Cyclotomic cosets and minimal polynomials

LetEbe a finite extension field ofFq. ThenEis a vector space overFqand soE=Fq^tfor some positive integert. By Theorem 3.3.2, each elementαofEis a root of the polynomial x^qt −x. Thus there is a monic polynomialM_α(x) inFq[x] of smallest degree which hasα as a root; this polynomial is called theminimal polynomial ofαoverFq. In the following theorem we collect some elementary facts about minimal polynomials.

Theorem 3.7.1 LetFq^t be an extension field of Fq and letαbe an element of Fq^t with minimal polynomial M_α(x)inFq[x]. The following hold:

(i) M_α(x)is irreducible overFq.

(ii) If g(x)is any polynomial inFq[x]satisfying g(α)=0,then M_α(x)|g(x).

113 3.7 Cyclotomic cosets and minimal polynomials

(iii) M_α(x)is unique;that is,there is only one monic polynomial inFq[x]of smallest degree which hasαas a root.

Exercise 183 Prove Theorem 3.7.1.

If we begin with an irreducible polynomial f(x) overFq of degreer, we can adjoin a root of f(x) toFqand obtain the fieldFq^r. Amazingly, all the roots of f(x) lie inFq^r. Theorem 3.7.2 Let f(x)be a monic irreducible polynomial overFq of degree r . Then:

(i) all the roots of f(x)lie inFq^r and in any field containingFq along with one root of f(x),

(ii) f(x)=+r

i=1(x−αi),whereαi ∈Fq^r for1≤i ≤r,and (iii) f(x)|x^qr −x.

Proof: Letαbe a root of f(x) which we adjoin toFqto form a fieldE_αwithq^relements. If βis another root of f(x), not inE_α, it is a root of some irreducible factor, overE_α, of f(x).

AdjoiningβtoEαforms an extension fieldEofEα. However, insideE, there is a subfieldEβ

obtained by adjoiningβtoFq.E_β must haveq^r elements as f(x) is irreducible of degree r overFq. SinceEα andEβ are subfields ofE of the same size, by Theorem 3.5.3(iii), E_α=E_β proving that all roots of f(x) lie inFq^r. Since any field containingFq and one root off(x) containsFq^r, part (i) follows. Part (ii) now follows from Exercise 159. Part (iii) follows from part (ii) and the fact thatx^qr−x=+

α∈Fqr(x−α) by Theorem 3.3.2.

In particular this theorem holds for minimal polynomialsM_α(x) overFq as such poly-nomials are monic irreducible.

Theorem 3.7.3 LetFq^t be an extension field of Fq and letα be an element ofFq^t with minimal polynomial M_α(x)inFq[x]. The following hold:

(i) M_α(x)|x^qt−x.

(ii) M_α(x)has distinct roots all lying inFq^t. (iii) The degree of M_α(x)divides t.

(iv) x^qt −x=+

αM_α(x),whereαruns through some subset of Fq^t which enumerates the minimal polynomials of all elements of Fq^t exactly once.

(v) x^qt −x=+

f f(x),where f runs through all monic irreducible polynomials overFq

whose degree divides t.

Proof: Part (i) follows from Theorem 3.7.1(ii), sinceα^qt −α=0 by Theorem 3.3.2. Since the roots ofx^qt−xare theq^t elements ofFq^t,x^qt −x has distinct roots, and so (i) and Theorem 3.7.2(i) imply (ii). By Theorem 3.4.1x^qt −x=+n

i=1pi(x), where pi(x) is ir-reducible overFq. Asx^qt −xhas distinct roots, the factors pi(x) are distinct. By scaling them, we may assume that each is monic asx^qt −x is monic. Sopi(x)=M_α(x) for any α∈Fq^t withpi(α)=0. Thus (iv) holds. But ifM_α(x) has degreer, adjoiningαtoFqgives the subfieldFq^r =Fp^mr ofFq^t =Fp^mt implyingmr|mtby Theorem 3.5.3 and hence (iii).

Part (v) follows from (iv) if we show that every monic irreducible polynomial overFq of degreerdividingtis a factor ofx^qt−x. But f(x)|(x^qr −x) by Theorem 3.7.2(iii). Since

mr|mt, (x^qr−x)|(x^qt −x) by Lemma 3.5.2.

Two elements of Fq^t which have the same minimal polynomial inFq[x] are called conjugate overFq. It will be important to find all the conjugates ofα∈Fq, that is, all the roots ofM_α(x). We know by Theorem 3.7.3(ii) that the roots ofM_α(x) are distinct and lie inFq^t. We can find these roots with the assistance of the following theorem.

Theorem 3.7.4 Let f(x)be a polynomial inFq[x]and letα be a root of f(x)in some extension fieldFq^t. Then:

(i) f(x^q)= f(x)^q,and

(ii) α^q is also a root of f(x)inFq. Proof: Let f(x)=n

i=0aixⁱ. Sinceq = p^m, wherepis the characteristic ofFq, f(x)^q = n

i=0a^q_ix^{i q} by applying Exercise 152 repeatedly. However,a_i^q =ai, becauseai ∈Fqand elements ofFqare roots ofx^q−xby Theorem 3.3.2. Thus (i) holds. In particular, f(α^q)=

f(α)^q =0, implying (ii).

Repeatedly applying this theorem we see that α, α^q, α^q2, . . . are all roots of M_α(x).

Where does this sequence stop? It will stop afterr terms, whereα^qr =α. Suppose now thatγ is a primitive element ofFq^t. Thenα=γ^s for somes. Henceα^qr =αif and only ifγ^sqr−s=1. By Theorem 3.3.1(ii),sq^r ≡s (modq^t−1). Based on this, we define the q-cyclotomic coset of s modulo q^t−1 to be the set

Cs = {s,sq, . . . ,sq^r−1}(modq^t−1),

whereris the smallest positive integer such thatsq^r ≡s (modq^t−1). The setsCspartition the set{0,1,2, . . . ,q^t−2}of integers into disjoint sets. When listing the cyclotomic cosets, it is usual to listCsonly once, wheresis the smallest element of the coset.

Example 3.7.5 The 2-cyclotomic cosets modulo 15 areC0= {0},C1= {1,2,4,8},C3= {3,6,12,9},C5= {5,10}, andC7= {7,14,13,11}.

Exercise 184 Compute the 2-cyclotomic cosets modulo:

(a) 7, (b) 31,

(c) 63.

Exercise 185 Compute the 3-cyclotomic cosets modulo:

(a) 8,

(b) 26.

Exercise 186 Compute the 4-cyclotomic cosets modulo:

(a) 15, (b) 63.

Compare your answers to those of Example 3.7.5 and Exercise 184.

We now know that the roots ofM_α(x)=M_γ^s(x) include{γⁱ |i ∈Cs}. In fact these are all of the roots. So if we know the size ofCs, we know the degree ofM_γ^s(x), as these are the same.

115 3.7 Cyclotomic cosets and minimal polynomials

Theorem 3.7.6 Ifγ is a primitive element of Fq^t,then the minimal polynomial ofγ^sover Fqis

M_γ^s(x)=

i∈C_s

(x−γⁱ).

Proof: This theorem is claiming that, when expanded, f(x)=+

i∈C_s(x−γⁱ)=

j fjx^j is a polynomial inFq[x], not merelyFq^t[x]. Letg(x)= f(x)^q. Theng(x)=+

i∈C_s(x^q− γ^qi) by Exercise 152. AsCsis aq-cyclotomic coset,qi runs throughCs asi does. Thus g(x)= f(x^q)=

j fjx^{q j}. Butg(x)=(

j fjx^j)^q =

j f_j^qx^{q j}, again by Exercise 152.

Equating coefficients, we have f_j^q = fj and hence by Theorem 3.6.1(iii), f(x)∈Fq[x].

Exercise 187 Prove that the sizer of aq-cyclotomic coset moduloq^t−1 satisfiesr |t.

Example 3.7.7 The field F8 was constructed in Examples 3.4.2 and 3.4.3. In the table below we give the minimal polynomial overF2 of each element ofF8and the associated 2-cyclotomic coset modulo 7.

Roots Minimal polynomial 2-cyclotomic coset

0 x

1 x+1 {0}

α, α², α⁴ x³+x+1 {1,2,4}

α³, α⁵, α⁶ x³+x²+1 {3,5,6}

Notice thatx⁸−x =x(x+1)(x³+x+1)(x³+x²+1) is the factorization ofx⁸−xinto irreducible polynomials inF2[x] consistent with Theorem 3.7.3(iv). The polynomialsx³+ x+1 andx³+x²+1 are primitive polynomials of degree 3.

Example 3.7.8 In Exercise 173, you are asked to constructF16using the irreducible poly-nomial x⁴+x+1 over F2. With α as a root of this polynomial, we give the minimal polynomial overF2of each element ofF16and the associated 2-cyclotomic coset modulo 15 in the table below.

Roots Minimal polynomial 2-cyclotomic coset

0 x

1 x+1 {0}

α, α², α⁴, α⁸ x⁴+x+1 {1,2,4,8}

α³, α⁶, α⁹, α¹² x⁴+x³+x²+x+1 {3,6,9,12}

α⁵, α¹⁰ x²+x+1 {5,10}

α⁷, α¹¹, α¹³,α¹⁴ x⁴+x³+1 {7,11,13,14}

The factorization ofx¹⁵−1 into irreducible polynomials inF2[x] is (x+1)(x⁴+x+1)(x⁴+x³+x²+x+1)(x²+x+1)(x⁴+x³+1),

again consistent with Theorem 3.7.3(iv).

Exercise 188 Referring to Example 3.7.8:

(a) verify that the table is correct,

(b) find the elements ofF16that make up the subfieldsF2andF4, and

(c) find which irreducible polynomials of degree 4 are primitive.

Exercise 189 The irreducible polynomials of degree 2 overF3were found in Exercise 170.

(a) Which of the irreducible polynomials of degree 2 are primitive?

(b) Letαbe a root of one of these primitive polynomials. Construct the fieldF9by adjoining αtoF3and giving a table with each vector inF²₃associated to 0 and the powers ofα.

(c) In a table as in Examples 3.7.7 and 3.7.8, give the minimal polynomial overF3of each element ofF9and the associated 3-cyclotomic coset modulo 8.

(d) Verify that the product of all the polynomials in your table isx⁹−x.

Exercise 190 Without factoringx⁶³−1, how many irreducible factors does it have over F2 and what are their degrees? Answer the same question about x⁶³−1 overF4. See

Exercises 184 and 186.

Exercise 191 Without factoringx²⁶−1, how many irreducible factors does it have over

F3and what are their degrees? See Exercise 185.

Exercise 192 Let f(x)= f0+ f1x+ · · · + fax^a be a polynomial of degreea inFq[x].

Thereciprocal polynomialof f(x) is the polynomial f^∗(x)=x^af(x⁻¹)= fa+ fa−1x+ · · · + f₀x^a.

(We will study reciprocal polynomials further in Chapter 4.)

(a) Give the reciprocal polynomial of each of the polynomials in the table of Example 3.7.7.

(b) Give the reciprocal polynomial of each of the polynomials in the table of Example 3.7.8.

(c) What do you notice about the roots and the irreducibility of the reciprocal polynomials that you found in parts (a) and (b)?

(d) How can you use what you have learned about reciprocal polynomials in parts (a), (b), and (c) to help find irreducible factors ofx^q−xoverFpwhereq = p^mwithpa prime?