2 Bounds on the size of codes
3.4 Constructing finite fields
α∈Fq
α=0.
Exercise 165 What is the smallest field of characteristic 2 that contains a:
(a) primitive nineth root of unity?
(b) primitive 11th root of unity?
What is the smallest field of characteristic 3 that contains a:
(c) primitive seventh root of unity?
(d) primitive 11th root of unity?
3.4 Constructing finite fields
We are now ready to link the additive structure of a finite field arising from the vector space interpretation with the multiplicative structure arising from the powers of a primitive element and actually construct finite fields.
A nonconstant polynomial f(x)∈Fq[x] isirreducible overFqprovided it does not factor into a product of two polynomials inFq[x] of smaller degree. The irreducible polynomials inFq[x] are like the prime numbers in the ring of integers. For example, every integer greater than 1 is a unique product of positive primes. The same result holds inFq[x], makingFq[x]
a unique factorization domain.
Theorem 3.4.1 Let f(x)be a nonconstant polynomial. Then f(x)= p1(x)a1p2(x)a2· · ·pk(x)ak,
where each pi(x)is irreducible,the pi(x)s are unique up to scalar multiplication,and the ais are unique.
Not only isFq[x] a unique factorization domain, it is also a principal ideal domain. An ideal I in a commutative ring Ris a nonempty subset of the ring that is closed under subtraction such that the product of an element inI with an element inRis always inI.
The idealI isprincipalprovided there is ana ∈Rsuch thatI= {r a|r∈R}; this ideal will be denoted (a). Aprincipal ideal domainis an integral domain in which each ideal is principal. Exercises 153 and 166 show thatFq[x] is a principal ideal domain. (The fact that Fq[x] is a unique factorization domain actually follows from the fact that it is a principal ideal domain.)
Exercise 166 Show using the Division Algorithm that every ideal ofFq[x] is a principal
ideal.
107 3.4 Constructing finite fields
To construct a field of characteristicp, we begin with a polynomial f(x)∈Fp[x] which is irreducible overFp. Suppose that f(x) has degreem. By using the Euclidean Algorithm it can be proved that theresidue class ring
Fp[x]/(f(x))
is actually a field and hence a finite fieldFqwithq =pmelements; see Exercise 167. Every element of the residue class ring is a cosetg(x)+(f(x)), whereg(x) is uniquely determined of degree at mostm−1. We can compress the notation by writing the coset as a vector in Fmp with the correspondence
gm−1xm−1+gm−2xm−2+ · · · +g1x+g0+(f(x))↔gm−1gm−2· · ·g1g0. (3.2) This vector notation allows you to add in the field using ordinary vector addition.
Exercise 167 Let f(x) be an irreducible polynomial of degreeminFp[x]. Prove that Fp[x]/(f(x))
is a finite field withpmelements.
Example 3.4.2 The polynomial f(x)=x3+x+1 is irreducible over F2; if it were re-ducible, it would have a factor of degree 1 and hence a root inF2, which it does not. So F8=F2[x]/(f(x)) and, using the correspondence (3.2), the elements ofF8are given by
Cosets Vectors
0+(f(x)) 000 1+(f(x)) 001 x+(f(x)) 010 x+1+(f(x)) 011 x2+(f(x)) 100 x2+1+(f(x)) 101 x2+x+(f(x)) 110 x2+x+1+(f(x)) 111
As an illustration of addition, addingx+(f(x)) tox2+x+1+(f(x)) yieldsx2+1+ (f(x)), which corresponds to adding 010 to 111 and obtaining 101 inF32. How do you multiply? To multiply g1(x)+(f(x)) times g2(x)+(f(x)), first use the Division Algorithm to write
g1(x)g2(x)= f(x)h(x)+r(x), (3.3)
where degr(x)≤m−1 or r(x)=0. Then (g1(x)+(f(x)))(g2(x)+(f(x)))=r(x)+ (f(x)). The notation is rather cumbersome and can be simplified if we replace x byα and let f(α)=0; we justify this shortly. From (3.3),g1(α)g2(α)=r(α) and we extend our correspondence (3.2) to
gm−1gm−2· · ·g1g0↔gm−1αm−1+gm−2αm−2+ · · · +g1α+g0. (3.4)
So to multiply inFq, we simply multiply polynomials in αin the ordinary way and use the equation f(α)=0 to reduce powers of αhigher thanm−1 to polynomials inαof degree less thanm. Notice that the subset{0αm−1+0αm−2+ · · · +0α+a0|a0∈Fp} = {a0|a0∈Fp}is the prime subfield ofFq.
We continue with our example ofF8 adding this new correspondence. Notice that the groupF∗8 is cyclic of order 7 and hence all nonidentity elements ofF∗8 are primitive. In particularαis primitive, and we include powers ofαin our table below.
Example 3.4.3 Continuing with Example 3.4.2, using correspondence (3.4), we obtain Polynomials Powers
Vectors inα ofα
000 0 0
001 1 1=α0
010 α α
011 α+1 α3
100 α2 α2
101 α2+1 α6
110 α2+α α4
111 α2+α+1 α5
The column “powers ofα” is obtained by usingf(α)=α3+α+1=0, which implies that α3=α+1. So α4=αα3=α(α+1)=α2+α, α5=αα4=α(α2+α)=α3+α2=
α2+α+1, etc.
Exercise 168 In the fieldF8given in Example 3.4.3, simplify (α2+α6−α+1)(α3+α)/(α4+α).
Hint: Use the vector form of the elements to do additions and subtractions and the powers
ofαto do multiplications and divisions.
We describe this construction by saying thatFq is obtained fromFp by “adjoining” a rootαof f(x) toFp. This rootαis formally given byα=x+(f(x)) in the residue class ringFp[x]/(f(x)); thereforeg(x)+(f(x))=g(α) and f(α)= f(x+(f(x)))= f(x)+ (f(x))=0+(f(x)).
In Example 3.4.3, we were fortunate thatαwas a primitive element ofF8. In general, this will not be the case. We say that an irreducible polynomial overFpof degreemisprimitive provided that it has a root that is a primitive element ofFq =Fpm. Ideally we would like to start with a primitive polynomial to construct our field, but that is not a requirement (see Exercise 174). It is worth noting that the irreducible polynomial we begin with can be multiplied by a constant to make it monic as that has no effect on the ideal generated by the polynomial or the residue class ring.
It is not obvious, but either by using the theory of splitting fields or by counting the number of irreducible polynomials over a finite field, one can show that irreducible polynomials of any degree exist. In particular we have the following result.
109 3.4 Constructing finite fields
Theorem 3.4.4 For any prime p and any positive integer m,there exists a finite field,unique up to isomorphism,with q= pmelements.
Since constructing finite fields requires irreducible polynomials, we note that tables of irreducible and primitive polynomials overF2can be found in [256].
Remark:In the construction ofFq by adjoining a root of an irreducible polynomial f(x) toFp, the fieldFpcan be replaced by any finite fieldFr, whereris a power of pand f(x) by an irreducible polynomial of degreeminFr[x] for some positive integerm. The field constructed containsFras a subfield and is of orderrm.
Exercise 169
(a) Find all irreducible polynomials of degrees 1, 2, 3, and 4 overF2.
(b) Compute the product of all irreducible polynomials of degrees 1 and 2 inF2[x].
(c) Compute the product of all irreducible polynomials of degrees 1 and 3 inF2[x].
(d) Compute the product of all irreducible polynomials of degrees 1, 2, and 4 inF2[x].
(e) Make a conjecture based on the results of (b), (c), and (d).
(f) In part (a), you found two irreducible polynomials of degree 3. The roots of these polynomials lie inF8. Using the table in Example 3.4.3 find the roots of these two
polynomials as powers ofα.
Exercise 170 Find all monic irreducible polynomials of degrees 1 and 2 overF3. Then compute their product inF3[x]. Does this result confirm your conjecture of Exercise 169(e)?
If not, modify your conjecture.
Exercise 171 Find all monic irreducible polynomials of degrees 1 and 2 overF4. Then compute their product inF4[x]. Does this result confirm your conjecture of Exercise 169(e) or your modified conjecture in Exercise 170? If not, modify your conjecture again.
Exercise 172 In Exercise 169, you found an irreducible polynomial of degree 3 different from the one used to constructF8in Examples 3.4.2 and 3.4.3. Letβbe a root of this second polynomial and construct the fieldF8 by adjoiningβ toF2 and giving a table with each
vector inF32associated to 0 and the powers ofβ.
Exercise 173 By Exercise 169, the polynomial f(x)=x4+x+1∈F2[x] is irreducible overF2. Letαbe a root of f(x).
(a) Construct the fieldF16 by adjoiningαtoF2and giving a table with each vector inF42 associated to 0 and the powers ofα.
(b) Which powers ofαare primitive elements ofF16?
(c) Find the roots of the irreducible polynomials of degrees 1, 2, and 4 from Exercise 169(a).
Exercise 174 Let f(x)=x2+x+1∈F5[x].
(a) Prove that f(x) is irreducible overF5.
(b) By part (a)F25 =F5[x]/(f(x)). Letαbe a root of f(x). Show thatαis not primitive.
(c) Find a primitive element inF25of the formaα+b, wherea,b∈F5.
Exercise 175 By Exercise 169,x2+x+1, x3+x+1, andx4+x+1 are irreducible
overF2. Isx5+x+1 irreducible overF2?
Exercise 176 Define a functionτ :Fq →Fqbyτ(0)=0 andτ(α)=α−1forα∈F∗q. (a) Show thatτ(ab)=τ(a)τ(b) for alla,b∈Fq.
(b) Show that ifq =2, 3, or 4, thenτ(a+b)=τ(a)+τ(b).
(c) Show that ifτ(a+b)=τ(a)+τ(b) for alla,b∈Fq, thenq =2, 3, or 4. Hint: Let α∈Fq withα+α2=0. Then seta =αandb=α2.