field Element
a
2a  1
2a  2
a4
a5
a6
2
2a
a7
a8
a2
a 1
1
Adding in the element 0 and the set of 9 finite field elements are given by
Z 3 [ x] /( x 2  x  2)  {a, 2a  1, 2a  2, 2, 2a, a  2, a  1, 1, 0} .
Since we were able to generate the entire finite field, a is a primitive element and
p( x)  x 2  x  2 is a primitive polynomial.
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Every irreducible polynomial p (x ) used to construct a finite field Z p [ x] /( p ( x)) is not
guaranteed to be a primitive polynomial. We illustrate this in the next example.
Example 3: Illustrate why the polynomial p( x)  x 2  1 is irreducible over Z 3 but is not
a primitive polynomial when used to construct the finite field Z 3 [ x] /( x 2  1) .
Solution:
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Example 4: Construct the finite field Z 2 [ x] /( x 4  x  1) .
Solution:
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Exercises
1. For each of the following polynomials p (x ) , determine if p (x ) is irreducible in
Z 3 [ x] . If so, decide if p (x ) is primitive in Z 3 [ x] by attempting to construct the field
elements that correspond to powers of the root a in Z 3 [ x] /( p( x)) . If so, list the
elements of the finite field.
a. p( x)  x 2  2 x  2
b.
p( x)  x 2  2 x  1
c.
d.
p( x)  2 x 2  2
p ( x)  2 x  2
2. For each of the following polynomials p (x ) , determine if p (x ) is irreducible in
Z 2 [ x] . If so, decide if p (x ) is primitive in Z 2 [ x] by attempting to construct the field
elements that correspond to powers of the root a in Z 2 [ x] /( p( x)) . If so, list the
elements of the finite field.
a. p( x)  x 3  x  1
b.
p( x)  x 3  x 2  1
c.
p( x)  x 4  x 3  1
d.
p ( x)  x 4  x 3  x 2  1
e.
p( x)  x 4  x 3  x 2  x  1
3. Let p( x)  x 2  x  2
a. Show that p (x ) is primitive in Z 5 [ x] by constructing the field elements that
correspond to powers of the root root a in Z 5 [ x] /( p( x)) . If so, list the elements
of the finite field.
b. Show that p (x ) is not primitive in Z11[ x] by showing that p (x ) is not irreducible
in Z11[ x] .
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