On powers of polynomials

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On powers of polynomials

Rodney Coleman, Laurent Zwald

To cite this version:

Rodney Coleman, Laurent Zwald. On powers of polynomials. 2014. �hal-01001707�

On Powers of Polynomials

Rodney Coleman, Laurent Zwald Laboratoire Jean Kuntzmann

Domaine Universitaire de Saint-Martin-d’H` eres, France.

Abstract

The aim of this short note is to show that, under certain conditions, when the coefficients of a power of a polynomial over a field lie in a subfield, then the coefficients of the polynomial itself lie in the subfield.

LetKbe a field andka subfield ofK. IfP(X)∈K[X],m∈N^∗andP^m(X)∈k[X], then in general we cannot say that P(X) ∈ k[X]. For example, if P(X) = iX ∈ C[X], thenP²(X) = −X² ∈ Q[X], or if P(X) = √

2 + ^√1₂X ∈ R[X], then P²(X) = 2 + 2X+ ¹₂X² ∈Q[X]. In both cases the square of the polynomial P lies in a subfield of a field and the polynomial itself does not lie in this subfield. In this note we will show that this cannot occur if the coefficients of the polynomialP(X) satisfy a certain simple condition. To fix our ideas we will initially suppose that K=Candk=Q. Of course, if m= 1 there is nothing to prove, so we will suppose that this is not the case.

Let us writeP(X) =Pn

i=0aiXⁱ, with theai ∈C. Using the multinomial theorem we obtain P^m(X) = X

k0+k1+···+kn=m

m k0, k1, . . . , kn

Y

0≤t≤n

(atX^t)^kt,

where

m k0, k1, . . . , kn

= m!

k0!k1!· · ·kn!. We will write Q(X) =P^m(X) =Pmn

j=0bjX^j and at first suppose thata06= 0. Then the coefficientbs of X^s∈Q(X) may be written

bs=X

a^k₀⁰a^k₁¹· · ·a^k_nⁿ

m k0, k1, . . . , kn

,

where

k1+ 2k2+· · ·+nkn = s (1)

k0+k1+k2+· · ·+kn = m. (2)

Ifs= 0, then from the above equations we obtaink0=mandki= 0 fori6= 0, thusb0=a^m₀ . If s= 1, then, from equation (1) , k1 = 1 and ki = 0 ifi > 1. Now, using equation (2), we obtain k0=m−1 and sob1=ma^m₀⁻¹a1.

Now let us consider the cases= 2. From (1)ki = 0 ifi >2 and for (k1, k2) we have (0,1) or (2,0).

From (2) the corresponding values ofk0 arem−1 andm−2. Hence we have b2=ma^m₀⁻¹a2+a^m₀⁻²a²₁

m m−2,2

.

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Continuing, we now consider the cases= 3. From (1) we see that ki = 0 ifi >3 and for (k1, k2, k3) we have (0,0,1), (1,1,0) or (3,0,0). Using (2) we obtain the corresponding values ofk0, namely m−1, m−2 orm−3. Therefore

b3=ma^m₀⁻¹a3+a^m₀⁻²a1a2

m m−2,1,1

+a^m₀⁻³a³₁ m

m−3,3

.

We now turn to the general case. One possibility for the ki’s is k0 =m−1,ks= 1 and ki = 0 for i 6= 0, s. This gives us the term ma^m₀⁻¹as. All other possibilities haveks = 0 which from (1) and (2) above gives us

k1+ 2k2+· · ·+ (s−1)ks−1 = s (3)

k0+k1+k2+· · ·+ks−1 = m. (4)

We thus obtain

bs=ma^m₀⁻¹as+X

a^k₀⁰a^k₁¹· · ·a^k_s^s₋⁻₁¹

m k0, k1, . . . , ks−1

,

where the sum is over the s-tuples (k0, k1, . . . , ks−1) satisifying equations (3) and (4) above. (In passing, notice that in all the terms in the sum the power of a0is strictly less thanm−1.)

Lemma 1 Ifbj∈Qfor allj, then ma^m₀⁻¹ai∈Qfor all i≥1.

proof We use an induction argument. As b1 = ma^m₀⁻¹a1, the statement is true for s = 1. We now suppose that the statement is true up to a givensand consider the cases+ 1. We have

bs+1=ma^m₀⁻¹as+1+X

a^k₀⁰a^k₁¹· · ·a^k_s^s

m k0, k1, . . . , ks

,

where

k1+ 2k2+· · ·+sks = s+ 1 (5)

k0+k1+k2+· · ·+ks = m. (6)

We can write

a^k₀⁰a^k₁¹· · ·a^k_s^s = a^k₀⁰(ma^m₀⁻¹a1)^k1· · ·(ma^m₀⁻¹as)^ks m^k1+···+ksa^(m₀ ^{−1)(k1+···+ks)}

= (ma^m₀⁻¹a1)^k1· · ·(ma^m₀⁻¹as)^ks m^k1+···+ksa^(m₀ ^{−1)(k1+···+ks−1)−k0}

= (ma^m₀⁻¹a1)^k1· · ·(ma^m₀⁻¹as)^ks m^k1+···+ksa^m(k₀ ^{1+···+ks−1)} .

Asa^m₀ =b0∈Qandma^m₀⁻¹ai∈Q, fori= 1, . . . , s, by hypothesis, we havea^k₀⁰a^k₁¹· · ·a^k_s^s ∈Q. However, bs+1∈Q, so ma^m₀⁻¹as+1∈Q. This finishes the induction step. ✷

We can now prove our first result on powers of polynomials.

Theorem 1 Let P(X)∈ C[X] and m ∈ N^∗ and suppose that Q(X) = P^m(X) ∈ Q[X]. If there is a coefficient asof P(X)such that as∈Q^∗, thenP(X)∈Q(X].

proof Let us first suppose thata0 6= 0. We will show thata0 ∈Q^∗. Ifs= 0, then there is nothing to prove, so let us suppose that s6= 0. From what we have just seen, if P(X) =Pn

i=0aiXⁱ, thena^m₀ ∈Q and ma^m₀⁻¹ai∈Qfori= 1, . . . , n. Now

a0

mas

= a^m₀

ma^m₀⁻¹as ∈Q=⇒a0∈Q^∗, 2

because as∈Q^∗. Now suppose thati6= 0, s. Then

ma^m₀⁻¹ai∈Q, a0∈Q^∗=⇒ai∈Q.

ThusP(X)∈Q[X].

To finish we consider the case where the coefficienta0= 0. Ifau is the first non-zero coefficient, then we may writeP(X) =X^uP1(X), withP1(X) =Pⁿ−u

i=0 ciXⁱ andci=ai+u. ThenP^m(X) =X^umP₁^m(X) and applying the above argument to P1(X) gives us the desired result. ✷

It is not difficult to generalize the result we have proved to a more general situation.

Theorem 2 Let k be a subfield of a field K and m∈N^∗. If the polynomial P(X)∈K[X] is such that Q(X) =P^m(X)∈k[X]and there is a coefficient of P(X), which belongs tok^∗, thenP(X)∈k[X].

proofIfkis of characteristic 0, or of characteristicp >0 andmis not a multiple ofp, then we may use an argument analogous to that used in the preceding theorem. Suppose now that k has characteristic p >0 andm=p^αm^′, withα≥1 and (p, m^′) = 1. First we notice that

P^m(X) = (P^pα(X))^m′ =P^m′(X^pα).

Hence P^m′(X^pα)∈k[X]. However, ifP(X) =Pn

i=0aiXⁱ, then P(X^pα) =

n

X

i=0

aiX^pαi=

np^α

X

j=0

cjX^j,

with

cj=nai forj=p^αi, i= 0, . . . , n

0 otherwise .

If we writeP1(X) =Pnp^α

j=0cjX^j, then P₁^m′(X)∈k[X] and one of the coefficientscj belongs tok^∗; also, m^′ is not a multiple ofp. We thus deduce thatP1(X)∈k[X]. This completes the proof. ✷ Corollary 1 Let kbe a subfield of a fieldK andm∈N^∗. If the polynomial P(X)∈K[X]is monic and such that Q(X) =P^m(X)∈k[X], thenP(X)∈k[X].

We might be tempted to generalize Corollary 1 to products of distinct polynomials, i.e., ifP1(X), . . . , Ps(X)∈ C[X], each having a coefficient inQ, and P1(X)· · ·Ps(X)∈Q[X], then Pi(X)∈Q[X] for all i. How- ever, it is easy to find a counterexample; for instance, if P1(X) = −i+X and P2(X) = i+X, then P1(X)P2(X) = 1 +X²∈Z[X]⊂Q[X].

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