Mathematik.uni-wuerzburg.de

Contemporary Mathematics
Volume 00, 0000
Primitive Monodromy Groups of Polynomials
Abstract. For a polynomial f ∈ C[X], let G be the Galois group of theGalois closure of the field extension C(X)|C(f (X)). We classify the groupsG in the indecomposable case. For polynomials with rational coefficientsthere are, besides four infinite series, only three more “sporadic” exam-ples. In the Appendix we reprove the classical Theorems of Ritt aboutdecompositions of polynomials using the group-theoretic setup.
1. Introduction
Let f be a polynomial of degree n with complex coefficients. In a fixed alge- braic closure of the field of rational functions C(t) consider the field Ω which isgenerated over C by the n different elements xi fulfilling f (xi) = t. Then theGalois group G = Gal(Ω|C(t)) permutes transitively the elements xi. This groupG is called the monodromy group of f . It is natural to ask what groups G canoccur this way. A polynomial is called indecomposable if it cannot be written as acomposition of two non–linear polynomials. In section 2 we classify the possiblemonodromy groups for indecomposable polynomials, there are four infinite seriesand twelve more cases which do not belong to these series. Section 3 is aboutthe question of what groups occur as monodromy groups of polynomials withrational coefficients. The result is Theorem. Let f ∈ Q[X] be indecomposable and let G be its monodromy Then G is either alternating, symmetric, cyclic, or dihedral or G is PGL2(5), PΓL2(8), or PΓL2(9). In the latter three cases f is, up to compositionwith linear polynomials, uniquely given by X4(X2 + 6X + 25), 9X9 + 108X7 +72X6+486X5+504X4+1228X3+888X2+1369X or (X2−405)4(X2+50X+945). Some remarks about the appearance of monodromy groups are in order: Es- pecially M. Fried (see [7], [8], [9], and his papers cited in [11]) exhibited the
1991 Mathematics Subject Classification. Primary 11C08, 20B15, 20B20; Secondary 12E05.
Financial support from the DAAD is greatfully acknowledged.
This paper is in final form, and no version of it will be submitted for publication elsewhere.
importance of these groups in discussing several arithmetical questions about
polynomials. That is many questions depend merely on the monodromy group
rather than on the full information given by a polynomial. One of these problems
is a question of Davenport, which asks to classify the pairs of polynomials with
integer coefficients such that the value sets on Z are the same modulo all but
finitely many primes. See [11, 19.6], [9], and [20].
We merely classify the monodromy groups in the indecomposable case. By a Theorem of Ritt (see [21] or [3]) the study of arbitrary polynomials can be
reduced to these polynomials to some extent. For instance (over fields of char-
acteristic 0) any two decompositions of a polynomial into indecomposable poly-
nomials have the same number of factors, and the degrees are the same up to
a permutation. Ritt even gives an algorithm how to pass from one composi-
tion to the other one by interchanging and “twisting” consecutive factors. In
the Appendix, we give a concise account of this, employing the group-theoretic
setup.
olklein for drawing my attention to this question. I thank B. H. Matzat for informing me about [17] where he already computed
the polynomials for the groups PΓL2(8) and PΓL2(9). He also noted that I
erroneously excluded PΓL2(8) in an earlier version of this paper.
2. Primitive Monodromy Groups
2.1 Notation and Definitions. We retain the notation from the Introduc-
tion. For technical reasons we need a further description of the monodromygroup of f ∈ C[X] (deg f = n): Consider the branched n-fold covering f : P1 → P1. Let S = {p1, p2, . . . , pr} be the set of branch points, where pr is the point at infinity. Fix a point a inP1 \ S. Then π1 = π1(P1 \ S, a) acts on f−1(a) by lifting of paths. The homo-morphic image of π 1 in Sn = Sym(f −1(a)) will also be denoted by G, as this group can be identified with the monodromy group defined in the Introduction,with G acting in the same way on the elements xi as on the points of the fiberf −1(a). This identification relies on the isomorphism between the group of cov-ering transformations of a Galois cover of compact connected Riemann surfacesand the Galois group of the corresponding extension of fields of meromorphicfunctions on these surfaces.
In this section we use the geometric description of G from above, i.e. we view G as a subgroup of Sn via identification of {1, 2, . . . , n} with f −1(a).
Pick r generators λi of π1(P1 \ S, a) such that λi winds only around pi and λ1λ2 · · · λr = 1 (this is a so–called “standard homotopy basis”). These r gener-ators of π1(P1 \ S, a) then yield generators σ1, σ2, . . . , σr of G with σ1σ2 · · · σr = 1 . As pr = ∞ ramifies completely, σr is an n-cycle.
PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS This tuple (σ1, . . . , σr) is called the branch cycle description of the cover For σ ∈ Sn denote by ind σ the quantity ‘n − the number of orbits of σ ’.
The main constraint is imposed by the Riemann Hurwitz genus formula ind σi = 2(n − 1) . For an elementary argument yielding this latter relation confer [7, Lemma 5].
Conversely, a finite permutation group having a set of generators fulfilling the
above restrictions is the monodromy group of a suitable polynomial by Riemann’s
existence Theorem.
So we are reduced to a completely group–theoretic question. The purpose of this section is to give a complete classification in the indecomposable case. The
corresponding question for rational functions instead of polynomials is much
tougher and still open, see [15] and [1].
2.2 Notations and Main Result. Let G be a permutation group acting
on n elements. We consider the following condition on G, which we later referto as (*).
Condition (*). G is generated by σ1, σ2, . . . , σs (σi = 1) such that σ1σ2 · · · σs It is obvious that the situation in (*) is equivalent to the configuration of 2.1.
Suppose (*) holds. Using ab = bab we see that we may assume |σ1| ≤ |σ2| ≤ · · · ≤ |σs|, where |σ| denotes the order of σ. Following Feit in [4] we say that G
is of type (|σ1|, |σ2|, · · · , |σs| : n).
Denote by Cp and Dp the cyclic and dihedral groups of degree p, respectively.
Let PGLk(q) be the projective linear group over the field with q elements, act-ing on the projective space of dimension k − 1. This group, together with thecomponent–wise action of Aut(Fq) on the projective space, generates the semi-linear group PΓLk(q). The Mathieu groups of degree n are labelled by Mn. Inthis section we prove Theorem. Let G be the monodromy group of a polynomial f ∈ C[X]. Then G is one of the following groups. Conversely, each of these groups occurs. (i) Cp of type (p : p), p a prime. Dp of type (2, 2 : p), p an odd prime. (ii) PSL2(11) of type (2, 3 : 11) PGL3(2) of types (2, 3 : 7), (2, 4 : 7), and (2, 2, 2 : 7)PGL3(3) of types (2, 3 : 13), (2, 4 : 13), (2, 6, 13), and (2, 2, 2 : 13)PGL4(2) of types (2, 4 : 15), (2, 6 : 15), and (2, 2, 2 : 15)PΓL3(4) of type (2, 4 : 21)PGL5(2) of type (2, 4 : 31) (iii) An (n odd) and Sn of many, not reasonably classifiable types. M11 of type (2, 4 : 11)M23 of type (2, 4 : 23)PGL2(5) of type (2, 4 : 6)PGL2(7) of type (2, 3 : 8)PΓL2(8) of types (2, 3 : 9) and (3, 3 : 9)PΓL2(9) of type (2, 4 : 10) Remark. In [19, 2.6.10] we get, as a side product to the Hilbert–Siegel prob-
lem, a classification of the monodromy groups of the rational functions f (X)/Xwhere f is an arbitrary polynomial f ∈ C[X] with f (0) = 0. The list is asfollows: AGL1(p) with p ∈ {2, 3, 5, 7}, AΓL1(8), AGL3(2), AΓL2(4), AGL4(2),AGL5(2), AΓL1(9), AGL2(3), An (n even), Sn, PSL2(5), PGL2(5), PSL2(7),PGL2(7), PSL2(13), M11 with n = 12, M12, M24.
There are also results about monodromy groups of indecomposable polynomi- als with coefficients in a finite field or in an algebraically closed field of positive
characteristic. In [14] is a classification of the primitive groups which meet a
necessary condition for being the monodromy group of a polynomial. The main
constraint comes from the ramification at infinity. The genus condition however
is hardly to use.
2.3 About the Proof. Let f ∈ C[X] be a polynomial and G its monodromy
uroth’s Theorem, f is indecomposable if and only if G is a primitive group (i.e. G does not act on a non–trivial partition of the
underlying set), see [12, 3.4].
Now let f be indecomposable and σ1, . . . , σs a generating system of G fulfilling (*). Set Z = σ1σ2 · · · σs , then Z is a transitive cyclic subgroup of G. Together
with the primitivity of G, we get by classical results of Schur and Burnside (see
[22, 11.7 and 25.2]) that G ≤ AGL1(p) (p a prime) or G is doubly transitive.
Let A be a minimal normal subgroup of G. If A is elementary abelian, then G ≤ AGL1(p) or G ≤ S4, see the proof of [16, Satz 5]. The non–solvable doubly
transitive groups with a cyclic transitive subgroup are known by the classification
of the finite simple groups and listed in [5, 4.1]. Thus we have to investigate
the following groups G.
Cp ≤ G ≤ AGL1(p), An (n odd), Sn, M11, M23, PSL2(11) of degree11, and (not all) groups between PSLm(q) and PΓLm(q) (with m ≥ 2,q a prime power) in its action on the projective space.
The cases G = PSL2(11) with n = 11 and the semi–projective linear groups with m ≥ 3 have been investigated by Feit in [4]. (These are just the cases
where G admits two inequivalent doubly transitive representations with Z acting
transitively in both of them.) However, his proof needs to be modified, as [4,
3.4] is wrong. In the following we supply an alternative treatment in the case
when [4, 3.4] does not work. I thank Feit for a discussion about this.
PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS We will discuss the different classes of groups separately. The case Cp ≤ G ≤ AGL1(p) is quite easy and left to the reader.
2.4 Counting Orbits. Some more notation: For σ ∈ G denote by o(σ)
the number of cycles of σ (thus n = o(σ) + ind σ). Let f (σ) be the number ofelements fixed by σ. For later use we derive an elementary relation between o(σ)and the number of fixed points of powers of σ: For this denote by ai the numberof i-cycles of σ. Clearly µ( )f (σk) . µ(t) = ϕ(m) and o(σ) = a 1 + a2 + . . . we get the basic relation 2.5 The Mathieu Groups. For the two candidates M11 and M23 we use
the Atlas of the finite simple groups [2] and its notation. Besides other things
the character tables in this source allow us to compute the ind–function. Let
us start with G = M11: First we get ind σi ≥ 4, hence s = 2 by (*). We see
that σ1 ∈ 2A and σ2 ∈ {3A, 4A}. Let χ1, χ2,. . . ,χh be the irreducible characters
of G and C1, C2,. . . ,Ch be the conjugacy classes of G. For an x ∈ Ck denote
by N (i, j; k) the number of solutions of x = uv with u ∈ Ci and v ∈ Cj. It is
well–known (see e.g. [13, 4.2.12]) that
χm(Ci)χm(Cj)χm(Ck) We want to exclude the case σ2 ∈ 3A: Pick an element g ∈ G of order 11. Using
(2) we see that there are exactly 11 solutions of g = uv with u ∈ 2A, v ∈ 3A.
Since g is transitive and abelian g does not centralize u (and v). Thus g acts
fixed–point–freely on the pairs (u, v) with g = uv. So there is essentially one
solution to g = uv with u ∈ 2A and v ∈ 3A. Now [2] tells us that G contains
a transitive subgroup isomorphic to H ∼
= PSL2(11). Again using (2) and [2] we
see that there are elements g, u, and v in H of orders 11, 2, and 3 respectivelywith g = uv. The previous consideration shows that a conjugate of σ1, σ2 is asubgroup of H, therefore σ1 and σ2 do not generate G.
Now consider the case σ2 ∈ 4A. Set H = σ1, σ2 . If H = G then H ≤ PSL2(11), as PSL2(11) is the only maximal and transitive subgroup of M11 by
[2]. However, PSL2(11) does not contain an element of order 4. Thus σ1 and σ2
generate G. An explicit example is
σ1 = (4, 5)(6, 7)(8, 9)(11, 11), σ2 = (1, 11, 2, 9)(8, 3, 5, 7).
We treat the case G = M23 quite similarly: Here we get σ1 ∈ 2A, σ2 ∈ 4A, and H = σ1, σ2 with σ1σ2 an 23-cycle. The only transitive and maximal subgroup
of G has order 253, see [2]. Thus σ1 and σ2 generate G. Again we give one
explicit example:
σ1 = (8, 9)(10, 11)(12, 13)(14, 15)(16, 17)(18, 19)(20, 21)(22, 23),
σ2 = (2, 14)(17, 19)(1, 10, 16, 4)(8, 13, 15, 6)(12, 20, 11, 7)(3, 9, 5, 22)
2.6 PSL2(q) ≤ G ≤ PGL2(q). The distinction between the projective case
and the non–projective semilinear case simplifies the somewhat tedious waythrough the estimations. We assume q ≥ 5 (because PGL2(4) ∼ For the case q = 11 and G of degree 11 see [4, 4.3]. Thus assume from now
on that G acts naturally on the projective line.
Pick a σ ∈ G. There are three cases: (i) σ has at least 2 fixed points. Then ind σ = (q − 1)(1 − 1 (ii) σ has exactly one fixed point. Then ind σ = q(1 − 1 |σ| ) and |σ| = p.
(iii) σ has no fixed points. Denote by ˆ σ a preimage of σ in GL2(q). Then ˆ σ] is a quadratic field extension of Fq by Schur’s Lemma. We deduce that σ acts fixed–point–freely on P1(q),
thus ind σ = (q + 1)(1 − 1
In all three cases we obtain ind σ ≥ (q − 1)(1 − 1 s = 2 by (*). As G contains the non–solvable group PSL2(q), σ2 cannot be aninvolution (recall the monotony of the orders of σi). This shows (again using(*)) (q − 1)(1 − 1 + 1 − 1 ) ≤ q, hence q ≤ 7.
Suppose q = 5. We have ind σ ∈ {2, 3}, ind σ = 4, or ind σ ∈ {3, 4} if σ is of type (i), type (ii), or type (iii) respectively. Thus both σ1 and σ2 are of type (i)with |σ1| = 2, |σ2| = 4. One readily checks (see the arguments in the Mathieugroup case) that σ1 and σ2 generate a group containing PSL2(5). Since σ2 isan odd permutation, they actually generate PGL2(5). An explicit example isσ1 = (1, 2)(3, 4), σ2 = (3, 2, 5, 6).
Now suppose q = 7. Similarly as above we get |σ1| = 2, |σ2| = 3, and f (σ1) = f (σ2) = 2. So σ1 is an odd permutation, hence G = PGL2(7). This case occursas well, an example is provided by σ1 = (1, 2)(3, 4)(5, 6), σ2 = (1, 3, 6)(2, 7, 8).
2.7 PSLm(q) ≤ G ≤ PΓLm(q). Set q = pe with a prime p. This case is even
for m = 2 more complicated than the projective linear case since there are manymore types of possible cycle decompositions. In particular a fixed–point–freeelement need not generate a fixed–point–free group.
Write ΓLm(q) = GLm(q) Γ with Γ = Aut(Fq). Let ΓLm(q) act from the left σ a preimage of σ in ΓLm(q). Feit [6] told
me a special case of the following lemma.
σ = gγ with g ∈ GLm(q) and γ ∈ Γ. Let e/i be the order of PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS γ. Then f (σ) ≤ pim−1 . σe/i = h ∈ GLm(q). If ˆ σv = αv for v ∈ Fm \ {0}, α ∈ F hv = N (α)v, where N : Fq → Fpi denotes the norm.
First suppose that h is a scalar. If ˆ σw = βw for some w ∈ Fm \ {0}, β ∈ F then hw = N (β)w and therefore N (α) = N (β). By Hilbert’s Theorem 90 thereis a ζ ∈ F× with α/β = ζγ/ζ. From ˆ σw = ζγ βw = αζw we conclude σ–invariant Fq–line L contains an element u = 0 with ˆ are exactly pi − 1 such points on L. One easily sees that a basis of the Fpi–space {v ∈ Fm| ˆσv = αv} is linearly independent over F q . This proves the assertion.
Now suppose that h is not a scalar.
Let η1, . . . , ηs be those elements in Fpi which are eigenvalues of h. Let Vk be the eigenspace of h with eigenvalueηk. Note that h|V is the smallest power of ˆ σ–invariant line lies in one of these subspaces. The From (xδ1 − 1) + (xδ2 − 1) ≤ xδ1+δ2 − 1 for x ≥ 1 and δ1, δ2 ≥ 0 and σ = gγ with g ∈ GLm(q) and γ ∈ Γ. Let e/i be the order of γ. Set r = 1 if e = i. Otherwise let r be the smallest prime divisor of e/i.
Then σ| )(qm−1 − 1) + (1 − 1 Proof. We use the formula in 2.4, together with the well–known relation ϕ(t) = m. If σk / m(q), then we estimate the number of fixed points with the preceding lemma. Note that γk has the order (a, b) denotes the greatest common divisor of a and b. If however 1 = σk ∈PGLm(q), then clearly f (σ) ≤ 1 (qm−1 − 1 + q − 1) = qm−1−1 + 1. − qm−1 − 1 − 1)ϕ(1)+ + 1 − qm/r − 1 ) · ϕ( + 1 − qm/r − 1 ) · ϕ(t) + If i < e, then we get, using 2 ≤ r ≤ e/i ≤ |σ|, Corollary. Let σ ∈ PΓLm(q) \ PGLm(q).
If m ≥ 4, then σ| )(qm−1 − 1) + 2 If m ∈ {2, 3}, then |σ|)(qm−1 − q1/2). Now we are prepared to discuss condition (*). Let σ , . . . , σ ∈ G ≤ PΓL If not all the σ are involutions, then assume without loss that σ is not an involution, and set σ1 = σ σ . . . σ If all the σ are involutions, then s ≥ 3, as G is not dihedral. By conjugation and operations of the kind . . . , a, b, . . . → . . . , b, ab, . . . we may assume that PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS and σ do not commute. Then σ σ is not an involution, and we set In either case we have σ1, σ2 ∈ G which not both are involutions, such that ind σ1+ind σ2 ≤ n−1 (with n = (qm−1)/(q−1)) and such that σ1σ2 is an n–cycle.
(As to the inequality for the index note that ind σ is also the minimal number oftranspostions required to write σ as a product with. Thus ind στ ≤ ind σ +ind τ .
From this we actually get ind σ1 + ind σ2 = n − 1.) If σ1, σ2 ∈ PGL2(q), then q = 5 or 7 as in section 2.6. If σ1, σ2 ∈ PGLm(q) for m ≥ 3, then proceed as in [4]. The key tool [4, 3.4] is correct in this case.
From now on suppose that one of the elements σ1, σ2 is not contained in PGLm(q). As a consequence of Zsigmondy’s Theorem and Schur’s Lemma, we
get that the n–cycle σ1σ2 is contained in PGLm(q) except possibly for m = 2, q =
8, see the proof of [4, 5.1]. The case m = 2, q = 8 is excluded until otherwise
stated.
Thus σ1 and σ2 have the same order e/i ≥ 2 modulo PGLm(q). In particular |σ1| and |σ2| have a common divisor > 1.
First suppose m ≥ 4. Using n − 1 ≥ ind σ1 + ind σ2 and the Corollary we get ≥ ((1 − 1) + (1 − 1))(qm−1 − 1) + ( − qm/2 − 1 + 1). As the last summand on the right side is positive, we get ≥ 5(qm−1 − 1), hence q = 4. Now we use q = 4 in (2), and easily get the contradiction m ≤ 3.
Now suppose m = 3. Similarly as above we get q(q + 1) ≥ 5 (q2 − q1/2), hence q = 4. For a treatment of G ≤ PΓL3(4) confer [4].
From now on suppose m = 2. Without loss we assume |σ1| ≤ |σ2|. As above q ≥ 5 (q − q1/2), If q = 25, then |σ1| = 2, |σ2| = 4, and σ2 ∈ PGL 2(25), hence f (σ2) ≤ f (σ2 2. From 2.4 we get ind σ2 ≥ 18, hence ind σ1 ≤ 7, contrary to the Corollary.
Now suppose q = 16. We quickly get |σ1| = 2 and |σ2| = 4. As σ1 and σ2 have the same order modulo PGL2(16), we get σ2 ∈ PGL one fixed point, hence so does σ2. It follows ind σ2 = 12, hence ind σ1 = 4,contrary to the Corollary.
Now suppose q = 9. We get s = 2, for if s ≥ 3 then s = 3 and the σi are fixed–point–free involutions. However PGL2(9) does not contain fixed–point–freeinvolutions, contrary to σ1σ2σ3 ∈ PGL2(9).
So we get |σ1| = 2 and |σ2| = 4. An explicit example is σ1 = (2, 7)(5, 6)(8, 10), σ2 = (1, 4, 9, 2)(3, 5, 7, 10).
For the last case G = PΓL2(8) we get from the index estimations s = 2 and (|σ1|, |σ2|) = (2, 3) or (3, 3). Explicit examples are σ1 = (2, 3)(4, 5)(6, 7)(8, 9),σ2 = (7, 5, 8)(1, 9, 3) and σ1 = (4, 5, 6)(7, 8, 9), σ2 = (5, 9, 2)(6, 3, 1).
= S5 we do not discuss q = 4.
3. Rationality Questions
3.1. Using a special case of the so–called branch cycle argument (for a short
proof see [10]), we get the following
Lemma. Let f ∈ Q[X] be a polynomial of degree n. Let G be the monodromy group of f , and let σ be an n–cycle as in (*). Let ˆ Sn. Then any two generators of Z = σ are conjugate in ˆ We now prove the Theorem from the Introduction.
We note that f is indecomposable even over C by [12, 3.5]. Thus we apply
our result from section 2.2. As for type (i), consider the polynomials f (x) = xpor f ∈ Q[X] defined by f (z + 1 ) = zp + 1 to get the cyclic group or the dihedral If f is of type (ii), then Fried showed ([8, Section 3]) that f /
remark that Fried did this without actually knowing the occurring groups (evento prove that there are only finitely many examples seems to require the classi-fication of the finite simple groups).
Now we discuss the type (iii). The group Sn is in some sense the generic case.
To get this group take for instance f (X) = Xn −X. The discriminant of f (X)−tis a polynomial of degree n−1 in t, and the roots of the discriminant are preciselythe finite branch points of f : P1 → P1. In this case the discriminant has n − 1different simple roots, therefore σi is a transposition for i = 1, . . . , r − 1, see 2.1.
Thus G is a transitive group generated by transpositions, and such a group issymmetric.
Similarly we get the alternating group. Choose f such that its derivative equals (Xm − 1)2, thereby 2m + 1 = n. Denote by S the set of mth roots ofunity. Then the discriminant of f (X) − t equals, up to a multiplicative constant,∆(t) = (t − f (ζ))2. Note that ∆ has m different roots, each of multiplicity 2. Now f (ζ) = f (ζ) = 0 = f (ζ) for each ζ ∈ S. This shows that (in the
notation of 2.1) r = m + 1 and each σi (i = 1, . . . , r − 1) is a 3-cycle. So the
transitive group G is generated by 3-cycles, thus G = An (see [18, lemme 1 in
4.]).
Next we exclude M11, M23 and PGL2(7) using the Lemma from above. Ob- serve that every automorphism of these groups is inner, hence ˆ PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS cases. To exclude the two Mathieu groups it suffices to see that no element of
order 11 (resp. 23) is conjugate to its inverse. This can be deduced from [2].
Suppose that PGL2(7) meets the conclusion of the Theorem. As Z has 4 generators, 8 · 4 = |NG(Z)| does divide 7 · 48 = |G|, a contradiction.
For the group PΓL2(8) we got two different types of branch cycle descriptions.
We show that the case with |σ1| = 2, |σ2| = 3 does not occur. Of courseσ1 ∈ PGL2(8) and σ2 / ∈ PGL2(8). Thus the 9–cycle σ = σ1σ2 is not contained in PGL2(8). Thus σ has order 3 modulo PGL2(8), and therefore cannot beconjugate to its inverse.
It remains to show that PGL2(5), PΓL2(8) of type (3, 3 : 9), and PΓL2(9) are monodromy groups of polynomials with rational coefficients and to exhibit thecorresponding polynomials.
3.2 The group PGL2(5). We know from our result in section 2, that there
is a polynomial f ∈ C[X] with monodromy group PGL2(5). We just compute it,and it will turn out that it can be chosen with rational coefficients. Recall thedefinition of the generators σ1, . . . , σs of the monodromy group. In our case wehave (up to simultaneous conjugation with elements in S6 and reordering the σ’s)σ1 = (1, 2)(3, 4) and σ2 = (3, 2, 5, 6); that is a consequence of the considerationsin 2.6.
Let f be monic, and let 0 be the branch point corresponding to σ2. Without loss, above 0 lies the 4-fold point 0, and the simple points κ1 and κ2 (κ1, κ2 = 0).
We have κ1 + κ2 = 0, for otherwise f were a composition with a quadraticpolynomial. We may assume κ1 + κ2 = −6. Then f (X) = X4(X2 + 6X + p)with p ∈ C. The finite branch points of f are the zeroes of f . We have f (X) =2X3(3X2 + 15X + 2p). Let λ1 and λ2 be the zeroes of h(X) = 3X2 + 15X + 2p.
They are different, and have the same images under f . Write f = q · h + rwith polynomials q and r, such that deg r ≤ 1. Then f (λi) = r(λi), hencer(λ1) = r(λ2). Thus r is a constant. On the other hand, by dividing thepolynomials, we get that the coefficient of X in r is 8/3(p − 75/8)(p − 25). Thechoice p = 75/8 yields 3125/128 as the second finite branch point. However,f (X) − 3125/128 = 1/128(16X3 − 24X2 + 30X − 25)(2X + 5)3 shows that theramification above this point is the wrong one. Thus p = 25.
3.3 The groups PΓL2(8) and PΓL2(9). The polynomials have been com-
puted by Matzat. See [17, 8.5] for PΓL2(8) and [17, 8.7] for PΓL2(9).
Appendix: The Theorems of Ritt
The setup from section 2.1 allows for short proofs of the classical Theorems of Ritt about decompositions of polynomials. Throughout this section we deal with
polynomials with complex coefficients. Via model theory the assertions hold for
any algebraically closed field of characteristic 0. Using [12, 3.5] one readily gets
that R.1 and R.2 hold for arbitrary fields of characteristic 0. By a maximal
decomposition of a polynomial f we mean a decomposition f = f1 ◦ f2 ◦ · · · ◦ frwhere the fi are non–linear and indecomposable.
Theorem R.1 (Ritt). Let f = f1 ◦ · · · ◦ fr = g1 ◦ · · · ◦ gs be two maximal decompositions of a polynomial f ∈ C[X]. Then r = s and the degrees of thefi’s are a permutation of the degrees of the gi’s. Furthermore, one can pass fromone decomposition to the other one by altering two adjacent polynomials in eachstep. From the latter part of this Theorem the question arises when a◦b = c◦d with indecomposable polynomials a, b, c, and d. Recall that the Cebychev polynomialTn is defined by Tn(Z + 1/Z) = Zn + 1/Zn.
Theorem R.2 (Ritt). Let a, b, c, and d be non–linear indecomposable poly- nomials such that a ◦ b = c ◦ d. Assume without loss deg a ≥ deg c. Then thereexist linear polynomials L1, L2, L3, and L4 such that one of the following holds. a = c ◦ L1, b = L−1 ◦ d, (the uninteresting case). L1 ◦ a ◦ L−1 = Xk · t(X)m, L L1 ◦ c ◦ L−1 = Xm, L 4 ◦ d ◦ L3 = X k · t(X m) L1 ◦ a ◦ L−1 = T L2 ◦ b ◦ L3 = Tn L1 ◦ c ◦ L−1 = T L4 ◦ d ◦ L3 = Tm . for Cebychev polynomials Tm and Tn. Let f be a polynomial with complex coefficients, and let x be a transcendental over C. Set t = f (x), and let Ω be the Galois closure of C(x)|C(t). Let G =Gal(Ω|C(t)) be the monodromy group of f , and let U be stabilizer of x. We viewtwo decomposition of f as equivalent, if they differ just by linear twists (like (1)in R.2). As an easy consequence of L¨ uroth’s theorem, we see that the equivalence classes of maximal decompositions of f correspond bijectively to the maximalchains of subgroups from U to G. If f = f1 ◦ · · · ◦ fr is such a decomposition,then the associated chain of subgroups is U = U0 < U1 < . . . < Ur−1 < Ur = G,where Ui is the stabilizer of f1(f2(· · · (fi(x) · · · )). Then Theorem R.1 is a directconsequence of Theorem R.3. Let G be a finite group with subgroups U and C such that G = U C and C is abelian. Then the maximal chains of subgroups from U to G haveequal lengths and (up to permutation) the same relative indices. Furthermore,one can pass from one chain to an other one just by changing one group in thechain in each step. Proof of R.3. Choose a minimal counter–example subject to U + [G : U ]
being minimal. Let U = A0 < A1 < . . . < Ar = G and U = B0 < B1 < . . . < PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS Bs = G be two chains failing the assertion. Then A1 = B1 and coreG(U ) = 1,where coreG(U ) means the maximal normal subgroup of G, which is contained inU . Set NA = coreG(A1) and NB = coreG(B1). These groups are non–trivial, asG = A1C, hence 1 = A1 ∩ C ≤ in A1, and NA ≤ U , we get A1 = U NA. Likewise B1 = U NB. Set D = A1, B1 .
The assumptions of the Theorem are fulfilled if we replace U by A1 or B1. Weconsider, in addition to the given chains, a maximal chain from D to G. Thus weare done once we know that A1 and B1 are maximal in D, [D : A1] = [B1 : U ],and [D : B1] = [A1 : U ]. Note that D = NAU, B1 = NAB1. Suppose there is agroup X properly between A1 and D = NAB1. Then X = NA(X ∩ B1), henceU < X ∩ B1 < B1, a contradiction. By symmetry B1 is maximal in D as well.
Finally we have NA ∩B1 ≤ A1 ∩B1 = U , hence NA ∩B1 = NA ∩U . This yields [D : B1] = [NA : NA ∩ B1] = [NA : NA ∩ U ] = [A1 : U ]. Again by symmetry weget [D : A1] = [B1 : U ].
Proof of R.2. As in 2.1, let G be the Galois group of the Galois closure of
C(X)|C(t), where a(b(X)) = c(d(X)) = t.
Let A, B, and U be the fix groups of b(X), d(X), and X, respectively. The case A = B yields (1) of the Theorem. From now on assume A = B. ThenU = A ∩ B is core–free in G, and the chains of subgroups U ⊂ A ⊂ G andU ⊂ B ⊂ G are maximal. Let Z be a cyclic complement of U in G (c.f. 2.1).
Set NA = coreG(A) (= 1, see the proof of R.3), NB = coreG(B).
Then, by the maximality of the chains above, G = ANB = BNA, A = U NA, Set m = [G : A] = [B : U ], n = [G : B] = [A : U ].
Claim 1. The monodromy groups of b and c are the same, as well as the ones Proof. Set N = coreB(U ). Of course B ∩ NA = U ∩ NA ≤ N . On the other hand, the set of G-conjugates of A is the same as the set of B-conjugatesof A. Therefore N ≤ NA, hence N ≤ B ∩ NA. This shows B ∩ NA = N . NowG/N A = BNA/NA = B/B ∩ NA = B/N yields the assertion.
Now we are going to study three different permutation representations of G.
First let G act on the cosets of U . Then the set of cosets of A provides a systemof imprimitivity, and so does the set of cosets of B. The intersection of a cosetof A and a coset of B is a coset of U : Without loss consider A and Bg. We mayassume g ∈ NA ⊆ A (as G = BNA). Then U g ≤ A ∩ Bg. If U h ≤ A ∩ Bg, thenhg−1 ∈ A ∩ B = U , hence U h = U g. Therefore U g = A ∩ Bg.
Denote by πA the canonical homomorphism G −→ G/NA ≤ Sym(G/A) of permutation groups, likewise for B. Note that πA(G) and πB(G) act primitivelyby the maximality of A and B in G.
Let ind(g), o(g), and f(g) be the index of g, the number of cycles of g, and the number of fixed–points of g. Define indA(g), o(g), and fA(g) analogously forπA(g), likewise for B. From the considerations above we get f(g) = fA(g) · fB(g).
For a fixed g ∈ G let [ν1, ν2, · · · , νk] be the cycle type of πA(g) (i.e. πA(g) hascycles of lengths ν1, ν2, . . . ) and [µ1, µ2, · · · , µl] be the cycle type of πB(g). Then k = oA(g) = m − indA(g) l = oB(g) = n − indB(g) (νi, µj) = o(g) = mn − ind(g) . ind(g) ≥ n · indA(g) ind(g) ≥ m · indB(g) . Let g1, g2,. . . ,gs be a generating system of G according to 2.2(*).
A(gu) = m − 1, B (gu) = n − 1. indA(gu) ≥ m − 1 (for otherwise the elements πA(gu) were a branch cycle description of a cover X → P1 with X having negative genus).
On the other hand, (mn − 1) = m − 1 . This proves the assertion. Here and in the following we use implicitly the symmetryof certain assertions in A and B.
Claim 3. If πA(G) is not cyclic, then ind(g) ≥ m · indB(g) + indA(g) for all g ∈ {g1, g2, . . . , gs}. Proof. Assume the contrary, which implies o(g) > m · oB(g) + oA(g) − m for some g ∈ {g1, g2, . . . , gs}. Assign to this g the ν’s and µ’s as above. Then Thus there is an index i, without loss i = 1, such that Let T be the number of j’s such that ν1 does not divide µj. Then (ν1 − (ν1, µj)) < ν1 − 1 , hence T ≤ 1. Thus there is at most one j0 such that ν1 does not divide µj . But (ν1, µj ) = 1 yields the contradiction 0 < 0. Therefore the µ’s have a common divisor δ > 1. From Claim 2 we know that the elements πA(g1), . . . , πA(gs)provide a branch cycle description of a polynomial. A common divisor δ of theµ’s means, that this polynomial has the form h(X)δ + e for some polynomial h PRIMITIVE MONODROMY GROUPS OF POLYNOMIALS and a constant e. However, this polynomial is decomposable, contrary to πA(G)being primitive.
Claim 4. If πA(G) is not cyclic, then ind(g) = m · indB(g) + indA(g) for all g ∈ {g1, g2, . . . , gs}. Proof. Suppose wrong. Then Claim 2 and Claim 3 yield the contradiction indA(gu) = m(n−1)+m−1 = mn−1 . Claim 5. Exactly one of the following holds. (1) πA(G) or πB(G) is cyclic.
(2) G, πA(G), and πB(G) are dihedral and act naturally (i.e. the cyclic group of index 2 acts regularly in each case). Proof. Suppose that (1) doesn’t hold.
Choose any g ∈ {g1, g2, . . . , gs}.
Assign to g the cycle lengths νi and µj of πA(g) and πB(g) as in the proof ofClaim 3.
νi + 1 − νj for each i = 1, . . . , k . Furthermore, also by this proof, the following holds: For each j there is at mostone i such that µj does not divide νi. In particular, for fixed j, νi ≥ µj besidesat most one index i. Thus (oA(g) − 1)µj + 1 ≤ m . Now, for gu in {g1, g2, . . . , gs}, let wu be the maximal associated cycle lengthµj. Then (m − indA(gu) − 1)wu + 1 ≤ m . Dividing by wu and adding for u = 1, 2, . . . , s yields Therefore all besides two w’s are 1, and these two exceptions are 2. By the choiceof the w’s, this implies the existence of two indices u1 and u2 such that πB(gu ) and πB(gu ) are involutions, and the other π B (gu)’s are trivial. The same holds for the images of gu in πA(G) for the same indices u, as can be seen (for instance)by Claim 4. Finally, as G −→ πA(G) × πB(G) is an injective homomorphism,the involutions gu and g are the only non–trivial elements in {g We get the assertion about the action as follows: gu g (neither of which is contained in g a dihedral group of order 2mn.
The final step is to formulate our results in terms of polynomials: Suppose that (1) in Claim 5 holds. Then, by Claim 1, we need to study the decompositionp(Xm) = q(X)m for some polynomials p and q. Set p(X) = Xk · r(X) with r apolynomial such that r(0) = 0. Then Xkm · r(Xm) = q(X)m. Thus Xk dividesq(X), hence q(X) = Xk · s(X) with a polynomial s such that s(0) = 0. We getr(Xm) = s(X)m. Now every zero of r occurs with a multiplicity divisible by m,hence r(X) = t(X)m with a polynomial t. But then s(X) = ζt(Xm) for somem-th root ζ of 1. Substituting back we get our result.
Now assume that case (2) of Claim 5 holds. Without loss of generality we assume a◦b = c◦d = Tmn = Tm◦Tn. Then the fix groups of Tn(X) and of b(X) inG have the same order, thus they are equal (every group M between U and G isuniquely determined by its order, as M = U Z ∩M = U (M ∩Z) and subgroups ofcyclic groups are determined by their order). Thus C(Tn(X)) = C(b(X)), henceb = L−1 ◦ T n for some linear polynomial L2. Then Tn ◦ Tm = a ◦ b = a ◦ L−1 Analogously express c and d in terms of Tm and Tn. The assertion follows.
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E-mail address: mueller@mi.uni-erlangen.de

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