The concept of q-cycle is investigated for its properties and applications. Connections
with irreducible polynomials over a finite field are established with emphases on the
notions of order and degree. The results are applied to deduce new results about
primitive and self-reciprocal polynomials.
1. Introduction
Let 𝔽q denote the finite field of q elements and let N∈ℕ, gcd(q,N)=1. Let a0,a1,…,aℓ-1 be ℓdistinct numbers chosen from ℤN:={0,1,…,N-1}. If
(1)aiq≡ai+1modN(i=0,1,…,ℓ-2),(a0qℓ)≡aℓ-1q≡a0modN,
then we say that (a0,a1,…,aℓ-1) forms a q-cycle mod N with leading element a0, abbreviated by q(a0)-cycle(N) or q-cycle(N) when the leading element in the cycle is immaterial, and call ℓ the length of this q(a0)-cycle(N). The notion of q-cycles was introduced by Wan in his book [1, page 203]. Since qoN(q)≡1modN, where oN(q) is the order of q in ℤN*:=ℤN∖{0} (the multiplicative group of nonzero integers modulo N), it clearly follows that each q(a)-cycle(N) always has a unique length ℓ which is the least positive integer ℓ for which a≡aqℓmodN. Observe that q-cycles are nothing but ℤ-orbits qℤa,a∈ℤN; that is, 1∈ℤ acts on ℤN by multiplication with q. The concept of q-cycles is important because of the following connections with irreducible polynomials.
(A) (see [1, Theorem 9.11]) Let α be a primitive Nth root of unity (if the order of q in ℤN* is m, then there exists a primitive Nth root of unity in 𝔽qm). If (a0,a1,…,aℓ-1) is a q-cycle(N), then
(2)h(x)=(x-αa0)(x-αa1)⋯(x-αaℓ-1)
is a monic irreducible factor of xN-1 in 𝔽q[x]. Conversely, if h(x) is a monic irreducible factor of xN-1 in 𝔽q[x], then all the roots of h(x) are powers of α whose exponents form a q-cycle(N). We henceforth refer to these two facts as the α-correspondence.
(B) (see [1, Corollary 9.12]) The number of distinct irreducible polynomials dividing xN-1 in 𝔽q[x] is equal to the number of q-cycles(N) formed with the leading N elements taken from {0,1,…,N-1}.
Our objectives here are to illustrate the versatility of q-cycles by using them to prove new results about irreducible polynomials. General properties of q-cycles are given in the next section; specific details in two special cases corresponding to N=qn-1,qm+1 are worked out for applications to primitive and self-reciprocal polynomials in the last section. Section 3 deals with results about the order of a polynomial, while Section 4 does the same for the degree of a polynomial. Section 5 shows that knowing a q-cycle is equivalent to knowing all coefficients of the corresponding polynomial (2). The last section provides applications of q-cycles to primitive and self-reciprocal polynomials.
Notation and Terminology. Throughout, we fix the following symbols and their meanings.
pis a fixed prime, q is a power of p, and 𝔽q is the finite field with q elements.
N is a fixed positive integer such that gcd(q,N)=1.
d:=d(N) is the number of (positive integer) divisors of N.
N=N1>N2>⋯>Nd=1 (all divisors of N).
For n∈ℕ, the Möbius function μ is defined by
(3)μ(n)={1,ifn=1,(-1)k,ifnisasquarefreeproductofkdistinctprimes,0,ifnisnotsquarefree.
For n∈ℕ, the Euler's function φ(n) is the number of integers k∈{1,2,…,n} with gcd(k,n)=1.
For M∈ℕ, denote by oM(q) the order of q in ℤM*:=ℤM∖{0}; that is,
(4)qoM(q)≡1modM,butqi≢1modM∀i∈ℕ,i<oM(q).
For r∈{1,2,…,d}, set 𝒞Nr={q-cycles(Nr)}, the set of all q-cycles modNr.
2. Properties of q-Cycles
Let (a0,a1,…,aℓ-1) be a q(a0)-cycle(N). It is easy to see that ℤN can be decomposed into a finite union of disjoint q-cycles(N), namely,
(5)ℤN=(0)∪(1,q,q2,…,qoN(q)-1)⋃a∈ℤN∖{0,1}(a,aq,aq2,…).
Since qoN(q)≡1modN and aqoN(q)≡amodN for all a∈ℤN, each q-cycle(N) is of length ≤oN(q) with the one containing 1 having the largest length. We collect in our first theorem further properties of q-cycles, whose straightforward proofs are omitted.
Theorem 1.
Let a, b∈ℤN, N0:=ℕ∪{0}.
The element b∈q(a)-cycle(N) if and only if b≡aqmmodN for some m∈ℕ0.
Two q-cycles, q(a)-cycle(N) and q(b)-cycle(N), are identical if and only if b∈q(a)-cycle(N).
For b∈ℤN*, each q(b)-cycle(N) has length ℓ if and only if N/
gcd
(b,N) divides qℓ-1 but does not divide q-1,q2-1,…,qℓ-1-1 (if ℓ=1, only the first divisibility needs to be checked). The q(0)-cycle(N) has length 1.
The length of each q-cycle(N) divides oN(q).
If b∈ℤN is such that
gcd
(b,N)=Nr, then the q(b)-cycle(N) has length oN/Nr(q), and there are altogether φ(N/Nr)/oN/Nr(q) distinct such q-cycles. In particular, if b∈ℤN is such that
gcd
(b,N)=1, then the q(b)-cycle(N) has length oN(q), and there are altogether φ(N)/oN(q) distinct q-cycles with length oN(q).
The length of any q-cycle(N) is of the form oNr(q) for some r∈{1,…,d}.
The number of q-cycles(N) of length ℓ is (1/ℓ)∑1≤r≤d,oNr(q)=ℓφ(Nr).
The total number of q-cycles(N) is |𝒞N|=∑1≤r≤d(φ(Nr)/oNr(q)) (|·| signifies the number of elements in the set).
2.1. Two Special Cases
There are two particular cases N=qn-1 and N=qm+1, which are closely related to primitive and self-reciprocal and will be needed in the last section.
2.1.1. The Case N=qn-1
When N=qn-1(n∈ℕ), more precise information is now derived.
Theorem 2.
Let n∈ℕ. Then
oqn-1(q)=n;
n is the largest length among q-cycles(qn-1) and any other length of q-cycles(qn-1) divides n;
the number of q-cycles(qn-1) having length ℓ is equal to
(6)1ℓ∑1≤r≤doNr(q)=ℓφ(Nr)={q-1ifℓ=11ℓ∑k∣ℓμ(k)qℓ/kifℓ>1.
Proof.
Assertions (i) and (ii) are immediate from the definition and Theorems 1(iii)-(iv). To verify (iii), observe that if ℓ=1, then it is easily checked that all the q-cycles(qn-1) of length 1 are those starting with 0, A:=qn-1+qn-2+⋯+q+1,2A,…,(q-2)A whose total number is q-1. Assume now that ℓ>1 and that the assertion holds up to ℓ-1. From Theorem 1(vii), the number of q-cycles(qn-1) having length ℓ is (1/ℓ)∑1≤r≤d,oNr(q)=ℓφ(Nr). Thus, to prove the desired result, we only need to verify that
(7)∑1≤r≤doNr(q)=ℓφ(Nr)=∑k∣ℓμ(k)qℓ/k.
Consider
(8)∑1≤r≤doNr(q)=ℓφ(Nr)=∑Nr∣(qℓ-1)Nr∤(qs-1)∀s∣ℓ,1≤s<ℓφ(Nr)=∑Nr∣(qℓ-1)φ(Nr)-∑Nr∣(qs-1)∀s∣ℓ,1≤s<ℓφ(Nr)=qℓ-1-∑Nr∣(qs-1)∀s∣ℓ,1≤s<ℓφ(Nr).
Using the induction hypothesis for the second term on the right-hand side of (8) with an extra -1 to take care of the case ℓ=1, we get
(9)∑Nr∣(qs-1)∀s∣ℓ,1≤s<ℓφ(Nr)=∑s∣ℓ1≤s<ℓ(∑t∣sμ(t)qs/t)-1=∑k∣ℓ1≤k<ℓqk∑ik∣ℓ1≤ik<ℓμ(i)-1=∑k∣ℓ1≤k<ℓqk(∑i∣(ℓ/k)μ(i)-μ(ℓk))-1=-∑k∣ℓ1≤k<ℓqkμ(ℓk)-1=-∑k∣ℓ1≤k<ℓμ(k)qℓ/k-1.
The result now follows by substituting this into (8).
Implicit in the proof of Theorem 2 is the following nice identity.
Corollary 3.
With the same notation as in Theorem 2, one has
(10)∑1≤r≤doNr(q)=ℓφ(Nr)=∑d∣ℓμ(d)qℓ/d-⌊1ℓ⌋.
2.1.2. The Case N=qm+1
When N=qm+1(m∈ℕ), more precise information about q-cycles is now derived.
Theorem 4.
Let m∈ℕ. Then
oqm+1(q)=2m;
2m is the largest length among q-cycles(qm+1) and any other length divides 2m;
the number of q-cycles(qm+1) having length ℓ is (1/ℓ)∑1≤r≤d,oNr(q)=ℓφ(Nr).
Proof.
Since q2m≡1 mod qm+1, we have oqm+1(q)∣2m. Assertion (i) thus follows from the observation that oqm+1(q)>m. Assertions (ii) and (iii) are direct consequences of Theorems 1(iii)–(vii).
In [2], Yucas and Mullen introduced the following set:
(11)Dm:={(qk){-1}n∈ℕ;n∣(qm+1),dn∤(qk+1),k∈{0,1,…,m-1}}.
We show next that elements in this set can also be described through order.
Lemma 5.
Let m, n(>2)∈ℕ. Then
n∈Dm⇔ the multiplicative order ofqmodnis2m, that is, on(q)=2m, and there is a positive integer t dividing m such that n∣(qt+1).
Proof.
If n∈Dm, then n∣(q2m-1). If n∣(qs-1)(s∈ℕ), then [2, Proposition 1] tells us that s must be an even multiple of m. Thus, on(q)=2m. Conversely, assume that on(q)=2m and t is the least positive integer for which n∣(qt+1). Thus, [2, Proposition 1] tells us that 2m is an even multiple of t, and so t∣m. If t<m, the divisibility n∣(q2t-1) contradicts on(q)=2m. Hence, t=m.
The next proposition, which is of independent interest, connects the sum in Theorem 4(iii) with one involving the Möbius function.
Proposition 6.
One has
(12)∑n∈Dmφ(n)={qm-1ifmisapowerof2,qisodd,qmifmisapowerof2,qiseven,∑i∣mioddμ(i)qm/iifmisnotapowerof2.
To prove Proposition 6, we need some arithmetical facts about greatest common divisors.
Lemma 7.
Let e,f,s,m∈ℕ. Then
(13)
gcd
(qe+1,qf-1)={1,if
gcd
(2e,f)=
gcd
(e,f),2∣q2,if
gcd
(2e,f)=
gcd
(e,f),2∤qq
gcd
(e,f)+1,if
gcd
(2e,f)=2
gcd
(e,f),(14)
gcd
(qs+1,qm+1)={1,ifν(m)≥ν(m-s),2∣q2,ifν(m)≥ν(m-s),2∤qq
gcd
(m,s)+1,ifν(m)<ν(m-s),
where ν(j)=ν2(j) denotes the highest power of 2 that divides j∈ℕ.
Proof.
To prove (13), we consider two separate cases.
Case 1 (gcd(2e,f)=gcd(e,f)). Observe that
(15)gcd(qe+1,qf-1)∣gcd(q2e-1,qf-1)=qgcd(2e,f)-1=(qgcd(e,f)-1)∣(qe-1).
If 2∣q, then gcd(qe+1,qe-1)=1, and we infer that gcd(qe+1,qf-1)=1.
If 2∤q, then gcd(qe+1,qe-1)=2, and we infer that gcd(qe+1,qf-1)=2.
Case 2 (gcd(2e,f)=2gcd(e,f)). Observe that
(16)gcd(qe+1,qf-1)∣gcd(q2e-1,qf-1)=qgcd(2e,f)-1=q2gcd(e,f)-1=(qgcd(e,f)-1)(qgcd(e,f)+1).
If 2∣q, then gcd(qe+1,qgcd(e,f)-1)=1, and from (16), we have
(17)gcd(qe+1,qf-1)∣(qgcd(e,f)+1).
On the other hand,
(18)(qgcd(e,f)+1)∣(q2gcd(e,f)-1)=(qgcd(2e,f)-1)∣(qf-1).
Since e/gcd(e,f) is odd, we have (qgcd(e,f)+1)∣(qe+1). This last relation together with (17) and (18) yields gcd(qe+1,qf-1)=qgcd(e,f)+1, as required.
If 2∤q, then gcd((qe+1)/2,(qgcd(e,f)-1)/2)=1, and so we infer from (16) that
(19)gcd(qe+12,qf-12)∣(qgcd(e,f)+1).
Since (18) holds, we have
(20)qgcd(e,f)+12∣qf-12.
Since e/gcd(e,f) is odd, we also have
(21)qgcd(e,f)+12∣qe+12.
The relations (20) and (21) show that
(22)gcd(qe+12,qf-12)=qgcd(e,f)+12gcd(k1,k2)
for some k1, k2∈ℕ. Taking (19) into account, we deduce that gcd(k1,k2)=1 or 2. To complete the proof, it suffices to verify that gcd(k1,k2)=1. To this end, suppose that gcd(k1,k2)=2. Thus, (22) yields
(23)2(qgcd(e,f)+1)∣(qe+1).
Since e/gcd(e,f) is odd, we see that qe+1=(qgcd(e,f)+1)G, where G∈ℕ is odd. Combining this last relation with (23), we arrive that 2∣G, which is untenable.
To prove (14), we first observe that
(24)gcd(2e,f)=gcd(e,f)⟺ν(e)≥ν(f),gcd(2e,f)=2gcd(e,f)⟺ν(e)<ν(f).
Putting m=e, m-s=f in (13), the desired result (14) follows at once if we can show that
(25)gcd(qm+1,qm-s-1)=gcd(qm+1,qs+1).
To this end, let d1=gcd(qm+1,qm-s-1), d2=gcd(qm+1,qs+1). It is easily checked that d1=d2 because d2∣d1⇔d2∣(qm-s-1) and d1∣d2⇔d1∣(qs+1).
We are now ready to prove Proposition 6.
Case 1 (m is a power of 2).
Clearly, in this case ν(m)≥ν(m-s) for 0≤s<m. For brevity, let
(26)Sm≔{n∈ℕ;n∣(qs+1)dforsomes∈{0,1,…,m-1},n∣(qm+1)}.
Thus, ∑n∈Dmφ(n)=∑n∣(qm+1)φ(n)-∑n∈Smφ(n). If 2∣q, then using (14), we see that those s∈{0,1,…,m-1} for which n∣(qs+1) must satisfy gcd(qs+1,qm+1)=1, showing that n=1. On the other hand, if 2∤q, then using (14), we see those s∈{0,1,…,m-1} for which n∣(qs+1) must satisfy gcd(qs+1,qm+1)=2, showing that n=1,2. Combining the two possibilities, we get
(27)∑n∈Dmφ(n)=∑n∣(qm+1)φ(n)-{φ(1),if2∣q,φ(1)+φ(2),if2∤q,={qm,if2∣q,qm-1,if2∤q.
Case 2 (m is not a power of 2).
Then m=2νu, where ν:=ν(m) and u(≥3) is odd. From (14), we need to find those s∈{0,1,…,m-1} for which ν(m-s)>ν(m). These are s such that m-s=2v+1w(w∈ℕ); that is, s=2v(u-2w). Thus, all such s are of the form
(28)s=2ν·1,2ν·3,…,2ν·(u-2)
and correspondingly, gcd(s,m)=2νc, for some c∣u, 1≤c<u. We proceed now by induction on u, noting that the case u=1 (and u=2) is contained in Case 1. We have
(29)∑n∈Dmφ(n)=∑n∣(qm+1)φ(n)-∑n∈Smφ(n)=qm+1-∑c∣u1≤c<u∑n∈D2νcφ(n)+{φ(1)if2∣qφ(1)+φ(2)if2∤q=qm-∑c∣u1<c<u∑i∣cμ(i)q2νc/i-{0+q2νif2∣q1+q2ν-1if2∤qdddddddddddddddddddddddd(usinginduction)=qm-∑c∣u1≤c<u∑i∣cμ(i)q2νc/i=qm-∑k∣u1≤k<uq2νk∑i∣(u/k)1≤i<u/kμ(i)=qm+∑k∣u1≤k<uq2νkμ(uk)=∑j∣mjoddμ(j)qm/j.
3. Order of a Polynomial
Following [3, page 84], for f(x)∈𝔽q[x]∖{0} with f(0)≠0, the least e∈ℕ for which f(x) divides xe-1 is called the order of f(x), denoted by ord(f). If f(0)=0, then f(x)=xhg(x), where h∈ℕ and g∈𝔽q[x] with g(0)≠0 are uniquely determined, and ord(f) is defined to be ord(g). The order of a polynomial is also called the period or the exponent of that polynomial. The following facts are well-known.
(see [3, Theorem 3.3]) Let f(x)∈𝔽q[x] be irreducible with deg(f)=m∈ℕ, f(0)≠0. Then ord(f) is equal to the order of any root of f in the multiplicative group 𝔽qm*.
(see [3, Lemma 3.6]) Let c∈ℕ. Then f∈𝔽q[x] with f(0)≠0 divides xc-1 if and only if ord(f)∣c.
(see [3, Theorem 3.8]) Let g∈𝔽q[x] be irreducible over 𝔽q with g(0)≠0 and ord(g)=e, and let f=gb with b∈ℕ. Let t be the smallest integer with pt≥b, where p is the characteristic of 𝔽q. Then ord(f)=ept.
(see [3, Theorem 3.9]) Let g1,…,gk be pairwise relatively prime nonzero polynomials over 𝔽q, and let f=g1⋯gk. Then ord(f)=lcm{ord(g1),…,ord(gk)}.
3.1. Counting Formulae
Working with q-cycles, apart from the known result, [3, Theorem 3.5], that the number of monic irreducible f∈𝔽q[x], f(0)≠0 for which ord(f)=Nr(r∈{1,…,d}) and deg(f)=oNr(q) is φ(Nr)/oNr(q), some new information about the number of monic irreducible polynomials is now derived.
Theorem 8.
(i) The polynomial f∈𝔽q[x] constructed via (2) from a q(a)-cycle(N) with
gcd
(a,N)=Nr(r∈{1,…,d}) is a monic irreducible polynomial with f(0)≠0,
ord
(f)=N/Nr and deg(f)=oN/Nr(q). Conversely, each irreducible polynomial f∈𝔽q[x] with f(0)≠0,
ord
(f)=Nr and deg(f)=oNr(q) is an irreducible factor of xN-1 arisen, through (2), from an q(a)-cycle(N) with
gcd
(a,N)=N/Nr.
(ii) The set of monic, irreducible f∈𝔽q[x], f(0)≠0 for which
ord
(f)∣N is identical with the set of irreducible factors of xN-1.
(iii) The number of q-cycles(N), the number of monic irreducible factors of xN-1, and the number of monic irreducible f∈𝔽q[x] with f(0)≠0 and
ord
(f)∣N are all equal to ∑r=1dφ(Nr)/oNr(q).
Proof.
From Theorem 1(v), each q(a)-cycle(N) with gcd(a,N)=Nr has length oN/Nr(q) and it thus gives rise, through the α-correspondence, to a monic irreducible f∈𝔽q[x], f(0)≠0, of degree oN/Nr(q). Each root of f is of the form αaqi(i∈{0,1,…,oN/Nr(q)-1}) where α is a primitive Nth root of unity, and so its order is N/Nr. Part (i) is thus a consequence of [3, Theorem 3.3] and the α-correspondence.
To prove (ii), note that if the monic irreducible f∈𝔽q[x] satisfies f(0)≠0 and ord(f)∣N, then ord(f)=Nr for some r∈{1,…,d}. Thus, f∣(xNr-1), and so f∣(xN-1). On the other hand, if f is an irreducible factor of xN-1, the α-correspondence assures us that f is constructed from a q-cycle(N) and the desired conclusion follows from (i).
Part (iii) follows (ii), the α-correspondence, and Theorem 1(viii).
There is another straightforward technique to derive a formula for the number of monic irreducible polynomials of fixed order using q-cycles based on the inclusion-exclusion principle. Although the formula so obtained is not easy to use, we give a proof to illustrate this different approach. For r∈{1,2,…,d}, set (keeping the earlier notation in the first section)
(30)ℐNr={monicirreduciblepolynomialsarisingfromdtheq-cycles(Nr)asconstructedin(2)},𝒪Nd=𝒪1=|𝒞1|,𝒪Nr=|𝒞Nr|-∑M∣Nr,M<Nr𝒪M.
For a given s∈ℕ and distinct primes p1,p2,…,pk, define
(31)〈s〉≔{p1n1p2n2⋯pknk;n1,n2,…,nk∈ℕ,dn1+n2+⋯+nk=s}.
Proposition 9.
Let N=p1n1p2n2⋯pknk, the unique prime factorization of N. The number of monic, irreducible polynomials f∈𝔽q[x]∖𝔽q, f(0)≠0 having order N is equal to
(32)𝒪N=|ℐN1|-|⋃r=2dℐNr|(33)=|𝒞N1|-{∑r=2d|𝒞Nr|ssssss-∑x1,x2∈{N2,N3,…,Nd},x1<x2|𝒞
gcd
(x1,x2)|ssssss+⋯+(-1)dssssss×∑x1,x2,…,xd-1∈{N2,N3,…,Nd},x1<x2<⋯<xd-1|𝒞
gcd
(x1,x2,…,xd-1)|∑r=2d}(34)=|𝒞N1|-∑m∈〈s-1〉|𝒞m|+∑m∈〈s-2〉|𝒞m|+⋯+(-1)k∑m∈〈s-k〉|𝒞m|.
Proof.
Since the number of distinct irreducible factors of xN-1 in 𝔽q[x] is equal to the number of q-cycles(N) [1, Corollary 9.12], from its definition, the α-correspondence and [3, Lemma 3.6], 𝒪Nr is simply the number irreducible factors (in 𝔽q[x] constructed through (2)) of xNr-1 which are not irreducible factors of xM-1 for any M∣Nr, M<Nr, which is in turn equivalent to saying that 𝒪Nr is the number of monic, irreducible polynomials f∈𝔽q[x], f(0)≠0, having order Nr, and this proves (32).
Again using [1, Corollary 9.12] and [3, Lemma 3.6] to translate the number of q-cycles to those of corresponding irreducible polynomials, the inclusion-exclusion formulae on the right-hand sides of (33) and (34) clearly yield the number of irreducible factors of xN-1 which are not irreducible factors of xNr-1 for all r∈{2,3,…,d}, and this verifies (33) and (34).
Remark 10.
Let us remark that there are no irreducible polynomials in 𝔽q[x] of order n with gcd(q,n)>1. This is seen as follows: for f(x)∈𝔽q[x] with f(0)≠0, since ord(f) is the least positive integer for which f(x) divides xord(f)-1, monic irreducible polynomials of order n are contained in the set of irreducible factors of xn-1 in 𝔽q[x]. Putting n=Npe(e∈ℕ,gcd(q,N)=1), since xn-1=(xN-1)pe, the set of all irreducible factors of xn-1 is identical with the set of all irreducible factors of xN-1, and so each irreducible factor of xn-1 is of order ≤N<n.
In contrast to the preceding remark, there always exists a (reducible) polynomial in 𝔽q[x] of order n with gcd(q,n)>1. This is seen as follows: using the terminology of the last remark, since gcd(q,N)=1, Theorem 8 assures us that there always exists a monic, irreducible polynomial g(x)∈𝔽q[x], g(0)≠0, having ord(g)=N. Consequently, Theorem 3.8 in [3] tells us that ord(g(x)pe)=Npe=n.
Our next task is to compute the number of monic polynomials in 𝔽q[x] (both reducible and irreducible) having order N. Let f(x)∈𝔽q[x] be a factor of the polynomial xN-1 with ord(f)=N. Since gcd(q,N)=1, the polynomial xN-1 has no multiple factors. Assume that f(x) is decomposed into k distinct monic irreducible factors (in 𝔽q[x]), say, f(x)=f1(x)⋯fk(x). Since ord(f)=N=lcm{ord(f1),…,ord(fk)}, each ord(fi)∈{N=N1>N2>⋯>Nd=1} (the set of all divisors of N). Thus, the number of all such f's is equal to ∑e1+e2+⋯+ed=k,er≥0,lcm(Nr;er≠0)=N(𝒪N1e1)⋯(𝒪Nded). We have thus proved.
Theorem 11.
The total number of monic polynomials (both reducible and irreducible) f∈𝔽q[x], f(0)≠0, having order N is
(35)∑k=1m∑e1+e2+⋯+ed=k,er≥0
lcm
(Nr;er≠0)=N(𝒪N1e1)⋯(𝒪Nded),m:=∑r=1d𝒪Nr.
Having determined the number of polynomials with fixed order, it is natural to find out how many of them have orders dividing a fixed n∈ℕ.
Theorem 12.
Let n=peN∈ℕ with e∈ℕ0 and
gcd
(q,N)=1. Then the product of all monic irreducible polynomials f∈𝔽q[x], f(0)≠0 for which
ord
(f)∣n is equal to xN-1 and their number is equal to |𝒞N|=∑r=1dφ(Nr)/oNr(q).
Proof.
Since xn-1=(xN-1)pe, the set of all monic irreducible factors of xN-1 is identical with the set of all monic irreducible factors of xn-1. From [3, Theorem 3.6], for f∈𝔽q[x], f(0)≠0, we know that ord(f)∣n if and only if f(x)∣(xn-1). Thus, the set of all monic irreducible polynomials f∈𝔽q[x], f(0)≠0 for which ord(f)∣n is identical with the set of all irreducible factors of xn-1 which is also equal to the set of all irreducible factors of xN-1. Since xN-1 has no multiple root, the product of all monic irreducible polynomials f∈𝔽q[x], f(0)≠0 for which ord(f)∣n is xN-1 and the first assertion is established. The second assertion follows immediately from the first assertion and Theorem 8(iv).
3.2. Some Explicit Shapes
Having counted the number of monic irreducible polynomials of fixed order, we proceed to determine their explicit shapes. From [1, Corollary 9.12], we know that the number of irreducible factors of xN-1 in 𝔽q[x] is equal to the number of q-cycles(N) formed by the N numbers 0,1,…,N-1. To determine explicit shapes of all monic, irreducible polynomials f∈𝔽q[x], f(0)≠0, having order N, we consider the polynomial
(36)f(x)=(x-αa0)(x-αa1)⋯(x-αaℓ-1)∈𝔽q[x]
associated with the q-cycle(N)(a0,a1,…,aℓ-1), where α is a primitive Nth root of unity. By definition, [3, Lemma 3.6] and [1, Theorem 9.11], among all such polynomials, those which do not divide xNr-1, r∈{2,3,…,d}, are all the sought after polynomials having order N.
Although, the procedure just described is satisfactory in principle, in certain cases, more precise shapes can be given. For a positive integer n not divisible by p, recall that the nth cyclotomic polynomial [3, page 64] is defined as
(37)Qn(x)=∏i=1gcd(i,n)=1n(x-ζi)=∏i∣n(xi-1)μ(n/i)∈𝔽q[x],
where ζ is a primitive nth root of unity and μ is the Möbius function. It is known that
(38)xN-1=∏r=1dQNr(x).
Proposition 13.
Let d=d(N) be the number of divisors of N.
If the number of q-cycles(N) is equal to d, then in (38) each QNr(x) is irreducible in 𝔽q[x] and
ord
(QNr)=Nr. In particular, QN(x)∈𝔽q[x] is irreducible and of order N. The same conclusion holds if q is a primitive root mod Nr for all r∈{1,2,…,d}.
If QNr(x) is irreducible in 𝔽q[x] for all r∈{1,2,…,d}, then the number of q-cycles(N) is equal to d.
Proof.
We prove only part (i), for the other part is trivial from (38). If the number of q-cycles(N) is equal to d, the α-correspondence shows that the number of monic irreducible factors of xN-1 is also d. The first assertion that each QNr(x)∈𝔽q[x] is irreducible follows at once from (38). This is also the case if q is a primitive root mod Nr for all r∈{1,2,…,d} which is a direct consequence of [3, Theorem 2.47(ii)].
From (38), we see that QNr divides xNr-1. If ord(QNr)=j<Nr, then QNr divides xj-1. Further, Lemma 3.6 of [3] shows that j∣N, and so xj-1 divides xN-1. The first assertion then implies that each cyclotomic factor of xNr-1 and of xj-1 (as in (38)) is irreducible. Since gcd(Nr,q)=1, the polynomial xNr-1 contains no multiple root and so all its irreducible factors are distinct. But the cyclotomic factorization of xj-1 (as in (38)) does not contain the irreducible factor QNr (because Nr∤j), which is a contradiction.
Our next proposition, which shows how the number of q-cycles can be used to generate irreducible polynomials, is based on the following known results.
(Berlekamp's factorization theorem, [3, Theorem 4.1]). If f∈𝔽q[x] is monic and h∈𝔽q[x] is such that hq≡hmodf, then f(x)=∏c∈𝔽qgcd(f(x),h(x)-c).
(see [1, Theorem 9.13]) For each q-cycle(N)(a0,a1,…,aℓ-1), if g(x)=xa0+xa1+⋯+xaℓ-1, then g(x)q≡g(x)mod(xN-1).
Combining these two results, we get the following.
Proposition 14.
Let (a0,a1,…,aℓ-1) be a q-cycle(N), and g(x)=xa0+xa1+⋯+xaℓ-1. If the number of q-cycles(N) is equal to q, and if
gcd
(xN-1,g(x)-c)≠1 for all c∈𝔽q, then each polynomial
gcd
(xN-1,g(x)-c) is irreducible over 𝔽q.
4. Degree of a Polynomial
We begin by listing (without proofs) known facts involving degrees of irreducible polynomials, which can be proved using q-cycles.
(see [3, Theorem 3.20]) The product of all monic irreducible f∈𝔽q[x], f(0)≠0 for which deg(f)∣n is equal to xqn-1-1.
(see [3, Theorem 3.25]) The number of monic irreducible polynomials in 𝔽q[x] of degree ℓ is equal to (1/ℓ)∑k∣ℓμ(k)qℓ/k.
(see [3, Corollary 3.4]) The set of monic irreducible f∈𝔽q[x], f(0)≠0 for which deg(f)∣n is identical with the set of monic irreducible f∈𝔽q[x], f(0)≠0 for which ord(f)∣(qn-1). Moreover, for f∈𝔽q[x], f(0)≠0, we have deg(f)∣n if and only if ord(f)∣(qn-1).
Using q-cycles, that is, Theorem 1(viii) and the assertion [3, Theorem 3.20] the following new result is immediate.
Theorem 15.
The number of monic irreducible polynomials f(x)∈𝔽q[x], f(0)≠0 for which deg(f)∣n is equal to |𝒞qn-1|=∑M∣qn-1φ(M)/oM(q), where the definition of 𝒞qn-1 can be found in the heading “Notation and Terminology”.
Combining Theorems 15 and 2(iii), we obtain a nice arithmetical identity.
Corollary 16.
One has ∑M∣(qn-1)(φ(M)/oM(q))=∑M∣n((1/M)∑k∣Mμ(k)qM/k)-1.
5. Coefficients and q-Cycle
Since a given q-cycle gives rise, through the α-correspondence, to a unique polynomial, we illustrate now how to uniquely determine all coefficients of its corresponding polynomial (2) and conversely.
Theorem 17.
Let α be a primitive Nth root of unity, let m be the order of qmodN, and let ξ be a generator of 𝔽qm.
If the q(a0)-cycle(N)(a0,a1,…,aℓ-1) is given, then the coefficients of its corresponding polynomial in (2) h(x)=(x-αa0)(x-αa1)⋯(x-αaℓ-1):=∑k=0ℓ(-1)kckxℓ-k∈𝔽q[x] are given by
(39)c0=1,ck=∑0≤i0<i1<⋯<ik-1≤ℓ-1αai0αai1⋯αaik-1(k=1,…,ℓ).
If the coefficients of h(x) are given, then the corresponding q-cycle is uniquely determined through its leading element which is found by solving for s, A, and a0, respectively, from the system
(40)ξs=(-1)ℓcℓ,(41)(qm-1N)A≡smodqm-1,(42)a0(1+q+⋯+qℓ-1)≡AmodN.
Proof.
(i) The relations in (39) are simply the symmetric polynomial relations between the roots and coefficients of a polynomial.
(ii) It is trivial that each q-cycle is uniquely determined from its leading element. Next, note that (40) is simply a representation of element in 𝔽qm. Through the α-correspondence, the q-cycle (a0,a1,…,aℓ-1) associated with h(x) must satisfy
(43)ξs=(-1)ℓcℓ=αt,t∶=a0+a1+⋯+aℓ-1≡a0(1+q+⋯+qℓ-1)modN.
The conclusion follows from the facts that α=ξ(qm-1)/N and that solving (41) and (42) yields a0 uniquely mod N.
6. Primitive and Self-Reciprocal Polynomials
We now apply preceding results to two special cases, corresponding to N=qn-1 and N=qm+1, to deduce a number of results about primitive and self-reciprocal polynomials.
6.1. Primitive Polynomials
Recall from [3, Definition 3.15] that a primitive polynomial of degree n∈ℕ over 𝔽q is the minimal polynomial over 𝔽q of a primitive element of 𝔽qn. We mention, without proofs, some known results provable using q-cycles.
(see [3, Theorem 3.16]) An irreducible polynomial f(x)∈𝔽q[x] of degree n is primitive if and only if f is monic, f(0)≠0 and ord(f)=qn-1. In other words, each primitive polynomial in 𝔽q[x] of degree n can be constructed from a q(a)-cycle(qn-1) with gcd(a,qn-1)=1 through (2). Conversely, for each q(a)-cycle(qn-1) with gcd(a,qn-1)=1, the polynomial (2) is a primitive polynomial in 𝔽q[x] of degree n.
(see [1, Theorem 7.7]) The number of primitive polynomials in 𝔽q[x] of degree n is φ(qn-1)/n, which equals to the number of q(a)-cycle(qn-1) with gcd(a,qn-1)=1.
6.2. Self-Reciprocal Polynomials
Let f(x)=anxn+an-1xn-1+⋯+a1x+a0∈𝔽q[x] with an≠0. The reciprocal polynomial f* of f is defined by f*(x)=xnf(1/x)=a0xn+a1xn-1+⋯+an-1x+an. A nonzero polynomial f∈𝔽q[x] is called self-reciprocal if f(x)=f*(x):=xnf(1/x). We confine ourselves here to the study of self-reciprocal irreducible monic (srim) polynomials in 𝔽q[x]. Since there is only one first degree srim-polynomial, namely, x+1, throughout the rest of this section we treat only srim-polynomials of degree ≥2. We next mention a known characterization and results about srim-polynomials provable via q-cycles.
(see [2, page 275]) Let f∈𝔽q[x] be irreducible and monic of degree ≥2. Then f is self-reciprocal if and only if its set of roots (each of which is evidently non-zero) is closed under inversion (and so its degree must be even).
(see [2, Proposition 3]) If f∈𝔽q[x] is a srim-polynomial with deg(f)=2m, then ord(f)∣(qm+1) but ord(f)∤(qk+1) for all k∈{0,1,…,m-1}.
(see [4, Lemma 2.3], [5, Theorem 1(i)], [2, Corollary 5]) For a∈ℤqm+1 with gcd(a,qm+1)=Nr(r∈{1,…,d}), each q(a)-cycle(qm+1), which is of length oN/Nr(q)=2m, gives rise through (2) to a srim-polynomial in 𝔽q[x] of degree 2m and order dividing qm+1. Conversely, each srim-polynomial f∈𝔽q[x] of degree 2m arises via (2) from a q(a)-cycle(qm+1) with length o(qm+1)/Nr(q)=2m and ord(f)∣(qm+1), where Nr=gcd(a,qm+1).
(see [5, Theorem 1(ii)]) Each irreducible factor f(x) of xqm+1-1 with deg(f)≥2 is a srim-polynomial with deg(f)=2k, where k∣m and m/k is odd.
We next prove two new results using q-cycles.
Theorem 18.
(i) For each divisor Nr of qm+1, the number of all srim-polynomial's f∈𝔽q[x] for which
ord
(f)=Nr and deg(f)=oNr(q) is φ(Nr)/oNr(q).
(ii) The number of srim-polynomials of degree 2m in 𝔽q[x] is equal to (1/2m)∑1≤r≤d,oNr(q)=2mφ(Nr).
Proof.
(i) and (ii) follow from [4, Lemma 2.3], [5, Theorem 1(ii)], Theorem 4, [3, Theorem 3.5], and Theorem 8(ii).
Combining [2, Proposition 3], [5, Theorem 1(i)-(ii)], and Theorem 18 with results in Section 2, we obtain the following known results about the set Dm as defined in (11).
(see [2, Proposition 4]) Assume that f is a srim-polynomial over 𝔽q of degree 2m and let α∈𝔽q2m be a root of f. Then α is a primitive Dth root of unity for some D∈Dm.
(see [3, Exercise 3.15, page 141]) If f is a srim-polynomial in 𝔽q[x] of degree >1 and order e, then every monic irreducible polynomial in 𝔽q[x] of degree >1 whose order divides e is self-reciprocal.
(see [2, Corollary 5], [6, Theorem 2]) If f is a srim-polynomial of degree 2m over 𝔽q, then ord(f)∈Dm.
(see [2, Proposition 7]) Let D∈Dm and let β be a primitive Dth root of unity. If fβ(x)=∏i=02m-1(x-βqi), then the polynomial fβ is an srim-polynomial of degree 2m and order D.
(see [2, Theorem 8]) For a monic irreducible polynomial f of degree 2m over 𝔽q, the following statements are equivalent.
f is self-reciprocal.
ord(f)∈Dm.
f=fβ for some primitive Dth root of unity β, with D∈Dm.
(see [2, Theorem 9])
There are φ(D)/2m srim-polynomials in 𝔽q[x] of degree 2m and order D for each D∈Dm.
The number of srim-polynomials in 𝔽q[x] of degree 2m is (1/2m)∑D∈Dmφ(D).
(see [2, Theorem 11]) For D∈Dm, the Dth cyclotomic polynomial QD factors into the product of all srim-polynomials over 𝔽q of degree 2m and order D.
(see [5, Theorem 3]) Let Sq(m) denote the number of srim-polynomials of degree 2m over 𝔽q. Then
(44)Sq(m)={qm-12m,ifqisodd,m=2s;12m∑d∣mdoddμ(d)qm/d,otherwise.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
Supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
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