ON A CERTAIN GROUP STRUCTURE FOR POLYNOMIALS

proceedings of the american mathematical

Volume 82, Number 2, June 1981

society

ON A CERTAIN GROUP STRUCTURE FOR POLYNOMIALS T. t. moh1 Abstract. A problem of C. C. Yang concerning level sets in complex analysis is generalized and algebraized to a problem about a group structure of polynomials. The group structure is established and several partial solutions of Yang's problem follow at once.

1. Introduction. In [1] C. C. Yang has raised the following interesting problem about level sets in complex analysis. Problem 1. Let p(z) and q(z) be two polynomials of the same degree over the complex numbers C such that

p(z)(p(z)

- 1) = 0«* q(z)(q(z) - 1) = 0.

Prove (or disprove) that either p =q or p + q = 1.

Let us complete the square. We have p(z)(p(z)-l)

= {p(z)-\)2-\.

In fact the above problem is to determine if the zero set of a function F = f2 — ¿ completely determines the function / up to ± sign. We may assume that the function F is monic and replace \ by any nonzero constant. Hence we have the following problem which is equivalent to Problem 1. Problem 1*. Let F(z) =/2 — a2, G(z) = g2 — ß2 be two nonconstant monic polynomials of the same degree with a, ß nonzero constants. Furthermore

F(z) = 0«»G(z) = 0. Prove (or disprove) that F(z) = G(z). We may raise the following "global" problem. Problem 2. Given a set S. Determine all polynomials F(z) = f2 — a2 with f monic, a nonzero constant and F(z) does not vanish outside S. An even more general problem follows. Problem 3. Given a set S. Determine the algebraic structure of the set of polynomials F(z) = f2 — a2 with f monic, a a polynomial (/, a) = 1, deg / — deg a > a constant and F(z) does not vanish outside S.

We give the following definition.

Received by the editors August 6, 1978 and, in revised form, May 1, 1980. 1980 Mathematics Subject Classification. Primary 12D05, 12D10 Key words and phrases. Level set, generator, rank, abelian group, Hurwitz formula.

'This work was partially supported by the National Science Foundation under grant no. 0029-501395, at Purdue University. © 1981 American

Mathematical

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0002-9939/81 /0000-02 S4/$02.2 S

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184

T. T. MOH

Definition. Let A- be a field of characteristic =7*= 2. Let F = f2 — a2 with /, a G k[x] and (/, a) = 1. Let Fa denote the pair (F, a). Let « > 1 and P„ = {Fa: F and/monic, deg/ — deg a > «}. Remark. In the above definition Fa ^ F^ if a ¥= 0. Definition. Let S be a set. Let P„(S) = {Fa G P„: Fhas no zeros outside S). Definition. Let Fa = f2 - a2, Gß = g2 - ß2. Let

h = (fg + aß, Bf + ag) be monic, then we shall define * as follows

* Gß = j(fg + aß)^

\(ßf+ag)

Our main theorem is the following. Theorem. The group (Pn, *) is the torsion free abelian group. Given any set S, P„(S) is a subgroup of Pn. Furthermore P„(S) = {e) if card(S) < « and P„(S) is of rank < card(S) — « deg a = m. It follows from the Hurwitz formula that [2]

= «i + « =

-2 = -2(« + «I) + 2(e(p)-l) where p runs through all places of k(x) and e(p) is the ramification index of p.

We have e(p) = « if p = (x = oo) and 2(e(p) — 1) = 2(« + m) — d for p centered at a/f = ±1 with d the total number of district roots for/± a. Hence we have d > n + 1.

Lemma 6. Let Fa: F = f2 - a2 and Gß: G = g2 - ß2 G P„(S). Let b G S and let F and G have zeros of multiplicities m and I, respectively, at b. Suppose m < I. Then Fa * Gß and F_„ * Gß have zeros of mutiplicities m + I and I — mat b in some order.

Proof.

We assume without loss of generality that (x — b) is a factor of g — ß.

Then ß2F-

a2G = (ßf+

ag)(ßf - ag).

Suppose x — b is a factor of / — a. Then x — b is a factor of ßf — ag with multiplicity > min(«i, /) with equality if m ¥= I. Hence

(x - b)2m\(F G,(ßf-

ag)2)

and

(* - b)2m+x\(F-G,(ßf-

ag)2).

Let F_a * Gß = Hx = h2 —A2. Then 77 has a zero of multiplicity «i — / at b. Since x — b is a factor of ßf — ag, it cannot be a factor of / + g. Otherwise x — b would

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186

T. T. MOH

be a factor of 2ßf and 2ag. If it is a factor of F (i.e. »i > 0) then it cannot be a factor of either/ or a. Hence it must be a factor of g and ß, a contradiction. We

conclude that

(x-b)\(F-G,(ßf+ag)). Let Fa * Gß = í/e: U = u2 - e2. Then t/has multiplicity m + I at b. Suppose x — b is a factor f + a. Then replace a by -a in the above argument.

Q.E.D. Lemma 7. Suppose Fa: F = /2 — a2 G P„(.S) wAere F «as a zero of minimal positive multiplicity m at b G S. Let Gß: G = g2 - ß2 G Fn(S') vwï« multiplicity I at b S S. Then l = mp for some integer p. Furthermore either (Fp) * G or (Fp)(_cr) * G does not vanish at b.

Proof. Follows from Lemma 6 by Euclidean algorithm. Proof of the last part of Theorem 1. If there is no element in P„(S) with positive multiplicity at b G S then Pn(S) = P„(S \ {b}). The proof is completed by induction on the cardinality of S. Otherwise choose Fa: F = f2 — a2 with smallest possible positive multiplicity. Then P„(S) is generated by Fa and Pn(S \ {b}).

3. Discussion. Proposition 1. Let Fa G F„. If F = G' as ordinary power with G monic, then Gß G Pn for some ß.

Proof. Let F = f2 - a2 = (/ - a)(f + a) = Gl. Since (/ - a,f + a) = 1, then we have / - a = h[,

f+a

= ti2,

with («j, «2) = 1 and «„ «2 are monic. We have 2a = U(h2 - wlhx)

where w is a primitive /th root of unity. Let

deg «, = deg h2 = c,

deg(«2 - «,) = e.

Then we have deg(«, + h2) = c, deg/ = le,

deg a = (I — l)c 4- e,

n = le - [(/ - l)c + e] = c - e,

and G-M2-[!(*,

+ h2)]2-[\(hx

-h2)}2.

Choose/? =5(A, - «2). For Yang's problem the following Propositions 2 and 3 establish partial solutions. Proposition 2. If card S = n + 1, /Ae« Fn(5) « a çvch'c grot^p 0/ o«e generator. Hence the answer to Problem 1 in this case is affirmative.

Proof.

It follows trivially from our theorem.

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group structure

for polynomials

187

Proposition 3. If card 5 = 2«— 1, then there is at most one polynomial F with deg F = 2«, F vanishes and only vanishes at every point of S and Fa G P„(S) for some a.

Proof. Let G: G = g2 - ß2 be another one. Then Fa * Gß ¥* {e) = F_^ • Gß, since (ßf + ag, ßf — ag, FG) = 1 and we have

FG = (fg+aß)2-(ßf+ag)2 = (fg-otß)2-(ßf-ag)2, ß2F-a2G

= (ßf+

ag)(ßf - ag).

Suppose S = {bx,. . ., b2n_x) with (x — bx)F = (x — b^G. Then at least one of the two factors ßf + ag and ß — ag must have at least « — 1 of the remaining points as roots. Thus either Fa * Gß or F_„ * Gß will have at most « district roots which contradicts Lemma 5. We shall complete our article by listing several examples.

Example 1. The rank of PX(S) = card(S) - 1. Let S = {bx,. .., b„). Then FBl=(xbxXx - b,) = [xx2(bx+ b,)\2 - [\(bx - bid)]2 for i = 2, . . . , « clearly generate PX(S). Example 2. Let

S = {-I, 0, 1, f }, F„: x2(x -§)(* - l)2(x 4-1) = (x3 -\x2 +\f - (ff and

Gß:x(x-l)(x-l)(x+2-)

= (x>-x-l)2-(tf.

Then Fa and Gß generate P2(S) with rank P2(S) = 2.

References 1. Complex analysis, Proceedings of the SUNY Brockport Conference, Dekker, New York and Basel,

1978,p. 169. 2. W. Fulton, Algebraic curves, Benjamin, New York, 1969.

Department of Mathematics, Purdue University, West Lafayette, Indiana 47907

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