On irreducible polynomials of certain types in finite fields

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On irreducible polynomials of certain types in nite elds Stephen D. Cohen Mathematical Proceedings of the Cambridge Philosophical Society / Volume 66 / Issue 02 / September 1969, pp 335 - 344 DOI: 10.1017/S0305004100045023, Published online: 24 October 2008

Link to this article: http://journals.cambridge.org/abstract_S0305004100045023 How to cite this article: Stephen D. Cohen (1969). On irreducible polynomials of certain types in nite elds. Mathematical Proceedings of the Cambridge Philosophical Society, 66, pp 335-344 doi:10.1017/ S0305004100045023 Request Permissions : Click here

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Proc. Gamb. Phil. Soc. (1969), 66, 335 POPS 66-39 Printed in Great Britain

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335

On irreducible polynomials of certain types in finite fields B Y STEPHEN D. COHEN University of Glasgow [Received 25 April 1968) 1. Introduction. Let GF(q) be the finite field containing q = pl elements, where p is a prime and I a positive integer. Let P(x) be a monic polynomial in GF[q, x] of degree m. In this paper we investigate the nature and distribution of monic irreducible polynomials of the following types: (I) P(xr), where r is a positive integer (r-polynomials). (II) xmP(x + x~1). (Reciprocal polynomials.) These have the form Q(x)=a*" + aimr_1aP*-*+...+a1x+l, am_i = am+i (i^m-l). (1-1) (III) xrmP(xr+x-r). (r-reciprocal polynomials.) These have the form Q(xr), where Q(x) satisfies (1-1). Note that the degree of polynomials of types (I), (II) and (III) are rm, 2m and 2rm, respectively. Our principal results are the determinations of Lr(rm, q), M(2m, q) and Nr(2rm, q), the total number of monic, irreducible polynomials of types (I), (II) and (III), respectively. In general, they can be most concisely expressed in terms of 7r(m, q), the number of monic irreducible polynomials of degree m in GF(q, x). As is well known, we have i n(m,q) = - £ /i{s)qK (1-2) «1 st=m

Several authors have discussed the decomposition of r-polynomials into prime factors. See, for example (l), (5), (10) and (li). However, since we require only the criterion for P(xr) to be irreducible we outline a direct proof of the result concerned with this (Theorem 1 below) instead of deducing it from more complicated general results. In this way, we evaluate LT(rm q). Carlitz (4), has already determined M(2m, q). However he requires substantially different proofs for even and odd q. Using a somewhat different condition for a reciprocal polynomial to be irreducible, we obtain a shorter proof of his result applicable for all q. Our criteria for irreducibility are in a. suitable form to be combined to apply to r-reciprocal polynomials, which leads to the determination Nr(2rm, q). Each of the polynomials of type (I), (II) or (III) can be expressed in the form m g (x) P{f(x)/g(x)) for suitable relatively prime polynomials/and g. In each case, if K is a suitable algebraic extension of GF(q) and x has minimal polynomial/—^ over K(t) (where t is an indeterminate), then K(x) is a normal extension of GF(q,t), i.e. 9(x)f(y)—f(x)9(y) regarded as a function of y splits completely in K(x). It is in these circumstances that much is known about decomposition of primes in algebraic extensions. We illustrate this in section 6 by giving a further proof of the value of M(2m, q) using essentially these considerations.

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S. D. COHEN

Since if gm(x) P{f{x)jg(x)) is irreducible then so is P(x), we shall assume from now on that, unless otherwise stated, P(x) is an irreducible of degree m in GF[q, x\. We shall also assume that all polynomials considered are monic. 2. Outline of method. Our first lemma, in discussing the irreducibility of m g (x) P(f(x)/g(x)), is a particular case of a general theorem due to Capelli ((2), § 5). The proof given shows that the result is a simple consequence of the fact that if K c L £ M isatoweroffields,thendeg(i//^) = deg (MIL) deg {LjK). We remark that if P(A) = 0, GF(q, A) = GF(qm). m LEMMA 1. Iff{x), g(x) e GF[q, x], theng (x) P(f(x)/g(x)) is irreducible in GF[q, x] if and only iff(x) — Ag(x) is irreducible in GFlq™, x], where P(A) = 0. Proof. Let Q(x) = gm(x) P(f(x)jg{x)) and h = max (deg/, deg g) so that Q(x) has degree hm. Let y be a root of Q(x) (in a splitting field). Then clearly/(y) = Ag(y) for some root A of P(x), i.e. y is a root of/(x) — Ag(x), a polynomial of degree h in GF[q, A, x]. Evidently also GF(q, A) s GF(q, y), while, of course, deg (GF(q, A)/GF(q)) = m. Now Q(x) is irreducible in GF[q,x]odeg(GF(q,y)/VF(q)) = hm, i.e. o deg (GF(q, y)/GF(q, A)) = h, i.e. of(x) — Ag(x) is irreducible in GF[q, A, x]. The proof is complete. By Lemma 1, we require to find the number of irreducibles of the form

f(x)-Ag(x)

(2-1)

in GF[qm, x], where A runs through a set of roots of irreducibles of degree m in GF[q, x], there being exactly one root from each such polynomial in the set. We shall denote this number by n(flg, m, q). Thus, the number of irreducibles of the form (2-1) in GFlq™, x], where A runs through all elements of GF(qm) is n(f/g, l,qm). Evidently since GF(qm) comprises the roots of all irreducibles whose degree divides m, we have S sns(flg, m, q) = n(f/g, 1, q™),

(2-2)

s\m

where 7rs(flg, m, q) denotes the number of irreducibles P(x) of degree s, such that, if P(A) = 0, then (2-1) is irreducible in GF[q™,x]. Now an irreducible polynomial, F(x), of degree h in GF[q, x] is irreducible in the extension GF[ql,x] if and only if (h,t) = 1, ((5), §48). Thus, if (2-1) is irreducible in GF[qs, x] where A e GF(qs), s\m and deg (/— Ag) = h, it remains irreducible in GFlq™, x] if and only if (h, m/s) = 1. I t follows that s,q)

(h,m/s) = 1, otherwise.

Accordingly, we have, from (2-2) and (2-3), st = m

Hence to obtain n(fjg, m, q), we need only find v{f\g, 1, q).

(2-3)

Irreducible polynomials in

finite

fields

337

We shall use the concepts of the order of a non-zero element of GF(q) and the order of an irreducible polynomial (4= x) in GF[q,z]. The order of a non-zero element of GF(q) is its order as an element of the multiplicative group of GF(q). The order of an irreducible polynomial (4= x) in GF[q, x] is the least positive integer e such that x6 = l(modP(aj)). For convenience, we list some of the well-known elementary properties of the two concepts and the relationship between them in the next lemma. They are all immediate consequences of the facts that, if P(A) = 0, then GF(q,X) = GF[q,x] (mod P(x)) = GF(q™)

and that the multiplicative group of a finite field is cyclic. Indeed part (i) of Lemma 2 below is just a special case of part (ii). f

LEMMA 2. (i) Let A in GF(q) have order e. Then e\q— 1 and X = 1 if and only if e\f.

Moreover, for each e\q— 1, there are {e) elements of order e. (ii) Let P(x) have order e in GF[q, x\. Thene\qm— \andxf = 1 (mod P(x)) if and only if e\f. Furthermore, there exist irreducibles of order e and degree m if and only if q has order m(mode), when there are {e)lm such irreducibles. (iii) P(x) has order e if and only if all the roots of P(x) = Oin GF(qm) have order e.

3. r-polynomials. Since, ifp \ r,P(xr) = PP(XTIP), we can assume^) \r (^ 2).Fromnow on let r' denote the product of distinct primes in r. Necessary and sufficient conditions for the irreducibility of P(xr) are obtained as a particular case of a theorem stated (without proof) by Pellet, see ((5), § 59). We first state the result. THEOREM 1. Suppose r is a positive integer and P(x) an irreducible polynomial of degree m and order e in GF[q,x]. Then P(xr) is irreducible in GF[q, x~] if and only if r'K^-l), 4{(r,g m + 1) and {r,{qm-\)je) = 1. (In fact, we obtain by this theorem, every irreducible in GF(q, x) of degree rm and order re, ((5), §34).) We now outline a proof of Theorem 1. By Lemma 1, P(xr) is irreducible in GF(q, x) if and only if xr — A is irreducible in GF[qm,x~\. Lowe and Zelinsky(6), have shown that if r' \ q — 1 or 4|(r, q+ 1), xr—a. (aeGF(q)) is reducible in GF(q). Indeed, if r^is a prime such that r-\r but rx\q— 1, then a = /?ri for some ft in GF(q) and hence a;r'r» — fl\xr-a. Again, if q = — 1 (mod 4), it is easy to write down explicit quadratic factors of x4 - a. From the above, we deduce that P(xr) can be irreducible only if (r, q) = 1, r'\qm— 1 and 4f (r, q™ + 1). The important step in the proof of Theorem 1 is the following result given in Lemma 3 and due, in the case r odd, to MacCluer ((7), Lemma E), although when r is prime the result is well known. The argument used, of course, breaks down if r satisfies 4|(r, q+1).

LEMMA 3. / / 4-f(r,