Composite rational functions which are powers

Proceedings

of the Royal

Society of Edinbu.gh.

83A. 11-16.

1979

Composite rational functions which are powers * S. D. Cohen Department of Mathematics, University of Glasgow (Communicated by Professor R. A. Rankin) (MS received 2 May 1978. Read 30 October 1978)

SYNOPSIS

A description is given of the rational functions A (X), B (X) over a field n for which A (B (X» is an nth power in n(X).

1.

INTRODUCTION

An investigation concerning the extent to which the values of a rational function in a finite field determine the function has led the author to consider the following problem in the cases n = 2,3. Let n be a field of characteristic w ~ 0 and A, B rational functions in n(X). Given a positive integer n, can A(B) be an nth power in n(X) without A itself being an nth power? There is an obvious instance of this. Put A(X)=X'Aln(x), n ,r r, and B(X) = X". Then A (B(X)) =xmAln(xn). A less obvious example occurs when n=2 and w~2. Let 1'(~O)En and put A(X)=X2-1', B(X)= (X2+1')/2X. Then A(B(X)) = [(r-1')/2XJ2, (In fact, if l' is a square in n, this can be reduced to an example of the first case by a linear transformation.) If w > 0, there is another possibility. For example, if n is a perfect field and A is in n(X), then there exists A * in n(X) for which A(XW) = (A "(Xj)". We shall show that the above examples are essentially the only ones. In fact, the result when n is the complex field and A and B are polynomials follows from a more general theorem of Ritt [7] (to whose work a referee has kindly drawn the author's attention). Specifically, Ritt showed first, using the theory of Riemann surfaces, that, if f has functional decomposition f = f1 f2 o ..• fro where f1, ... , t. are indecomposable polynomials, then r and the degrees of the t. (in some order) are uniquely determined. Algebraic proofs of this fact for n a general field of characteristic 0 were given by Engstrom [2J and Fried and MacRae [5J (whose proof also applies in non-zero characteristic w provided w > deg f). Actually, in the case considered, Ritt took his analysis much further, thereby providing a full description of those polynomials f which possess distinct decompositions into indecomposables. An examination of this argument reveals that, at several points, the assumption that the t. are polynomials is crucial. In a preliminary investigation [8J into the situation when the t. are rational functions, Ritt admitted the difficultiesin obtaining any sort of complete classification. More 0

0

* This paper was assisted in publication by a grant from the Carnegie Trust for the Universities of Scotland.

S. D. Cohen

12

recent studies by Fried [e.g. 3, 4] seem to confirm this view. [For other related work see 1 and 9]. Accordingly, the self-contained elementary account which follows of the special case when f is an nth power may be of general interest. In the application which led to the present considerations, the author definitely requires f to be a rational function and 0 is a finite field (so is neither algebraically closed nor of characteristic 0), in which situation Ritt's result is inadequate. A brief description is given in §3. 2.

MAIN RESULT

For convenience we assume in the first place that the functions involved do not belong to O(XW). Indeed, if we call A separable if A(X)EO(X)-O(XW), then A is separable if and only if the formal derivative A'(X) is non-zero. Of course, if A(B) is separable and A(B) = C" then A, Band C are separable and w ,r n. We let L denote a non-singular, linear fractional transformation of 0 and prove the following result.

Let n be a positive integer with w ,r n and A, B rational functions in separable. Then A(B(X)) is an nth power in O(X) if and only if one of the following holds: (i) there exists an integer rand L, Al' R in O(X) such that THEOREM.

O(X) with A(B)

A(L(X))

=X'A~(X),

B =L(Rm),

where m = n/(r, n), (ii) n is even, w oF- 2 and there exists a non-square such that

y

in 0 and L, Al, R in O(X)

Proof. The proof is in three stages. Stage I. n a prime (oF- w). Put B = B1/B2 where B1 and B2 are eo-prime polynomials with degrees b1 and b2. Assume »«> b2. (Otherwise consider A(L -1) and L(B) for a suitable L.) Next, by replacing A and B by A(~-l) and oB for some o(:f: 0) in 0, we may assume that B is monic (i.e. is the quotient of monic polynomials). Moreover, we may then assume that A is monic, for if A = OAb where A1 is rnonic, then oA1(B) = (eCt, say, where C is monic. Hence 0 = en and we need only consider Al(B) = C", As a further simplification, by multiplying A by a suitable monic nth power, we may take A to be a monic n-free polynomial of degree a, say, such that, in fact, the multiplicities of all its irreducible factors lie between 1 and n -1 (inclusive). Now it is easy to see that B~ is an nth power and that (1) . where C is a polynomial eo-prime with Bz. Put n1 = n or 1 according to whether Bz is an nth power or not; then (2)

Composite Rational Functions

which are Powers

13

and, using , to denote a formal derivative with respect to X, we have

I (B2B1'

B21-1In,

(3)

- B1B2').

Now let Q be the monic square-free polynomial of degree q, say, obtained by taking the product of the distinct irreducible factors of A. In particular, if n = 2 then Q = A. Since n is prime, it follows from (1) that for some monic polynomial D of degree d, we have BiQ(B)

where (B2, D) = since b1 > b2 then

= D",

(4)

1 and, since Q is square-free (Bi-1Q'(B), d

D) = 1. Moreover,

= qb1/n.

(5)

Differentiating (4) formally and using (4) again, we obtain Bi-1Q'(B)(B2B1'

- B1B2' +qB2'Dn

But Q' and D' are non-zero, w~ nand D 1 I (B2B1' - B1B2') (~O). Indeed

=

(B2, D)

nB2Dn-1D'. =

(6)

1. Hence (6) implies that

n

-

B~-1In'D"-1I (B2B/

by (3), since (B2, D)

=

- B1B2'),

(7)

1. Comparing degrees in (7) we obtain

(n -l)d

~ b1 + (b~n1)-1

< b1(1 +nl1).

Thus, by (5), we see that q satisfies q

< (1 + nl1)/(1-

n-l).

(8)

It follows from (8) and (2) (and from q = a if n = 2) that q ~ 2 with equality possible only if n = 2 or if n > 2 and n I a. If, in fact, q = 1 then for some (polynomial) L, A (L(X» = X', 1~ r ~ n -1 and (i) must hold. Suppose q = 2. The first alternative is that for some linear polynomials L1 and L2 we have A = L~L~-', where 1~ r~ n -1. Thus A = (L1/L2YL~ and again (i) holds. The remaining possibility is that Q is an irreducible quadratic and so, since n is prime, n = 2 and A = Q. In this case, for some polynomial L, we have A (L) = X2 - "I, where "I is a non-square in fl. Replacing A and B by A(L) and L(B) we obtain from (1) B1+JYB2

= «(31+.j:r(32)(R1

+JYR2f,

say, where (31,(32Efl satisfy (3i-'Y(3~= 1 and RI' R2Efl[X]. Hence B = where R=R1/R2 and L1(X) = «(31X+'Y(32)/«(32X+(31)· = (X~ )/«(32X + (31)2and so (ii) holds.

L1«R2+'Y)/2R) Moreover, A (L1)

"I

Stage H. fl algebraically closed.

In this case, (ii) cannot hold. The proof is by induction on the number of (not necessarily distinct) primes in n. So assume that n is composite with least prime divisor p. Put q = nip> 1. Clearly we may also assume that the multiplicities of the irreducible factors appearing in the numerator or denominator of A are relatively prime. Then, by the induction hypothesis, there exists an integer r with (r, q1 = 1 and L, Al and R in fleX), with X absent from (the prime decomposition of) Ab such that A (L(X)=

X'Ai(X)

,

B=L(Rq).

(9)

14

S. D. Cohen

Put (10) Then Ao(R) is a pth power in n(X). Accordingly, there exists an integer s not divisible by p and LI' A2' S in n(x) with X absent from A2 for which (11) We now compare (10) and (11) and distinguish two cases. First suppose LI(X) = «X or ol X, where a(,eO)En. Then R=Ri where Rl=(a-l/PS)±l. Moreover, Al = A~, say, where, by (9), p ,(' r. It follows from (9) and (10) that A(L(X))

=X'A~(X),

B

= L(R7)

and (i) holds. Alternatively, if LI(X) ¥ aX or «[X, then Al is not a pth power, for if it were, then so would LHX)/X' which is not so since p ,r s. Write A, = P'A4, where P is an irreducible polynomial present in Al but not in A4 and t is an integer not divisible by p. Certainly, P(X) ¥ x. By (10) and (11), the irreducible factors of P(xq) all appear in L1(X) which (since q> 1) is impossible unless q = 2 and X does not appear in LI. Now certainly this would mean that r is odd. However, p was defined as the least prime dividing n so that, in fact, we must have p = 2 which implies that in (11), X appears to an even power contradicting the fact that r is odd. Stage Ill. n, n arbitrary. Again assume that the multiplicities of the irreducible factors of A are eo-prime. Let n denote an algebraic closure of n. By Il, we can suppose there exists an r with (r, n) = 1 and L, A, R in n(X) with X absent from Al such that A(X)

= L'(X)A~(L(X)),

B = L -l(Rn).

(12)

Suppose L = 5Lb R = SRl (5, s (¥ 0) En), where LI and RI are manic. Since n is algebraically closed we can replace L by L" R by R, and Al by Ai, where A'fn=yi'A;t(TJ), TJ =Sle", and assume that, in (12), Land R are manic. We distinguish two cases. Case (a). L(X) =X +a(a En). Let (T be any automorphism in the Galois group of n over n. Applying (T to (12) we get A (X)

= (T(A(X)) = (X + (T(a))'A7(X + (T(a)).

(13)

It follows that (T(a) = a, for otherwise (X +a) appears to the rth power in (12) but to a power divisible by n in (13), which is impossible since (r, n) = 1. Hence a En and, in (12), R" En(x) which implies that RE n(X), since R is manic. Thus (i) holds. Case (b). L(X)=(X+a)/(X+{3) (a,{3En). Here X +a and X + (3 are not both present in AI(L). Suppose that, in fact X + a is absent. Apply o to (12) as in case (a). We obtain A (X)

= (X + (T(a))'(X

+(T({3))-'A~«X +a-(a))'(X +a-{(3))-').

(14)

Comparing (12) and (14) we see, as in case (a), that either (T fixes a and {3or interchanges them. If the former is always the case, then (i) holds as in case (a). If

Composite Rational

Functions which are Powers

15

the alternative, that for some (T, (T(a) = (3 and (T«(3) = a, is valid, then Q(X) = (X +a)(X + (3) is an irreducible quadratic in O[X]. Hence A1(L(X))/(X + (3)2E O(X). But X +a is absent from Al(L). Hence A1(L) contains X +(3 to precisely the power 2 and so n must equal 2 and A (X)

= Q(X)A~(X)

in O,(X). Thus, by the argument at the end of stage I, we conclude that (ii) holds. This completes the proof. Next, we consider briefly the modifications necessary when A(B) is inseparable. Suppose w>O and for any polynomial R(X)=L,PiXi in O[X], put R(r)(x)= L,PiwXi. Extend this definition to R=R1/R2 in O(X) by R(r)=Rr)IR~). That nothing particularly new arises in the inseparable case can be seen from the following lemma. LEMMA. Suppose that A (B(X)) = CW"(XW'), where rand s are non-negative integers and C in o'(X) is separable. Then there exist non-negative integers a, b with a + b = r + s for which

Proof. Since CW"(XW')

=

c(r)(xw"+'),

it suffices to show that, if C is separable

and A(B)

= C(XW),

(15)

then either A or B is separable. This is essentially a standard result; in any case formal differentiation of (15) yields A'(B)B' = 0 and the result follows. Accordingly, if A(B) = CW"", where w.{ n, put C = Cl(XW'), where Cl is separable and apply the Lemma. The situation is then reduced to one similar to that of the Theorem (particularly when 0 is perfect) and a solution can be obtained. We omit the details. 3. REMARKS In some circumstances the hypothesis of the Theorem that A(B) be an nth power in O(X) may be weakened. For example, suppose 0 is a global field containing all nth roots of unity. Then theorems of Grosch and Ribenboim [6] may be applied. In particular, if 0 is an algebraic number field, then it suffices to assume that A(B(m)) is an nth power in 0 for almost all rational integers m. Similarly, if 0, is a finite separable extension of a simple transcendental extension F(t) of a finite field F, then it suffices to assume that A(B(m(t))) is an nth power for almost all polynomials m(t) in F[t]. Finally, we give an application of the Theorem. Let f = fl/f2' g be rational functions in o'(X) with deg j'(s= max Ideg f., degf2))=m. Let Df(y)(EO(y)) denote the discriminant of fl(X)- Yf2(X) (as a polynomial in X). Suppose that the Galois group of fl(X)-g(y)f2(X) over O(y) is a subgroup of the alternating group Am. Then Df(g(Y)) is a square in O(Y). We deduce from the Theorem that, in fact, Df(Y) is a perfect square apart from a factor of degree at most 2, which yields important information about f. Moreover, if Df(Y) itself is not a

16

1

S. D. Cohen

square, the shape of g is immediate. Specifically,suppose that n is a (large) finite field and that f and g have bounded degree and are such that the values of f in n include those of g. It turns out that, if deg f ~4, then essentially the situation described above prevails and the general shape of f and g can be deduced. Using also the case n = 3, the author has obtained a complete classification of such pairs (f, g) with deg f ~ 4. A full account of this work is in preparation. REFERENCES t

1 H. D. Block and H. P. Thielman. Commutative polynomials. Quart. J. Math. Oxford Ser. 2 (1951), 241-243. 2 H. T. Engstrom, Polynomial substitutions. Amer. J. Math. 63 (1941), 249-255. 3 M. Fried. Arithmetical properties of function fields (m. Acta Arith. 25 (1974), 225-258. 4 M. Fried. On a theorem of Ritt and related Diophantine problems. J. Reine Angew. Math. 264 (1973), 40-55. 5 M. D. Fried and R. E. MacRae. On the invariance of chains of fields. Rlinois J. Math. 13 (1969), 165-171. 6 P. Ribenboim. Polynomials whose values are powers. J. Reine Angew. Math. 268/269 (1974), 34-40. 7 J. R. Ritt. Prime and compositive polynomials. Trans. Amer. Math. Soc. 23 (1922), 51--{j6. 8 J. F. Ritt. Permutable rational functions. TraIlS. Amer. Math. Soc. 25 (1923), 399-448. 9 A. G. Walker. Commutative functions. Quart. J. Math. Oxford Ser. 17 (1946), 65-92. (Issued 12 June 1979)

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